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Velocity and Speed: Solutions to Problems

Solutions to the problems on velocity and speed of moving objects. More tutorials can be found in this website.

A man walks 7 km in 2 hours and 2 km in 1 hour in the same direction. a) What is the man's average speed for the whole journey? b) What is the man's average velocity for the whole journey? Solution to Problem 1: a)

speed and velocity - Problem 1

You start walking from a point on a circular field of radius 0.5 km and 1 hour later you are at the same point. a) What is your average speed for the whole journey? b) What is your average velocity for the whole journey? Solution to Problem 3: a) If you walk around a circular field and come back to the same point, you have covered a distance equal to the circumference of the circle.

John drove South 120 km at 60 km/h and then East 150 km at 50 km/h. Determine a) the average speed for the whole journey? b) the magnitude of the average velocity for the whole journey? Solution to Problem 4: a)

speed and velocity - Problem 4

If I can walk at an average speed of 5 km/h, how many miles I can walk in two hours? Solution to Problem 5: distance = (average speed) * (time) = 5 km/h * 2 hours = 10 km using the rate of conversion 0.62 miles per km, the distance in miles is given by distance = 10 km * 0.62 miles/km = 6.2 miles

A train travels along a straight line at a constant speed of 60 mi/h for a distance d and then another distance equal to 2d in the same direction at a constant speed of 80 mi/h. a)What is the average speed of the train for the whole journey?

A car travels 22 km south, 12 km west, and 14 km north in half an hour. a) What is the average speed of the car? b) What is the final displacement of the car? c) What is the average velocity of the car?

Solution to Problem 7: a)

speed and velocity - Problem 1

Solution to Problem 8: a)

velocity and speed - Problem 8

More References and links

  • Velocity and Speed: Tutorials with Examples
  • Velocity and Speed: Problems with Solutions
  • Acceleration: Tutorials with Examples
  • Uniform Acceleration Motion: Problems with Solutions
  • Uniform Acceleration Motion: Equations with Explanations

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Solved Speed, Velocity, and Acceleration Problems

Simple problems on speed, velocity, and acceleration with descriptive answers are presented for the AP Physics 1 exam and college students. In each solution, you can find a brief tutorial. 

Speed and velocity Problems: 

Problem (1): What is the speed of a rocket that travels $8000\,{\rm m}$ in $13\,{\rm s}$?

Solution : Speed is defined in physics  as the total distance divided by the elapsed time,  so the rocket's speed is \[\text{speed}=\frac{8000}{13}=615.38\,{\rm m/s}\]

Problem (2): How long will it take if you travel $400\,{\rm km}$ with an average speed of $100\,{\rm m/s}$?

Solution : Average speed is the ratio of the total distance to the total time. Thus, the elapsed time is \begin{align*} t&=\frac{\text{total distance}}{\text{average speed}}\\ \\ &=\frac{400\times 10^{3}\,{\rm m}}{100\,{\rm m/s}}\\ \\ &=4000\,{\rm s}\end{align*} To convert it to hours, it must be divided by $3600\,{\rm s}$ which gives $t=1.11\,{\rm h}$.

Problem (3): A person walks $100\,{\rm m}$ in $5$ minutes, then $200\,{\rm m}$ in $7$ minutes, and finally $50\,{\rm m}$ in $4$ minutes. Find its average speed. 

Solution : First find its total distance traveled ($D$) by summing all distances in each section, which gets $D=100+200+50=350\,{\rm m}$. Now, by definition of average speed, divide it by the total time elapsed $T=5+7+4=16$ minutes.

But keep in mind that since the distance is in SI units, so the time traveled must also be in SI units, which is $\rm s$. Therefore, we have\begin{align*}\text{average speed}&=\frac{\text{total distance} }{\text{total time} }\\ \\ &=\frac{350\,{\rm m}}{16\times 60\,{\rm s}}\\ \\&=0.36\,{\rm m/s}\end{align*}

Problem (4): A person walks $750\,{\rm m}$ due north, then $250\,{\rm m}$ due east. If the entire walk takes $12$ minutes, find the person's average velocity. 

Solution : Average velocity , $\bar{v}=\frac{\Delta x}{\Delta t}$, is displacement divided by the elapsed time. Displacement is also a vector that obeys the addition vector rules. Thus, in this velocity problem, add each displacement to get the total displacement . 

In the first part, displacement is $\Delta x_1=750\,\hat{j}$ (due north) and in the second part $\Delta x_2=250\,\hat{i}$ (due east). The total displacement vector is $\Delta x=\Delta x_1+\Delta x_2=750\,\hat{i}+250\,\hat{j}$ with magnitude of  \begin{align*}|\Delta x|&=\sqrt{(750)^{2}+(250)^{2}}\\ \\&=790.5\,{\rm m}\end{align*} In addition, the total elapsed time is $t=12\times 60$ seconds. Therefore, the magnitude of the average velocity is \[\bar{v}=\frac{790.5}{12\times 60}=1.09\,{\rm m/s}\]

Problem (5): An object moves along a straight line. First, it travels at a velocity of $12\,{\rm m/s}$ for $5\,{\rm s}$ and then continues in the same direction with $20\,{\rm m/s}$ for $3\,{\rm s}$. What is its average speed?

Solution: Average velocity is displacement divided by elapsed time, i.e., $\bar{v}\equiv \frac{\Delta x_{tot}}{\Delta t_{tot}}$.

Here, the object goes through two stages with two different displacements, so add them to find the total displacement. Thus,\[\bar{v}=\frac{x_1 + x_2}{t_1 +t_2}\] Again, to find the displacement, we use the same equation as the average velocity formula, i.e., $x=vt$. Thus, displacements are obtained as $x_1=v_1\,t_1=12\times 5=60\,{\rm m}$ and $x_2=v_2\,t_2=20\times 3=60\,{\rm m}$. Therefore, we have \begin{align*} \bar{v}&=\frac{x_1+x_2}{t_1+t_2}\\ \\&=\frac{60+60}{5+3}\\ \\&=\boxed{15\,{\rm m/s}}\end{align*}

Problem (6): A plane flies the distance between two cities in $1$ hour and $30$ minutes with a velocity of $900\,{\rm km/h}$. Another plane covers that distance at $600\,{\rm km/h}$. What is the flight time of the second plane?

Solution: first find the distance between two cities using the average velocity formula $\bar{v}=\frac{\Delta x}{\Delta t}$ as below \begin{align*} x&=vt\\&=900\times 1.5\\&=1350\,{\rm km}\end{align*} where we wrote one hour and a half minutes as $1.5\,\rm h$. Now use again the same kinematic equation above to find the time required for another plane \begin{align*} t&=\frac xv\\ \\ &=\frac{1350\,\rm km}{600\,\rm km/h}\\ \\&=2.25\,{\rm h}\end{align*} Thus, the time for the second plane is $2$ hours and $0.25$ of an hour, which converts to minutes as $2$ hours and ($0.25\times 60=15$) minutes.

Problem (7): To reach a park located south of his jogging path, Henry runs along a 15-kilometer route. If he completes the journey in 1.5 hours, determine his speed and velocity.

Solution:  Henry travels his route to the park without changing direction along a straight line. Therefore, the total distance traveled in one direction equals the displacement, i.e, \[\text{distance traveled}=\Delta x=15\,\rm km\]Velocity is displacement divided by the time of travel \begin{align*} \text{velocity}&=\frac{\text{displacement}}{\text{time of travel}} \\\\ &=\frac{15\,\rm km}{1.5\,\rm h} \\\\ &=\boxed{10\,\rm km/h}\end{align*} and by definition, its average speed is \begin{align*} \text{speed}&=\frac{\text{distance covered}}{\text{time interval}}\\\\&=\frac{15\,\rm km}{1.5\,\rm h}\\\\&=\boxed{10\,\rm km/h}\end{align*} Thus, Henry's velocity is $10\,\rm km/h$ to the south, and its speed is $10\,\rm km/h$. As you can see, speed is simply a positive number, with units but velocity specifies the direction in which the object is moving. 

Problem (8): In 15 seconds, a football player covers the distance from his team's goal line to the opposing team's goal line and back to the midway point of the field having 100-yard-length. Find, (a) his average speed, and (b) the magnitude of the average velocity.

Solution:  The total length of the football field is $100$ yards or in meters, $L=91.44\,\rm m$. Going from one goal's line to the other and back to the midpoint of the field takes $15\,\rm s$ and covers a distance of $D=100+50=150\,\rm yd$. 

average speed and velocity at football field

Distance divided by the time of travel gets the average speed, \[\text{speed}=\frac{150\times 0.91}{15}=9.1\,\rm m/s\] To find the average velocity, we must find the displacement of the player between the initial and final points. 

The initial point is her own goal line and her final position is the midpoint of the field, so she has displaced a distance of $\Delta x=50\,\rm yd$ or $\Delta x=50\times 0.91=45.5\,\rm m$. Therefore, her velocity is calculated as follows \begin{align*} \text{velocity}&=\frac{\text{displacement}}{\text{time elapsed}} \\\\ &=\frac{45.5\,\rm m}{15\,\rm s} \\\\&=\boxed{3.03\quad \rm m/s}\end{align*} Contrary to the previous problem, here the motion is not in one direction, hence, the displacement is not equal to the distance traveled. Accordingly, the average speed is not equal to the magnitude of the average velocity.

Problem (9): You begin at a pillar and run towards the east (the positive $x$ direction) for $250\,\rm m$ at an average speed of $5\,\rm m/s$. After that, you run towards the west for $300\,\rm m$ at an average speed of $4\,\rm m/s$ until you reach a post. Calculate (a) your average speed from pillar to post, and (b) your average velocity from pillar to post. 

Solution : First, you traveled a distance of $L_1=250\,\rm m$ toward east (or $+x$ direction) at $5\,\rm m/s$. Time of travel in this route is obtained as follows \begin{align*} t_1&=\frac{L_1}{v_1}\\\\ &=\frac{250}{5}\\\\&=50\,\rm s\end{align*} Likewise, traveling a distance of $L_2=300\,\rm m$ at $v_2=4\,\rm m/s$ takes \[t_2=\frac{300}{4}=75\,\rm s\]  (a) Average speed is defined as the distance traveled (or path length) divided by the total time of travel \begin{align*} v&=\frac{\text{path length}}{\text{time of travel}} \\\\ &=\frac{L_1+L_2}{t_1+t_2}\\\\&=\frac{250+300}{50+75} \\\\&=4.4\,\rm m/s\end{align*} Therefore, you travel between these two pillars in $125\,\rm s$ and with an average speed of $4.4\,\rm m/s$. 

(b) Average velocity requires finding the displacement between those two points. In the first case, you move $250\,\rm m$ toward $+x$ direction, i.e., $L_1=+250\,\rm m$. Similarly, on the way back, you move $300\,\rm m$ toward the west ($-x$ direction) or $L_2=-300\,\rm m$. Adding these two gives us the total displacement between the initial point and the final point, \begin{align*} L&=L_1+L_2 \\\\&=(+250)+(-300) \\\\ &=-50\,\rm m\end{align*} The minus sign indicates that you are generally displaced toward the west. 

Finally, the average velocity is obtained as follows: \begin{align*} \text{average velocity}&=\frac{\text{displacement}}{\text{time of travel}} \\\\ &=\frac{-50}{125} \\\\&=-0.4\,\rm m/s\end{align*} A negative average velocity indicating motion to the left along the $x$-axis. 

This speed problem better makes it clear to us the difference between average speed and average speed. Unlike average speed, which is always a positive number, the average velocity in a straight line can be either positive or negative. 

Problem (10): What is the average speed for the round trip of a car moving uphill at 40 km/h and then back downhill at 60 km/h? 

Solution : Assuming the length of the hill to be $L$, the total distance traveled during this round trip is $2L$ since $L_{up}=L_{down}=L$. However, the time taken for going uphill and downhill was not provided. We can write them in terms of the hill's length $L$ as $t=\frac L v$. 

Applying the definition of average speed gives us \begin{align*} v&=\frac{\text{distance traveled}}{\text{total time}} \\\\ &=\frac{L_{up}+L_{down}}{t_{up}+t_{down}} \\\\ &=\cfrac{2L}{\cfrac{L}{v_{up}}+\cfrac{L}{v_{down}}} \end{align*} By reorganizing this expression, we obtain a formula that is useful for solving similar problems in the AP Physics 1 exams. \[\text{average speed}=\frac{2v_{up} \times v_{down}}{v_{up}+v_{down}}\] Substituting the numerical values into this, yields \begin{align*} v&=\frac{2(40\times 60)}{40+60} \\\\ &=\boxed{48\,\rm m/s}\end{align*} What if we were asked for the average velocity instead? During this round trip, the car returns to its original position, and thus its displacement, which defines the average velocity, is zero. Therefore, \[\text{average velocity}=0\,\rm m/s\]

Acceleration Problems

Problem (9): A car moves from rest to a speed of $45\,\rm m/s$ in a time interval of $15\,\rm s$. At what rate does the car accelerate? 

Solution : The car is initially at rest, $v_1=0$, and finally reaches $v_2=45\,\rm m/s$ in a time interval $\Delta t=15\,\rm s$. Average acceleration is the change in velocity, $\Delta v=v_2-v_1$, divided by the elapsed time $\Delta t$, so \[\bar{a}=\frac{45-0}{15}=\boxed{3\,\rm m/s^2} \] 

Problem (10): A car moving at a velocity of $15\,{\rm m/s}$, uniformly slows down. It comes to a complete stop in $10\,{\rm s}$. What is its acceleration?

Solution:  Let the car's uniform velocity be $v_1$ and its final velocity $v_2=0$.   Average acceleration is the difference in velocities divided by the time taken, so we have: \begin{align*}\bar{a}&=\frac{\Delta v}{\Delta t}\\\\&=\frac{v_2-v_1}{\Delta t}\\\\&=\frac{0-15}{10}\\\\ &=\boxed{-1.5\,{\rm m/s^2}}\end{align*}The minus sign indicates the direction of the acceleration vector, which is toward the $-x$ direction.

Problem (11): A car moves from rest to a speed of $72\,{\rm km/h}$ in $4\,{\rm s}$. Find the acceleration of the car.

Solution: Known: $v_1=0$, $v_2=72\,{\rm km/h}$, $\Delta t=4\,{\rm s}$.  Average acceleration is defined as the difference in velocities divided by the time interval between those points \begin{align*}\bar{a}&=\frac{v_2-v_1}{t_2-t_1}\\\\&=\frac{20-0}{4}\\\\&=5\,{\rm m/s^2}\end{align*} In above, we converted $\rm km/h$ to the SI unit of velocity ($\rm m/s$) as \[1\,\frac{km}{h}=\frac {1000\,m}{3600\,s}=\frac{10}{36}\, \rm m/s\] so we get \[72\,\rm km/h=72\times \frac{10}{36}=20\,\rm m/s\] 

Problem (12): A race car accelerates from an initial velocity of $v_i=10\,{\rm m/s}$ to a final velocity of $v_f = 30\,{\rm m/s}$ in a time interval of $2\,{\rm s}$. Determine its average acceleration.

Solution:  A change in the velocity of an object $\Delta v$ over a time interval $\Delta t$ is defined as an average acceleration. Known: $v_i=10\,{\rm m/s}$, $v_f = 30\,{\rm m/s}$, $\Delta t=2\,{\rm s}$. Applying definition of average acceleration, we get \begin{align*}\bar{a}&=\frac{v_f-v_i}{\Delta t}\\&=\frac{30-10}{2}\\&=10\,{\rm m/s^2}\end{align*}

Problem (13): A motorcycle starts its trip along a straight line with a velocity of $10\,{\rm m/s}$ and ends with $20\,{\rm m/s}$ in the opposite direction in a time interval of $2\,{\rm s}$. What is the average acceleration of the car?

Solution:  Known: $v_i=10\,{\rm m/s}$, $v_f=-20\,{\rm m/s}$, $\Delta t=2\,{\rm s}$, $\bar{a}=?$. Using average acceleration definition we have \begin{align*}\bar{a}&=\frac{v_f-v_i}{\Delta t}\\\\&=\frac{(-20)-10}{2}\\\\ &=\boxed{-15\,{\rm m/s^2}}\end{align*}Recall that in the definition above, velocities are vector quantities. The final velocity is in the opposite direction from the initial velocity so a negative must be included.

Problem (14): A ball is thrown vertically up into the air by a boy. After $4$ seconds, it reaches the highest point of its path. How fast does the ball leave the boy's hand?

Solution : At the highest point, the ball has zero speed, $v_2=0$. It takes the ball $4\,\rm s$ to reach that point. In this problem, our unknown is the initial speed of the ball, $v_1=?$. Here, the ball accelerates at a constant rate of $g=-9.8\,\rm m/s^2$ in the presence of gravity.

When the ball is tossed upward, the only external force that acts on it is the gravity force. 

Using the average acceleration formula $\bar{a}=\frac{\Delta v}{\Delta t}$ and substituting the numerical values into this, we will have \begin{gather*} \bar{a}=\frac{\Delta v}{\Delta t} \\\\ -9.8=\frac{0-v_1}{4} \\\\ \Rightarrow \boxed{v_1=39.2\,\rm m/s} \end{gather*} Note that $\Delta v=v_2-v_1$. 

Problem (15): A child drops crumpled paper from a window. The paper hit the ground in $3\,\rm s$. What is the velocity of the crumpled paper just before it strikes the ground? 

Solution : The crumpled paper is initially in the child's hand, so $v_1=0$. Let its speed just before striking be $v_2$. In this case, we have an object accelerating down in the presence of gravitational force at a constant rate of $g=-9.8\,\rm m/s^2$. Using the definition of average acceleration, we can find $v_2$ as below \begin{gather*} \bar{a}=\frac{\Delta v}{\Delta t} \\\\ -9.8=\frac{v_2-0}{3} \\\\ \Rightarrow v_2=3\times (-9.8)=\boxed{-29.4\,\rm m/s} \end{gather*} The negative shows us that the velocity must be downward, as expected!

Problem (16): A car travels along the $x$-axis for $4\,{\rm s}$ at an average velocity of $10\,{\rm m/s}$ and $2\,{\rm s}$ with an average velocity of $30\,{\rm m/s}$ and finally $4\,{\rm s}$ with an average velocity $25\,{\rm m/s}$. What is its average velocity across the whole path?

Solution: There are three different parts with different average velocities. Assume each trip is done in one dimension without changing direction. Thus, displacements associated with each segment are the same as the distance traveled in that direction and is calculated as below: \begin{align*}\Delta x_1&=v_1\,\Delta t_1\\&=10\times 4=40\,{\rm m}\\ \\ \Delta x_2&=v_2\,\Delta t_2\\&=30\times 2=60\,{\rm m}\\ \\ \Delta x_3&=v_3\,\Delta t_3\\&=25\times 4=100\,{\rm m}\end{align*}Now use the definition of average velocity, $\bar{v}=\frac{\Delta x_{tot}}{\Delta t_{tot}}$, to find it over the whole path\begin{align*}\bar{v}&=\frac{\Delta x_{tot}}{\Delta t_{tot}}\\ \\&=\frac{\Delta x_1+\Delta x_2+\Delta x_3}{\Delta t_1+\Delta t_2+\Delta t_3}\\ \\&=\frac{40+60+100}{4+2+4}\\ \\ &=\boxed{20\,{\rm m/s}}\end{align*}

Problem (17): An object moving along a straight-line path. It travels with an average velocity $2\,{\rm m/s}$ for $20\,{\rm s}$ and $12\,{\rm m/s}$ for $t$ seconds. If the total average velocity across the whole path is $10\,{\rm m/s}$, then find the unknown time $t$.

Solution: In this velocity problem, the whole path $\Delta x$ is divided into two parts $\Delta x_1$ and $\Delta x_2$ with different average velocities and times elapsed, so the total average velocity across the whole path is obtained as \begin{align*}\bar{v}&=\frac{\Delta x}{\Delta t}\\\\&=\frac{\Delta x_1+\Delta x_2}{\Delta t_1+\Delta t_2}\\\\&=\frac{\bar{v}_1\,t_1+\bar{v}_2\,t_2}{t_1+t_2}\\\\10&=\frac{2\times 20+12\times t}{20+t}\\\Rightarrow t&=80\,{\rm s}\end{align*}

Note : whenever a moving object, covers distances $x_1,x_2,x_3,\cdots$ in $t_1,t_2,t_3,\cdots$ with constant or average velocities $v_1,v_2,v_3,\cdots$ along a straight-line without changing its direction, then its total average velocity across the whole path is obtained by one of the following formulas

  • Distances and times are known:\[\bar{v}=\frac{x_1+x_2+x_3+\cdots}{t_1+t_2+t_3+\cdots}\]
  • Velocities and times are known: \[\bar{v}=\frac{v_1\,t_1+v_2\,t_2+v_3\,t_3+\cdots}{t_1+t_2+t_3+\cdots}\]
  • Distances and velocities are known:\[\bar{v}=\frac{x_1+x_2+x_3+\cdots}{\frac{x_1}{v_1}+\frac{x_2}{v_2}+\frac{x_3}{v_3}+\cdots}\]

Problem (18): A car travels one-fourth of its path with a constant velocity of $10\,{\rm m/s}$, and the remaining with a constant velocity of $v_2$. If the total average velocity across the whole path is $16\,{\rm m/s}$, then find the $v_2$?

Solution: This is the third case of the preceding note. Let the length of the path be $L$ so \begin{align*}\bar{v}&=\frac{x_1+x_2}{\frac{x_1}{v_1}+\frac{x_2}{v_2}}\\\\16&=\frac{\frac 14\,L+\frac 34\,L}{\frac{\frac 14\,L}{10}+\frac{\frac 34\,L}{v_2}}\\\\\Rightarrow v_2&=20\,{\rm m/s}\end{align*}

Problem (19): An object moves along a straight-line path. It travels for $t_1$ seconds with an average velocity $50\,{\rm m/s}$ and $t_2$ seconds with a constant velocity of $25\,{\rm m/s}$. If the total average velocity across the whole path is $30\,{\rm m/s}$, then find the ratio $\frac{t_2}{t_1}$?

Solution: the velocities and times are known, so we have \begin{align*}\bar{v}&=\frac{v_1\,t_1+v_2\,t_2}{t_1+t_2}\\\\30&=\frac{50\,t_1+25\,t_2}{t_1+t_2}\\\\ \Rightarrow \frac{t_2}{t_1}&=4\end{align*} 

Read more related articles:  

Kinematics Equations: Problems and Solutions

Position vs. Time Graphs

Velocity vs. Time Graphs

In the following section, some sample AP Physics 1 problems on acceleration are provided.

Problem (20): An object moves with constant acceleration along a straight line. If its velocity at instant of $t_1 = 3\,{\rm s}$ is $10\,{\rm m/s}$ and at the moment of $t_2 = 8\,{\rm s}$ is $20\,{\rm m/s}$, then what is its initial speed?

Solution: Let the initial speed at time $t=0$ be $v_0$. Now apply average acceleration definition in the time intervals $[t_0,t_1]$ and $[t_0,t_2]$ and equate them.\begin{align*}\text{average acceleration}\ \bar{a}&=\frac{\Delta v}{\Delta t}\\\\\frac{v_1 - v_0}{t_1-t_0}&=\frac{v_2-v_0}{t_2-t_0}\\\\ \frac{10-v_0}{3-0}&=\frac{20-v_0}{8-0}\\\\ \Rightarrow v_0 &=4\,{\rm m/s}\end{align*} In the above, $v_1$ and $v_2$ are the velocities at moments $t_1$ and $t_2$, respectively. 

Problem (21): For $10\,{\rm s}$, the velocity of a car that travels with a constant acceleration, changes from $10\,{\rm m/s}$ to $30\,{\rm m/s}$. How far does the car travel?

Solution: Known: $\Delta t=10\,{\rm s}$, $v_1=10\,{\rm m/s}$ and $v_2=30\,{\rm m/s}$. 

Method (I) Without computing the acceleration: Recall that in the case of constant acceleration, we have the following kinematic equations for average velocity and displacement:\begin{align*}\text{average velocity}:\,\bar{v}&=\frac{v_1+v_2}{2}\\\text{displacement}:\,\Delta x&=\frac{v_1+v_2}{2}\times \Delta t\\\end{align*}where $v_1$ and $v_2$ are the velocities in a given time interval. Now we have \begin{align*} \Delta x&=\frac{v_1+v_2}{2}\\&=\frac{10+30}{2}\times 10\\&=200\,{\rm m}\end{align*}

Method (II) with computing acceleration: Using the definition of average acceleration, first determine it as below \begin{align*}\bar{a}&=\frac{\Delta v}{\Delta t}\\\\&=\frac{30-10}{10}\\\\&=2\,{\rm m/s^2}\end{align*} Since the velocities at the initial and final points of the problem are given so use the below time-independent kinematic equation to find the required displacement \begin{align*} v_2^{2}-v_1^{2}&=2\,a\Delta x\\\\ (30)^{2}-(10)^{2}&=2(2)\,\Delta x\\\\ \Rightarrow \Delta x&=\boxed{200\,{\rm m}}\end{align*}

Problem (22): A car travels along a straight line with uniform acceleration. If its velocity at the instant of $t_1=2\,{\rm s}$ is $36\,{\rm km/s}$ and at the moment $t_2=6\,{\rm s}$ is $72\,{\rm km/h}$, then find its initial velocity (at $t_0=0$)?

Solution: Use the equality of definition of average acceleration $a=\frac{v_f-v_i}{t_f-t_i}$ in the time intervals $[t_0,t_1]$ and $[t_0,t_2]$ to find the initial velocity as below \begin{align*}\frac{v_2-v_0}{t_2-t_0}&=\frac{v_1-v_0}{t_1-t_0}\\\\ \frac{20-v_0}{6-0}&=\frac{10-v_0}{2-0}\\\\ \Rightarrow v_0&=\boxed{5\,{\rm m/s}}\end{align*}

All these kinematic problems on speed, velocity, and acceleration are easily solved by choosing an appropriate kinematic equation. Keep in mind that these motion problems in one dimension are of the uniform or constant acceleration type. Projectiles are also another type of motion in two dimensions with constant acceleration.

Author:   Dr. Ali Nemati

Date Published: 9/6/2020

Updated: Jun 28,  2023

© 2015 All rights reserved. by Physexams.com

Speed and Velocity

Speed is how fast something moves.

Velocity is speed with a direction .

ariel running with ball

Saying Ariel the Dog runs at 9 km/h (kilometers per hour) is a speed.

But saying he runs 9 km/h Westwards is a velocity.

Imagine something moving back and forth very fast: it has a high speed, but a low (or zero) velocity.

Speed is measured as distance moved over time.

Speed = Distance Time

Example: A car travels 50 km in one hour.

Its average speed is 50 km per hour (50 km/h)

Speed = Distance Time = 50 km 1 hour

We can also use these symbols:

Speed = Δs Δt

Where Δ (" Delta ") means "change in", and

  • s means distance ("s" for "space")
  • t means time

runners

Example: You run 360 m in 60 seconds.

So your speed is 6 meters per second (6 m/s).

Speed is commonly measured in:

  • meters per second (m/s or m s -1 ), or
  • kilometers per hour (km/h or km h -1 )

A km is 1000 m, and there are 3600 seconds in an hour, so we can convert like this (see Unit Conversion Method to learn more):

1 m 1 s × 1 km 1000 m × 3600 s 1 h = 3600 m · km · s 1000 s · m · h = 3.6 km 1 h

So 1 m/s is equal to 3.6 km/h

Example: What is 20 m/s in km/h ?

20 m/s × 3.6 km/h 1 m/s = 72 km/h

Example: What is 120 km/h in m/s ?

120 km/h × 1 m/s 3.6 km/h = 33.333... m/s

Average vs Instantaneous Speed

The examples so far calculate average speed : how far something travels over a period of time.

But speed can change as time goes by. A car can go faster and slower, maybe even stop at lights.

So there is also instantaneous speed : the speed at an instant in time. We can try to measure it by using a very short span of time (the shorter the better).

Example: Sam uses a stopwatch and measures 1.6 seconds as the car travels between two posts 20 m apart. What is the instantaneous speed ?

Well, we don't know exactly, as the car may have been speeding up or slowing down during that time, but we can estimate:

20 m 1.6 s = 12.5 m/s = 45 km/h

It is really still an average, but is close to an instantaneous speed.

Constant Speed

When the speed does not change it is constant .

For constant speed, the average and instantaneous speeds are the same.

Because the direction is important velocity uses displacement instead of distance:

Velocity = Displacement Time in a direction.

Example: You walk from home to the shop in 100 seconds, what is your speed and what is your velocity?

Speed = 220 m 100 s = 2.2 m/s

Velocity = 130 m 100 s East = 1.3 m/s East

You forgot your money so you turn around and go back home in 120 more seconds: what is your round-trip speed and velocity?

The total time is 100 s + 120 s = 220 s:

Speed = 440 m 220 s = 2.0 m/s

Velocity = 0 m 220 s = 0 m/s

Yes, the velocity is zero as you ended up where you started.

Learn more at Vectors .

Motion is relative. When we say something is "at rest" or "moving at 4 m/s" we forget to say "in relation to me" or "in relation to the ground", etc.

Think about this: are you really standing still? You are on planet Earth which is spinning at 40,075 km per day (about 1675 km/h or 465 m/s), and moving around the Sun at about 100,000 km/h, which is itself moving through the Galaxy.

Next time you are out walking, imagine you are still and it is the world that moves under your feet. Feels great.

It is all relative!

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Speed, Velocity, and Acceleration Problems

  • by Abnurlion
  • December 26, 2022 November 2, 2023

Table of Contents

What is Speed?

Definition of Speed : Speed is the rate of change of distance with time . Speed is different from velocity because it’s not in a specified direction. In this article, you will learn how to solve speed, velocity, and acceleration problems.

Additionally, you need to know that speed is a scalar quantity and we can write its symbol as S. The formula for calculating the speed of an object is:

Speed, S = Distance (d) / Time (t)

Thus, s = d/t

Note: In most cases, we also use S as a symbol for distance.

The S.I unit for speed is meter per second (m/s) or ms -1

Non-Uniform or Average Speed : This is a non-steady distance covered by an object at a particular period of time. We can also define non-uniform speed as the type of distance that an object covered at an equal interval of time.

The formula for calculating non-uniform speed is

Average speed = Total distance covered by the object / Total time taken

Actual speed: This is also known as the instantaneous speed of an object which is the distance covered by an object over a short interval of time.

You may also like to read:

How to Find Displacement in Physics

also How to Find a Position in Physics

and How to Calculate Bearing in Physics

What is Velocity?

Definition of Velocity: Velocity is the rate of displacement with time. Velocity is the speed of an object in a specified direction. The unit of velocity is the same as that of speed which is meter per second (ms -1 ). We use V as a symbol for velocity.

Note: We often use U to indicate initial speed, and V to indicate final speed.

The formula for calculating Velocity (V) = displacement (S) / time (t)

The difference between velocity and speed is the presence of displacement and distance respectively. Because displacement is a measure of separation in a specified direction, while distance is not in a specified direction. Velocity is a vector quantity.

Uniform Velocity

Definition of uniform velocity: The rate of change of displacement is constant no matter how small the time interval may be. Also, uniform velocity is the distance covered by an object in a specified direction in an equal time interval.

The formula for uniform velocity = Total displacement / Total time taken

What is Acceleration?

Definition of Acceleration: Acceleration is the rate of change of velocity with time. Acceleration is measured in meters per second square (ms -2 ). The symbol for acceleration is a. Acceleration is also a vector quantity.

The formula for acceleration , a = change in velocity (v)/time taken (t)

Thus, a = v/t

We can also write acceleration as

a = change in velocity/time = ΔV/Δt = (v – u)/t

[where v = final velocity, u = initial velocity, and t = time taken]

Uniform Acceleration

In the case of uniform acceleration , the rate of change of velocity with time is constant.

Average velocity of the object = (Initial velocity + final velocity)/2

Average velocity = (v + u)/2

What is Retardation?

Retardation is the decreasing rate of change in velocity moved, covered, or traveled by an object.

The formula for calculating retardation is

Retardation = Change in a decrease in velocity/time taken

Equations of Motion

You can also apply the following equations of motion to calculate the speed, velocity, and acceleration of the body:

  • s = [ (v + u)/2 ]t
  • v 2 = u 2 + 2as

s = ut + (1/2)at 2

Where v = final velocity, u = initial velocity, t = time, a = acceleration, and s = distance

How to Calculate Maximum height

What are Prefixes in Physics

How to Calculate Dimension in Physics

Solved Problems of Speed, Velocity, and Acceleration

Here are solved problems to help you understand how to calculate speed, velocity, and acceleration:

A train moves at a speed of 54km/h for a one-quarter minute. Find the distance travelled by train.

Speed, Velocity, and Acceleration Problems

From the question above

Speed = 54 km/h = (54 x 1000)/(60 x 60) = 54,000/3,600 = 15 m/s

Time = one quarter minute = 1/4 minute = (1/4) x 60 = 15 seconds

Since we have

speed = distance/time

After cross-multiplication, we will now have

Distance = speed x time

We can now insert our data into the above expression

Distance = 15 m/s x 15 s = 225 m

Therefore, the distance travelled by train is 225 meters.

A car travelled a distance of 5km in 50 seconds. Find the speed in meters per second.

Speed, Velocity, and Acceleration Problems

Distance = 5km = 5 x 1000m = 5,000m

Time = 50 seconds

and the formula for speed

speed = distance/time = 5000/50 = 100m/s

A motorcycle starting from rest moves with a uniform acceleration until it attains a speed of 108 kilometres per hour after 15 seconds. Find its acceleration.

Initial velocity, u = 0 (because the motorcycle starts from rest)

Final velocity, v = speed = 108 km/h = (108 x 1000m) / (60 x 60s) = 108,000/3,600 = 30m/s

Time taken, t = 15 seconds

Therefore, we can now apply the formula that says

Acceleration = change in velocity/ time = (v-u)/t = (30-0)/15 =30/15 = 2ms -2

Therefore, the acceleration of the motorcycle is 2ms -2

A bus covers 50 kilometres in 1 hour. What is it is the average speed?

Total distance covered = 50 km = 50 x 1000m = 50,000m

Time taken = 1 hour = 1 x 60 x 60s =3,600s

Therefore, we can now calculate the average speed of the bus by substituting our data into the above formula

Average speed = 50,000/3,600 = 13.9 m/s

Therefore, the average speed of the bus is 13.9m/s or approximately 14 meters per second.

A car travels 80 km in 1 hour and then another 20 km in the next hour. Find the average velocity of the car.

Speed, Velocity, and Acceleration Problems

Initial displacement of the car = 80km

Final displacement of the car = 20km

The total displacement of the car = initial displacement of the car + final displacement of the car

The total displacement of the car = 80km + 20km = 100km

The time for 80km is 1hr

And the time for 20km is 1hr

Total time taken = The time for 80km (1hr) + The time for 20km (1hr)

Total time taken = 1hr + 1hr = 2hrs

Now, to calculate the average velocity of the car, we apply the formula that says

Average velocity = total displacement/total time taken = 100km/2hrs = 50km/h

We can further convert the above answer into meters per second

Average velocity = 50km/h = (50 x 1000m)/(1 x 60 x 60s) = 50,000/3,600 = 13.9m/s or 14ms -1

Therefore, the average velocity of the car is 14ms -1

How to Conduct Physics Practical

and How to Calculate Velocity Ratio of an Inclined Plane

A body falls from the top of a tower 100 meters high and hits the ground in 5 seconds. Find its acceleration.

Displacement = 100m

Time = 5 seconds

and we can apply the formula for acceleration that says

acceleration, a = Displacement/time 2 = 100/5 2 = 100/25 = 4ms -2

Therefore the acceleration due to the gravity of the body is 4ms -2

Note: The acceleration due to the gravity of a body on the surface of the earth is constant (9.8ms -2 ), even though there may be a slight difference due to the mass and altitude of the body.

An object is thrown vertically upward at an initial velocity of 10ms -1 and reaches a maximum height of 50 meters. Find its initial upward acceleration.

Initial velocity, u = 10ms -1

Final velocity, v = 0

maximum height = displacement = 50m

Initial upward acceleration, a =?

When we apply the formula that says a = (v 2 – u 2 )/2d we will have

a = (0 – 10 2 )/(2 x 50) = -100/100 = -1 ms -2

Hence, since our acceleration is negative, we can now say that we are dealing with retardation or deceleration.

Therefore, the retardation is -1ms -2

Note: Retardation is the negative of acceleration, thus it is written in negative form.

A car is traveling at a velocity of 8ms -1 and experiences an acceleration of 5ms -2 . How far does it travel in 4 seconds?

Initial velocity, u = 8ms -1

acceleration, a = 5ms -2

Distance, s =?

time, t = 4s

We can apply the formula that says

s = 8 x 4 + (1/2) x 5 x 4 2

s = 32 + (1/2) x 80 = 32 + 40 = 72m

Therefore, the distance covered by the car in 4s is 72 meters.

A body is traveling at a velocity of 10m/s and experiences a deceleration of 5ms -2 . How long does it take the body to come to a complete stop?

Initial velocity, u = 10m/s

acceleration , a = retardation = -5ms -2

time, t = ?

We already know that acceleration, a = change in velocity/time

This implies that

Time, t = change in velocity/acceleration

t = (v – u)/t = (0-10)/-5 = -10/-5 = 2s

Therefore, the time it takes the car to stop is 2 seconds .

A body is traveling at a velocity of 20 m/s and has a mass of 10 kg. How much force is required to change its velocity by 10 m/s in 5 seconds?

Change in velocity, v =10 m/s

mass of the object, m = 10kg

time, t = 5s

We can apply newton’s second law of motion which says f = ma

and since a = change in velocity/time

we will have an acceleration equal to

a = 10/5 = 2ms -2

Therefore, to find the force, we can now say

f = ma = 10 x 2 = 20N

Therefore, the force that can help us to change the velocity by 10 m/s in 5 seconds is 20-Newton.

Drop a comment if there is anything you don’t understand about speed, velocity, and acceleration Problems.

How to Calculate Cost of Electricity Per kwh

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  • 6.1 Solving Problems with Newton’s Laws
  • Introduction
  • 1.1 The Scope and Scale of Physics
  • 1.2 Units and Standards
  • 1.3 Unit Conversion
  • 1.4 Dimensional Analysis
  • 1.5 Estimates and Fermi Calculations
  • 1.6 Significant Figures
  • 1.7 Solving Problems in Physics
  • Key Equations
  • Conceptual Questions
  • Additional Problems
  • Challenge Problems
  • 2.1 Scalars and Vectors
  • 2.2 Coordinate Systems and Components of a Vector
  • 2.3 Algebra of Vectors
  • 2.4 Products of Vectors
  • 3.1 Position, Displacement, and Average Velocity
  • 3.2 Instantaneous Velocity and Speed
  • 3.3 Average and Instantaneous Acceleration
  • 3.4 Motion with Constant Acceleration
  • 3.5 Free Fall
  • 3.6 Finding Velocity and Displacement from Acceleration
  • 4.1 Displacement and Velocity Vectors
  • 4.2 Acceleration Vector
  • 4.3 Projectile Motion
  • 4.4 Uniform Circular Motion
  • 4.5 Relative Motion in One and Two Dimensions
  • 5.2 Newton's First Law
  • 5.3 Newton's Second Law
  • 5.4 Mass and Weight
  • 5.5 Newton’s Third Law
  • 5.6 Common Forces
  • 5.7 Drawing Free-Body Diagrams
  • 6.2 Friction
  • 6.3 Centripetal Force
  • 6.4 Drag Force and Terminal Speed
  • 7.2 Kinetic Energy
  • 7.3 Work-Energy Theorem
  • 8.1 Potential Energy of a System
  • 8.2 Conservative and Non-Conservative Forces
  • 8.3 Conservation of Energy
  • 8.4 Potential Energy Diagrams and Stability
  • 8.5 Sources of Energy
  • 9.1 Linear Momentum
  • 9.2 Impulse and Collisions
  • 9.3 Conservation of Linear Momentum
  • 9.4 Types of Collisions
  • 9.5 Collisions in Multiple Dimensions
  • 9.6 Center of Mass
  • 9.7 Rocket Propulsion
  • 10.1 Rotational Variables
  • 10.2 Rotation with Constant Angular Acceleration
  • 10.3 Relating Angular and Translational Quantities
  • 10.4 Moment of Inertia and Rotational Kinetic Energy
  • 10.5 Calculating Moments of Inertia
  • 10.6 Torque
  • 10.7 Newton’s Second Law for Rotation
  • 10.8 Work and Power for Rotational Motion
  • 11.1 Rolling Motion
  • 11.2 Angular Momentum
  • 11.3 Conservation of Angular Momentum
  • 11.4 Precession of a Gyroscope
  • 12.1 Conditions for Static Equilibrium
  • 12.2 Examples of Static Equilibrium
  • 12.3 Stress, Strain, and Elastic Modulus
  • 12.4 Elasticity and Plasticity
  • 13.1 Newton's Law of Universal Gravitation
  • 13.2 Gravitation Near Earth's Surface
  • 13.3 Gravitational Potential Energy and Total Energy
  • 13.4 Satellite Orbits and Energy
  • 13.5 Kepler's Laws of Planetary Motion
  • 13.6 Tidal Forces
  • 13.7 Einstein's Theory of Gravity
  • 14.1 Fluids, Density, and Pressure
  • 14.2 Measuring Pressure
  • 14.3 Pascal's Principle and Hydraulics
  • 14.4 Archimedes’ Principle and Buoyancy
  • 14.5 Fluid Dynamics
  • 14.6 Bernoulli’s Equation
  • 14.7 Viscosity and Turbulence
  • 15.1 Simple Harmonic Motion
  • 15.2 Energy in Simple Harmonic Motion
  • 15.3 Comparing Simple Harmonic Motion and Circular Motion
  • 15.4 Pendulums
  • 15.5 Damped Oscillations
  • 15.6 Forced Oscillations
  • 16.1 Traveling Waves
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Learning Objectives

By the end of this section, you will be able to:

  • Apply problem-solving techniques to solve for quantities in more complex systems of forces
  • Use concepts from kinematics to solve problems using Newton’s laws of motion
  • Solve more complex equilibrium problems
  • Solve more complex acceleration problems
  • Apply calculus to more advanced dynamics problems

Success in problem solving is necessary to understand and apply physical principles. We developed a pattern of analyzing and setting up the solutions to problems involving Newton’s laws in Newton’s Laws of Motion ; in this chapter, we continue to discuss these strategies and apply a step-by-step process.

Problem-Solving Strategies

We follow here the basics of problem solving presented earlier in this text, but we emphasize specific strategies that are useful in applying Newton’s laws of motion . Once you identify the physical principles involved in the problem and determine that they include Newton’s laws of motion, you can apply these steps to find a solution. These techniques also reinforce concepts that are useful in many other areas of physics. Many problem-solving strategies are stated outright in the worked examples, so the following techniques should reinforce skills you have already begun to develop.

Problem-Solving Strategy

Applying newton’s laws of motion.

  • Identify the physical principles involved by listing the givens and the quantities to be calculated.
  • Sketch the situation, using arrows to represent all forces.
  • Determine the system of interest. The result is a free-body diagram that is essential to solving the problem.
  • Apply Newton’s second law to solve the problem. If necessary, apply appropriate kinematic equations from the chapter on motion along a straight line.
  • Check the solution to see whether it is reasonable.

Let’s apply this problem-solving strategy to the challenge of lifting a grand piano into a second-story apartment. Once we have determined that Newton’s laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation. Such a sketch is shown in Figure 6.2 (a). Then, as in Figure 6.2 (b), we can represent all forces with arrows. Whenever sufficient information exists, it is best to label these arrows carefully and make the length and direction of each correspond to the represented force.

As with most problems, we next need to identify what needs to be determined and what is known or can be inferred from the problem as stated, that is, make a list of knowns and unknowns. It is particularly crucial to identify the system of interest, since Newton’s second law involves only external forces. We can then determine which forces are external and which are internal, a necessary step to employ Newton’s second law. (See Figure 6.2 (c).) Newton’s third law may be used to identify whether forces are exerted between components of a system (internal) or between the system and something outside (external). As illustrated in Newton’s Laws of Motion , the system of interest depends on the question we need to answer. Only forces are shown in free-body diagrams, not acceleration or velocity. We have drawn several free-body diagrams in previous worked examples. Figure 6.2 (c) shows a free-body diagram for the system of interest. Note that no internal forces are shown in a free-body diagram.

Once a free-body diagram is drawn, we apply Newton’s second law. This is done in Figure 6.2 (d) for a particular situation. In general, once external forces are clearly identified in free-body diagrams, it should be a straightforward task to put them into equation form and solve for the unknown, as done in all previous examples. If the problem is one-dimensional—that is, if all forces are parallel—then the forces can be handled algebraically. If the problem is two-dimensional, then it must be broken down into a pair of one-dimensional problems. We do this by projecting the force vectors onto a set of axes chosen for convenience. As seen in previous examples, the choice of axes can simplify the problem. For example, when an incline is involved, a set of axes with one axis parallel to the incline and one perpendicular to it is most convenient. It is almost always convenient to make one axis parallel to the direction of motion, if this is known. Generally, just write Newton’s second law in components along the different directions. Then, you have the following equations:

(If, for example, the system is accelerating horizontally, then you can then set a y = 0 . a y = 0 . ) We need this information to determine unknown forces acting on a system.

As always, we must check the solution. In some cases, it is easy to tell whether the solution is reasonable. For example, it is reasonable to find that friction causes an object to slide down an incline more slowly than when no friction exists. In practice, intuition develops gradually through problem solving; with experience, it becomes progressively easier to judge whether an answer is reasonable. Another way to check a solution is to check the units. If we are solving for force and end up with units of millimeters per second, then we have made a mistake.

There are many interesting applications of Newton’s laws of motion, a few more of which are presented in this section. These serve also to illustrate some further subtleties of physics and to help build problem-solving skills. We look first at problems involving particle equilibrium, which make use of Newton’s first law, and then consider particle acceleration, which involves Newton’s second law.

Particle Equilibrium

Recall that a particle in equilibrium is one for which the external forces are balanced. Static equilibrium involves objects at rest, and dynamic equilibrium involves objects in motion without acceleration, but it is important to remember that these conditions are relative. For example, an object may be at rest when viewed from our frame of reference, but the same object would appear to be in motion when viewed by someone moving at a constant velocity. We now make use of the knowledge attained in Newton’s Laws of Motion , regarding the different types of forces and the use of free-body diagrams, to solve additional problems in particle equilibrium .

Example 6.1

Different tensions at different angles.

Thus, as you might expect,

This gives us the following relationship:

Note that T 1 T 1 and T 2 T 2 are not equal in this case because the angles on either side are not equal. It is reasonable that T 2 T 2 ends up being greater than T 1 T 1 because it is exerted more vertically than T 1 . T 1 .

Now consider the force components along the vertical or y -axis:

This implies

Substituting the expressions for the vertical components gives

There are two unknowns in this equation, but substituting the expression for T 2 T 2 in terms of T 1 T 1 reduces this to one equation with one unknown:

which yields

Solving this last equation gives the magnitude of T 1 T 1 to be

Finally, we find the magnitude of T 2 T 2 by using the relationship between them, T 2 = 1.225 T 1 T 2 = 1.225 T 1 , found above. Thus we obtain

Significance

Particle acceleration.

We have given a variety of examples of particles in equilibrium. We now turn our attention to particle acceleration problems, which are the result of a nonzero net force. Refer again to the steps given at the beginning of this section, and notice how they are applied to the following examples.

Example 6.2

Drag force on a barge.

The drag of the water F → D F → D is in the direction opposite to the direction of motion of the boat; this force thus works against F → app , F → app , as shown in the free-body diagram in Figure 6.4 (b). The system of interest here is the barge, since the forces on it are given as well as its acceleration. Because the applied forces are perpendicular, the x - and y -axes are in the same direction as F → 1 F → 1 and F → 2 . F → 2 . The problem quickly becomes a one-dimensional problem along the direction of F → app F → app , since friction is in the direction opposite to F → app . F → app . Our strategy is to find the magnitude and direction of the net applied force F → app F → app and then apply Newton’s second law to solve for the drag force F → D . F → D .

The angle is given by

From Newton’s first law, we know this is the same direction as the acceleration. We also know that F → D F → D is in the opposite direction of F → app , F → app , since it acts to slow down the acceleration. Therefore, the net external force is in the same direction as F → app , F → app , but its magnitude is slightly less than F → app . F → app . The problem is now one-dimensional. From the free-body diagram, we can see that

However, Newton’s second law states that

This can be solved for the magnitude of the drag force of the water F D F D in terms of known quantities:

Substituting known values gives

The direction of F → D F → D has already been determined to be in the direction opposite to F → app , F → app , or at an angle of 53 ° 53 ° south of west.

In Newton’s Laws of Motion , we discussed the normal force , which is a contact force that acts normal to the surface so that an object does not have an acceleration perpendicular to the surface. The bathroom scale is an excellent example of a normal force acting on a body. It provides a quantitative reading of how much it must push upward to support the weight of an object. But can you predict what you would see on the dial of a bathroom scale if you stood on it during an elevator ride? Will you see a value greater than your weight when the elevator starts up? What about when the elevator moves upward at a constant speed? Take a guess before reading the next example.

Example 6.3

What does the bathroom scale read in an elevator.

From the free-body diagram, we see that F → net = F → s − w → , F → net = F → s − w → , so we have

Solving for F s F s gives us an equation with only one unknown:

or, because w = m g , w = m g , simply

No assumptions were made about the acceleration, so this solution should be valid for a variety of accelerations in addition to those in this situation. ( Note: We are considering the case when the elevator is accelerating upward. If the elevator is accelerating downward, Newton’s second law becomes F s − w = − m a . F s − w = − m a . )

  • We have a = 1.20 m/s 2 , a = 1.20 m/s 2 , so that F s = ( 75.0 kg ) ( 9.80 m/s 2 ) + ( 75.0 kg ) ( 1.20 m/s 2 ) F s = ( 75.0 kg ) ( 9.80 m/s 2 ) + ( 75.0 kg ) ( 1.20 m/s 2 ) yielding F s = 825 N . F s = 825 N .
  • Now, what happens when the elevator reaches a constant upward velocity? Will the scale still read more than his weight? For any constant velocity—up, down, or stationary—acceleration is zero because a = Δ v Δ t a = Δ v Δ t and Δ v = 0 . Δ v = 0 . Thus, F s = m a + m g = 0 + m g F s = m a + m g = 0 + m g or F s = ( 75.0 kg ) ( 9.80 m/s 2 ) , F s = ( 75.0 kg ) ( 9.80 m/s 2 ) , which gives F s = 735 N . F s = 735 N .

Thus, the scale reading in the elevator is greater than his 735-N (165-lb.) weight. This means that the scale is pushing up on the person with a force greater than his weight, as it must in order to accelerate him upward. Clearly, the greater the acceleration of the elevator, the greater the scale reading, consistent with what you feel in rapidly accelerating versus slowly accelerating elevators. In Figure 6.5 (b), the scale reading is 735 N, which equals the person’s weight. This is the case whenever the elevator has a constant velocity—moving up, moving down, or stationary.

Check Your Understanding 6.1

Now calculate the scale reading when the elevator accelerates downward at a rate of 1.20 m/s 2 . 1.20 m/s 2 .

The solution to the previous example also applies to an elevator accelerating downward, as mentioned. When an elevator accelerates downward, a is negative, and the scale reading is less than the weight of the person. If a constant downward velocity is reached, the scale reading again becomes equal to the person’s weight. If the elevator is in free fall and accelerating downward at g , then the scale reading is zero and the person appears to be weightless.

Example 6.4

Two attached blocks.

For block 1: T → + w → 1 + N → = m 1 a → 1 T → + w → 1 + N → = m 1 a → 1

For block 2: T → + w → 2 = m 2 a → 2 . T → + w → 2 = m 2 a → 2 .

Notice that T → T → is the same for both blocks. Since the string and the pulley have negligible mass, and since there is no friction in the pulley, the tension is the same throughout the string. We can now write component equations for each block. All forces are either horizontal or vertical, so we can use the same horizontal/vertical coordinate system for both objects

When block 1 moves to the right, block 2 travels an equal distance downward; thus, a 1 x = − a 2 y . a 1 x = − a 2 y . Writing the common acceleration of the blocks as a = a 1 x = − a 2 y , a = a 1 x = − a 2 y , we now have

From these two equations, we can express a and T in terms of the masses m 1 and m 2 , and g : m 1 and m 2 , and g :

Check Your Understanding 6.2

Calculate the acceleration of the system, and the tension in the string, when the masses are m 1 = 5.00 kg m 1 = 5.00 kg and m 2 = 3.00 kg . m 2 = 3.00 kg .

Example 6.5

Atwood machine.

  • We have For m 1 , ∑ F y = T − m 1 g = m 1 a . For m 2 , ∑ F y = T − m 2 g = − m 2 a . For m 1 , ∑ F y = T − m 1 g = m 1 a . For m 2 , ∑ F y = T − m 2 g = − m 2 a . (The negative sign in front of m 2 a m 2 a indicates that m 2 m 2 accelerates downward; both blocks accelerate at the same rate, but in opposite directions.) Solve the two equations simultaneously (subtract them) and the result is ( m 2 − m 1 ) g = ( m 1 + m 2 ) a . ( m 2 − m 1 ) g = ( m 1 + m 2 ) a . Solving for a : a = m 2 − m 1 m 1 + m 2 g = 4 kg − 2 kg 4 kg + 2 kg ( 9.8 m/s 2 ) = 3.27 m/s 2 . a = m 2 − m 1 m 1 + m 2 g = 4 kg − 2 kg 4 kg + 2 kg ( 9.8 m/s 2 ) = 3.27 m/s 2 .
  • Observing the first block, we see that T − m 1 g = m 1 a T = m 1 ( g + a ) = ( 2 kg ) ( 9.8 m/s 2 + 3.27 m/s 2 ) = 26.1 N . T − m 1 g = m 1 a T = m 1 ( g + a ) = ( 2 kg ) ( 9.8 m/s 2 + 3.27 m/s 2 ) = 26.1 N .

Check Your Understanding 6.3

Determine a general formula in terms of m 1 , m 2 m 1 , m 2 and g for calculating the tension in the string for the Atwood machine shown above.

Newton’s Laws of Motion and Kinematics

Physics is most interesting and most powerful when applied to general situations that involve more than a narrow set of physical principles. Newton’s laws of motion can also be integrated with other concepts that have been discussed previously in this text to solve problems of motion. For example, forces produce accelerations, a topic of kinematics , and hence the relevance of earlier chapters.

When approaching problems that involve various types of forces, acceleration, velocity, and/or position, listing the givens and the quantities to be calculated will allow you to identify the principles involved. Then, you can refer to the chapters that deal with a particular topic and solve the problem using strategies outlined in the text. The following worked example illustrates how the problem-solving strategy given earlier in this chapter, as well as strategies presented in other chapters, is applied to an integrated concept problem.

Example 6.6

What force must a soccer player exert to reach top speed.

  • We are given the initial and final velocities (zero and 8.00 m/s forward); thus, the change in velocity is Δ v = 8.00 m/s Δ v = 8.00 m/s . We are given the elapsed time, so Δ t = 2.50 s . Δ t = 2.50 s . The unknown is acceleration, which can be found from its definition: a = Δ v Δ t . a = Δ v Δ t . Substituting the known values yields a = 8.00 m/s 2.50 s = 3.20 m/s 2 . a = 8.00 m/s 2.50 s = 3.20 m/s 2 .
  • Here we are asked to find the average force the ground exerts on the runner to produce this acceleration. (Remember that we are dealing with the force or forces acting on the object of interest.) This is the reaction force to that exerted by the player backward against the ground, by Newton’s third law. Neglecting air resistance, this would be equal in magnitude to the net external force on the player, since this force causes her acceleration. Since we now know the player’s acceleration and are given her mass, we can use Newton’s second law to find the force exerted. That is, F net = m a . F net = m a . Substituting the known values of m and a gives F net = ( 70.0 kg ) ( 3.20 m/s 2 ) = 224 N . F net = ( 70.0 kg ) ( 3.20 m/s 2 ) = 224 N .

This is a reasonable result: The acceleration is attainable for an athlete in good condition. The force is about 50 pounds, a reasonable average force.

Check Your Understanding 6.4

The soccer player stops after completing the play described above, but now notices that the ball is in position to be stolen. If she now experiences a force of 126 N to attempt to steal the ball, which is 2.00 m away from her, how long will it take her to get to the ball?

Example 6.7

What force acts on a model helicopter.

The magnitude of the force is now easily found:

Check Your Understanding 6.5

Find the direction of the resultant for the 1.50-kg model helicopter.

Example 6.8

Baggage tractor.

  • ∑ F x = m system a x ∑ F x = m system a x and ∑ F x = 820.0 t , ∑ F x = 820.0 t , so 820.0 t = ( 650.0 + 250.0 + 150.0 ) a a = 0.7809 t . 820.0 t = ( 650.0 + 250.0 + 150.0 ) a a = 0.7809 t . Since acceleration is a function of time, we can determine the velocity of the tractor by using a = d v d t a = d v d t with the initial condition that v 0 = 0 v 0 = 0 at t = 0 . t = 0 . We integrate from t = 0 t = 0 to t = 3 : t = 3 : d v = a d t , ∫ 0 3 d v = ∫ 0 3.00 a d t = ∫ 0 3.00 0.7809 t d t , v = 0.3905 t 2 ] 0 3.00 = 3.51 m/s . d v = a d t , ∫ 0 3 d v = ∫ 0 3.00 a d t = ∫ 0 3.00 0.7809 t d t , v = 0.3905 t 2 ] 0 3.00 = 3.51 m/s .
  • Refer to the free-body diagram in Figure 6.8 (b). ∑ F x = m tractor a x 820.0 t − T = m tractor ( 0.7805 ) t ( 820.0 ) ( 3.00 ) − T = ( 650.0 ) ( 0.7805 ) ( 3.00 ) T = 938 N . ∑ F x = m tractor a x 820.0 t − T = m tractor ( 0.7805 ) t ( 820.0 ) ( 3.00 ) − T = ( 650.0 ) ( 0.7805 ) ( 3.00 ) T = 938 N .

Recall that v = d s d t v = d s d t and a = d v d t a = d v d t . If acceleration is a function of time, we can use the calculus forms developed in Motion Along a Straight Line , as shown in this example. However, sometimes acceleration is a function of displacement. In this case, we can derive an important result from these calculus relations. Solving for dt in each, we have d t = d s v d t = d s v and d t = d v a . d t = d v a . Now, equating these expressions, we have d s v = d v a . d s v = d v a . We can rearrange this to obtain a d s = v d v . a d s = v d v .

Example 6.9

Motion of a projectile fired vertically.

The acceleration depends on v and is therefore variable. Since a = f ( v ) , a = f ( v ) , we can relate a to v using the rearrangement described above,

We replace ds with dy because we are dealing with the vertical direction,

We now separate the variables ( v ’s and dv ’s on one side; dy on the other):

Thus, h = 114 m . h = 114 m .

Check Your Understanding 6.6

If atmospheric resistance is neglected, find the maximum height for the mortar shell. Is calculus required for this solution?

Interactive

Explore the forces at work in this simulation when you try to push a filing cabinet. Create an applied force and see the resulting frictional force and total force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. View a free-body diagram of all the forces (including gravitational and normal forces).

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Access for free at https://openstax.org/books/university-physics-volume-1/pages/1-introduction
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  • Publisher/website: OpenStax
  • Book title: University Physics Volume 1
  • Publication date: Sep 19, 2016
  • Location: Houston, Texas
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  • Section URL: https://openstax.org/books/university-physics-volume-1/pages/6-1-solving-problems-with-newtons-laws

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Speed and Velocity

Practice problem 1.

Average speed

This problem is deceptively easy. Averaging is taught in elementary school, which makes this an elementary problem. Right?

The add-and-divide method of averaging only works when averaging items of equal weight. The average age of the students in a classroom is the sum of their ages divided by the number of students only because each student is considered to have the same significance (a student, is a student, is a student, … ). In this problem, however, the two segments of the walk are significantly different. The second "half" was actually the majority of the walk. It carries more weight than the shorter first "half". Thus, the add-and-divide method won't work.

Let's return to our definition. Since speed is the rate of change of distance with time, we'll need both the distance traveled and the time it took to complete the walk. After we determine both of these numbers, the rest is easy.

Look closely at the calculations on the right side. Notice that the formula contains ∆ (delta) symbols and yet I added the distances in the numerator and the times in the denominator. That's because ∆ doesn't mean difference, it means change. During the walk my position didn't change from 6.0 km to 10 km, it changed first by 6.0 km and then by 10 km for a total change of 16 km.

Average velocity

Velocity is the rate of change of displacement with respect to time. Velocity is a vector, which means the problem should be solved graphically. Draw an arrow pointing toward the top of the page (north). Label it 6 km. Draw another arrow to the left (west) starting from the previous one (arranged head to tail). Make it slightly longer and label it 10 km. Draw a third arrow starting on the tail of the first and ending on the head of the second. Since north and west are at right angles to one another, the resultant displacement is the hypotenuse of a right triangle. Use Pythagorean theorem to find its magnitude and tangent to find its direction.

Divide displacement by time to get velocity.

practice problem 2

Notice that no numbers are stated in this problem. When a numerical value is needed to solve a problem and that number is not given, it could mean one of several things.

  • Look it up! It may appear somewhere in the textbook you are using — on the inside covers, in an appendix , or in the text of the chapter you are currently working on. It may be found in the reference table that some teachers distribute. Standardized exams usually also have their own reference table .
  • Know it! Some numbers are numbers that you should just know. In this problem, there is one relevant number that nearly everyone knows. You may also be expected to memorize certain numbers by an instructor or professor.
  • Calculate it! Maybe there's a way to find the number you need to know using other numbers given in the problem.
  • Forget about it! Maybe you don't really need the number you think you do. Maybe you are on the wrong track. Especially under test conditions, it is highly unlikely that you could be asked a question that requires a numerical value that you can not find, do not know, or can not calculate. Perhaps there is another method to solve this problem.

In order to calculate speed, you will need distance and time. What distance does a point on the equator move in a convenient period of time? Well, I hope you know that the Earth rotates once on its axis every day. You should also know how to calculate the length of a day in seconds. (A day is the period of the Earth's rotation, for which an upper case T is the symbol.) During a day, a point on the Earth's equator would have traveled a distance equal to the circumference of the Earth. The radius of the Earth is a number commonly found in textbooks and on reference tables. The problem can now be solved.

That's about 75% faster than the speed of sound in air at room temperature (343 m/s). An interesting problem to be dealt with later is that if the Earth is spinning so rapidly, how come things on the equator don't fly off into space?

practice problem 3

Astronomical distances are so large that using meters is cumbersome. For really large distances the light year is the best unit. A light year is the distance that light would travel in one year in a vacuum. Since the speed of light is fast, and a year is long, the light year is a pretty good unit for astronomy. One light year is about ten petameters (ten quadrillion meters) as the following calculation shows.

Start with the definition of speed and solve it for distance. The traditional symbol for the speed of light is c from the Latin word for swiftness — celeritas .

Numbers in, answer out.

Since both the speed of light and the year have exactly defined values in the International System of Units, the light year can be stated with an unnecessarily large number of significant digits.

Some distances in light years are provided below.

  • The distance to Proxima Centauri (the star nearest the Sun) is 4.3 light years.
  • The diameter of the Milky Way (the collection of stars that includes the Sun and all the stars visible to the naked eye) is about 100,000 light years.
  • The distance to Andromeda (the nearest galaxy outside the Milky Way) is about 2 million light years.
  • The distance to the edge of the universe (the observable part of it) is 13.77 billion light years .

practice problem 4

Everyone should know (or at least realize after a bit of thought) that there are…

∆ t  =  60 × 60 = 3,600 s

in an hour. Many Americans who are fans of track and field know that four laps around a 400 m outdoor track is almost one mile.

∆ s  =  1 mile ≈ 4 × 400 m = 1,600 m = 1.6 km

More precisely… actually, most precisely… actually, exactly by definition…

∆ s  =  1 mile = 1,609.344 m = 1.609344 km

The first answer…

For comparison, the speed limit on many Canadian highways is 100 km/h.

The second answer…

You should note that the number with International units is a little bit less than half the value of the number with British-American units. I've gotten used to mph, but I have to be conversant in m/s for my job. A good rule of thumb for comparing speeds is to…

divide by 2 and subtract a little when converting from mph to m/s

multiply by 2 and add a little when converting from m/s to mph.

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Chapter 8: Rational Expressions

8.8 Rate Word Problems: Speed, Distance and Time

Distance, rate and time problems are a standard application of linear equations. When solving these problems, use the relationship rate (speed or velocity) times time equals distance .

[latex]r\cdot t=d[/latex]

For example, suppose a person were to travel 30 km/h for 4 h. To find the total distance, multiply rate times time or (30km/h)(4h) = 120 km.

The problems to be solved here will have a few more steps than described above. So to keep the information in the problem organized, use a table. An example of the basic structure of the table is below:

The third column, distance, will always be filled in by multiplying the rate and time columns together. If given a total distance of both persons or trips, put this information in the distance column. Now use this table to set up and solve the following examples.

Example 8.8.1

Joey and Natasha start from the same point and walk in opposite directions. Joey walks 2 km/h faster than Natasha. After 3 hours, they are 30 kilometres apart. How fast did each walk?

The distance travelled by both is 30 km. Therefore, the equation to be solved is:

[latex]\begin{array}{rrrrrrl} 3r&+&3(r&+&2)&=&30 \\ 3r&+&3r&+&6&=&30 \\ &&&-&6&&-6 \\ \hline &&&&\dfrac{6r}{6}&=&\dfrac{24}{6} \\ \\ &&&&r&=&4 \text{ km/h} \end{array}[/latex]

This means that Natasha walks at 4 km/h and Joey walks at 6 km/h.

Example 8.8.2

Nick and Chloe left their campsite by canoe and paddled downstream at an average speed of 12 km/h. They turned around and paddled back upstream at an average rate of 4 km/h. The total trip took 1 hour. After how much time did the campers turn around downstream?

The distance travelled downstream is the same distance that they travelled upstream. Therefore, the equation to be solved is:

[latex]\begin{array}{rrlll} 12(t)&=&4(1&-&t) \\ 12t&=&4&-&4t \\ +4t&&&+&4t \\ \hline \dfrac{16t}{16}&=&\dfrac{4}{16}&& \\ \\ t&=&0.25&& \end{array}[/latex]

This means the campers paddled downstream for 0.25 h and spent 0.75 h paddling back.

Example 8.8.3

Terry leaves his house riding a bike at 20 km/h. Sally leaves 6 h later on a scooter to catch up with him travelling at 80 km/h. How long will it take her to catch up with him?

The distance travelled by both is the same. Therefore, the equation to be solved is:

[latex]\begin{array}{rrrrr} 20(t)&=&80(t&-&6) \\ 20t&=&80t&-&480 \\ -80t&&-80t&& \\ \hline \dfrac{-60t}{-60}&=&\dfrac{-480}{-60}&& \\ \\ t&=&8&& \end{array}[/latex]

This means that Terry travels for 8 h and Sally only needs 2 h to catch up to him.

Example 8.8.4

On a 130-kilometre trip, a car travelled at an average speed of 55 km/h and then reduced its speed to 40 km/h for the remainder of the trip. The trip took 2.5 h. For how long did the car travel 40 km/h?

[latex]\begin{array}{rrrrrrr} 55(t)&+&40(2.5&-&t)&=&130 \\ 55t&+&100&-&40t&=&130 \\ &-&100&&&&-100 \\ \hline &&&&\dfrac{15t}{15}&=&\dfrac{30}{15} \\ \\ &&&&t&=&2 \end{array}[/latex]

This means that the time spent travelling at 40 km/h was 0.5 h.

Distance, time and rate problems have a few variations that mix the unknowns between distance, rate and time. They generally involve solving a problem that uses the combined distance travelled to equal some distance or a problem in which the distances travelled by both parties is the same. These distance, rate and time problems will be revisited later on in this textbook where quadratic solutions are required to solve them.

For Questions 1 to 8, find the equations needed to solve the problems. Do not solve.

  • A is 60 kilometres from B. An automobile at A starts for B at the rate of 20 km/h at the same time that an automobile at B starts for A at the rate of 25 km/h. How long will it be before the automobiles meet?
  • Two automobiles are 276 kilometres apart and start to travel toward each other at the same time. They travel at rates differing by 5 km/h. If they meet after 6 h, find the rate of each.
  • Two trains starting at the same station head in opposite directions. They travel at the rates of 25 and 40 km/h, respectively. If they start at the same time, how soon will they be 195 kilometres apart?
  • Two bike messengers, Jerry and Susan, ride in opposite directions. If Jerry rides at the rate of 20 km/h, at what rate must Susan ride if they are 150 kilometres apart in 5 hours?
  • A passenger and a freight train start toward each other at the same time from two points 300 kilometres apart. If the rate of the passenger train exceeds the rate of the freight train by 15 km/h, and they meet after 4 hours, what must the rate of each be?
  • Two automobiles started travelling in opposite directions at the same time from the same point. Their rates were 25 and 35 km/h, respectively. After how many hours were they 180 kilometres apart?
  • A man having ten hours at his disposal made an excursion by bike, riding out at the rate of 10 km/h and returning on foot at the rate of 3 km/h. Find the distance he rode.
  • A man walks at the rate of 4 km/h. How far can he walk into the country and ride back on a trolley that travels at the rate of 20 km/h, if he must be back home 3 hours from the time he started?

Solve Questions 9 to 22.

  • A boy rides away from home in an automobile at the rate of 28 km/h and walks back at the rate of 4 km/h. The round trip requires 2 hours. How far does he ride?
  • A motorboat leaves a harbour and travels at an average speed of 15 km/h toward an island. The average speed on the return trip was 10 km/h. How far was the island from the harbour if the trip took a total of 5 hours?
  • A family drove to a resort at an average speed of 30 km/h and later returned over the same road at an average speed of 50 km/h. Find the distance to the resort if the total driving time was 8 hours.
  • As part of his flight training, a student pilot was required to fly to an airport and then return. The average speed to the airport was 90 km/h, and the average speed returning was 120 km/h. Find the distance between the two airports if the total flying time was 7 hours.
  • Sam starts travelling at 4 km/h from a campsite 2 hours ahead of Sue, who travels 6 km/h in the same direction. How many hours will it take for Sue to catch up to Sam?
  • A man travels 5 km/h. After travelling for 6 hours, another man starts at the same place as the first man did, following at the rate of 8 km/h. When will the second man overtake the first?
  • A motorboat leaves a harbour and travels at an average speed of 8 km/h toward a small island. Two hours later, a cabin cruiser leaves the same harbour and travels at an average speed of 16 km/h toward the same island. How many hours after the cabin cruiser leaves will it be alongside the motorboat?
  • A long distance runner started on a course, running at an average speed of 6 km/h. One hour later, a second runner began the same course at an average speed of 8 km/h. How long after the second runner started will they overtake the first runner?
  • Two men are travelling in opposite directions at the rate of 20 and 30 km/h at the same time and from the same place. In how many hours will they be 300 kilometres apart?
  • Two trains start at the same time from the same place and travel in opposite directions. If the rate of one is 6 km/h more than the rate of the other and they are 168 kilometres apart at the end of 4 hours, what is the rate of each?
  • Two cyclists start from the same point and ride in opposite directions. One cyclist rides twice as fast as the other. In three hours, they are 72 kilometres apart. Find the rate of each cyclist.
  • Two small planes start from the same point and fly in opposite directions. The first plane is flying 25 km/h slower than the second plane. In two hours, the planes are 430 kilometres apart. Find the rate of each plane.
  • On a 130-kilometre trip, a car travelled at an average speed of 55 km/h and then reduced its speed to 40 km/h for the remainder of the trip. The trip took a total of 2.5 hours. For how long did the car travel at 40 km/h?
  • Running at an average rate of 8 m/s, a sprinter ran to the end of a track and then jogged back to the starting point at an average of 3 m/s. The sprinter took 55 s to run to the end of the track and jog back. Find the length of the track.

Answer Key 8.8

Intermediate Algebra by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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solving speed and velocity problems

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UsingKinEqns1ThN.png

Check Your Understanding

Answer: d = 1720 m

Answer: a = 8.10 m/s/s

Answers: d = 33.1 m and v f = 25.5 m/s

Answers: a = 11.2 m/s/s and d = 79.8 m

Answer: t = 1.29 s

Answers: a = 243 m/s/s

Answer: a = 0.712 m/s/s

Answer: d = 704 m

Answer: d = 28.6 m

Answer: v i = 7.17 m/s

Answer: v i = 5.03 m/s and hang time = 1.03 s (except for in sports commericals)

Answer: a = 1.62*10 5 m/s/s

Answer: d = 48.0 m

Answer: t = 8.69 s

Answer: a = -1.08*10^6 m/s/s

Answer: d = -57.0 m (57.0 meters deep) 

Answer: v i = 47.6 m/s

Answer: a = 2.86 m/s/s and t = 30. 8 s

Answer: a = 15.8 m/s/s

Answer: v i = 94.4 mi/hr

Solutions to Above Problems

d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s 2 )*(32.8 s) 2

Return to Problem 1

110 m = (0 m/s)*(5.21 s)+ 0.5*(a)*(5.21 s) 2

110 m = (13.57 s 2 )*a

a = (110 m)/(13.57 s 2 )

a = 8.10 m/ s 2

Return to Problem 2

d = (0 m/s)*(2.60 s)+ 0.5*(-9.8 m/s 2 )*(2.60 s) 2

d = -33.1 m (- indicates direction)

v f = v i + a*t

v f = 0 + (-9.8 m/s 2 )*(2.60 s)

v f = -25.5 m/s (- indicates direction)

Return to Problem 3

a = (46.1 m/s - 18.5 m/s)/(2.47 s)

a = 11.2 m/s 2

d = v i *t + 0.5*a*t 2

d = (18.5 m/s)*(2.47 s)+ 0.5*(11.2 m/s 2 )*(2.47 s) 2

d = 45.7 m + 34.1 m

(Note: the d can also be calculated using the equation v f 2 = v i 2 + 2*a*d)

Return to Problem 4

-1.40 m = (0 m/s)*(t)+ 0.5*(-1.67 m/s 2 )*(t) 2

-1.40 m = 0+ (-0.835 m/s 2 )*(t) 2

(-1.40 m)/(-0.835 m/s 2 ) = t 2

1.68 s 2 = t 2

Return to Problem 5

a = (444 m/s - 0 m/s)/(1.83 s)

a = 243 m/s 2

d = (0 m/s)*(1.83 s)+ 0.5*(243 m/s 2 )*(1.83 s) 2

d = 0 m + 406 m

Return to Problem 6

(7.10 m/s) 2 = (0 m/s) 2 + 2*(a)*(35.4 m)

50.4 m 2 /s 2 = (0 m/s) 2 + (70.8 m)*a

(50.4 m 2 /s 2 )/(70.8 m) = a

a = 0.712 m/s 2

Return to Problem 7

(65 m/s) 2 = (0 m/s) 2 + 2*(3 m/s 2 )*d

4225 m 2 /s 2 = (0 m/s) 2 + (6 m/s 2 )*d

(4225 m 2 /s 2 )/(6 m/s 2 ) = d

Return to Problem 8

d = (22.4 m/s + 0 m/s)/2 *2.55 s

d = (11.2 m/s)*2.55 s

Return to Problem 9

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(2.62 m)

0 m 2 /s 2 = v i 2 - 51.35 m 2 /s 2

51.35 m 2 /s 2 = v i 2

v i = 7.17 m/s

Return to Problem 10

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(1.29 m)

0 m 2 /s 2 = v i 2 - 25.28 m 2 /s 2

25.28 m 2 /s 2 = v i 2

v i = 5.03 m/s

To find hang time, find the time to the peak and then double it.

0 m/s = 5.03 m/s + (-9.8 m/s 2 )*t up

-5.03 m/s = (-9.8 m/s 2 )*t up

(-5.03 m/s)/(-9.8 m/s 2 ) = t up

t up = 0.513 s

hang time = 1.03 s

Return to Problem 11

(521 m/s) 2 = (0 m/s) 2 + 2*(a)*(0.840 m)

271441 m 2 /s 2 = (0 m/s) 2 + (1.68 m)*a

(271441 m 2 /s 2 )/(1.68 m) = a

a = 1.62*10 5 m /s 2

Return to Problem 12

  • (NOTE: the time required to move to the peak of the trajectory is one-half the total hang time - 3.125 s.)

First use:  v f  = v i  + a*t

0 m/s = v i  + (-9.8  m/s 2 )*(3.13 s)

0 m/s = v i  - 30.7 m/s

v i  = 30.7 m/s  (30.674 m/s)

Now use:  v f 2  = v i 2  + 2*a*d

(0 m/s) 2  = (30.7 m/s) 2  + 2*(-9.8  m/s 2 )*(d)

0 m 2 /s 2  = (940 m 2 /s 2 ) + (-19.6  m/s 2 )*d

-940  m 2 /s 2  = (-19.6  m/s 2 )*d

(-940  m 2 /s 2 )/(-19.6  m/s 2 ) = d

Return to Problem 13

-370 m = (0 m/s)*(t)+ 0.5*(-9.8 m/s 2 )*(t) 2

-370 m = 0+ (-4.9 m/s 2 )*(t) 2

(-370 m)/(-4.9 m/s 2 ) = t 2

75.5 s 2 = t 2

Return to Problem 14

(0 m/s) 2 = (367 m/s) 2 + 2*(a)*(0.0621 m)

0 m 2 /s 2 = (134689 m 2 /s 2 ) + (0.1242 m)*a

-134689 m 2 /s 2 = (0.1242 m)*a

(-134689 m 2 /s 2 )/(0.1242 m) = a

a = -1.08*10 6 m /s 2

(The - sign indicates that the bullet slowed down.)

Return to Problem 15

d = (0 m/s)*(3.41 s)+ 0.5*(-9.8 m/s 2 )*(3.41 s) 2

d = 0 m+ 0.5*(-9.8 m/s 2 )*(11.63 s 2 )

d = -57.0 m

(NOTE: the - sign indicates direction)

Return to Problem 16

(0 m/s) 2 = v i 2 + 2*(- 3.90 m/s 2 )*(290 m)

0 m 2 /s 2 = v i 2 - 2262 m 2 /s 2

2262 m 2 /s 2 = v i 2

v i = 47.6 m /s

Return to Problem 17

( 88.3 m/s) 2 = (0 m/s) 2 + 2*(a)*(1365 m)

7797 m 2 /s 2 = (0 m 2 /s 2 ) + (2730 m)*a

7797 m 2 /s 2 = (2730 m)*a

(7797 m 2 /s 2 )/(2730 m) = a

a = 2.86 m/s 2

88.3 m/s = 0 m/s + (2.86 m/s 2 )*t

(88.3 m/s)/(2.86 m/s 2 ) = t

t = 30. 8 s

Return to Problem 18

( 112 m/s) 2 = (0 m/s) 2 + 2*(a)*(398 m)

12544 m 2 /s 2 = 0 m 2 /s 2 + (796 m)*a

12544 m 2 /s 2 = (796 m)*a

(12544 m 2 /s 2 )/(796 m) = a

a = 15.8 m/s 2

Return to Problem 19

v f 2 = v i 2 + 2*a*d

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(91.5 m)

0 m 2 /s 2 = v i 2 - 1793 m 2 /s 2

1793 m 2 /s 2 = v i 2

v i = 42.3 m/s

Now convert from m/s to mi/hr:

v i = 42.3 m/s * (2.23 mi/hr)/(1 m/s)

v i = 94.4 mi/hr

Return to Problem 20

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6.3: Solving Problems with Newton's Laws (Part 2)

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Newton’s Laws of Motion and Kinematics

Physics is most interesting and most powerful when applied to general situations that involve more than a narrow set of physical principles. Newton’s laws of motion can also be integrated with other concepts that have been discussed previously in this text to solve problems of motion. For example, forces produce accelerations, a topic of kinematics, and hence the relevance of earlier chapters.

When approaching problems that involve various types of forces, acceleration, velocity, and/or position, listing the givens and the quantities to be calculated will allow you to identify the principles involved. Then, you can refer to the chapters that deal with a particular topic and solve the problem using strategies outlined in the text. The following worked example illustrates how the problem-solving strategy given earlier in this chapter, as well as strategies presented in other chapters, is applied to an integrated concept problem.

Example 6.6: What Force Must a Soccer Player Exert to Reach Top Speed?

A soccer player starts at rest and accelerates forward, reaching a velocity of 8.00 m/s in 2.50 s. (a) What is her average acceleration? (b) What average force does the ground exert forward on the runner so that she achieves this acceleration? The player’s mass is 70.0 kg, and air resistance is negligible.

To find the answers to this problem, we use the problem-solving strategy given earlier in this chapter. The solutions to each part of the example illustrate how to apply specific problem-solving steps. In this case, we do not need to use all of the steps. We simply identify the physical principles, and thus the knowns and unknowns; apply Newton’s second law; and check to see whether the answer is reasonable.

  • We are given the initial and final velocities (zero and 8.00 m/s forward); thus, the change in velocity is \(\Delta\)v = 8.00 m/s . We are given the elapsed time, so \(\Delta\)t = 2.50 s. The unknown is acceleration, which can be found from its definition: $$a = \frac{\Delta v}{\Delta t} \ldotp$$Substituting the known values yields $$a = \frac{8.00\; m/s}{2.50\; s} = 3.20\; m/s^{2} \ldotp$$
  • Here we are asked to find the average force the ground exerts on the runner to produce this acceleration. (Remember that we are dealing with the force or forces acting on the object of interest.) This is the reaction force to that exerted by the player backward against the ground, by Newton’s third law. Neglecting air resistance, this would be equal in magnitude to the net external force on the player, since this force causes her acceleration. Since we now know the player’s acceleration and are given her mass, we can use Newton’s second law to find the force exerted. That is, $$F_{net} = ma \ldotp$$Substituting the known values of m and a gives $$F_{net} = (70.0\; kg)(3.20\; m/s^{2}) = 224\; N \ldotp$$

This is a reasonable result: The acceleration is attainable for an athlete in good condition. The force is about 50 pounds, a reasonable average force.

Significance

This example illustrates how to apply problem-solving strategies to situations that include topics from different chapters. The first step is to identify the physical principles, the knowns, and the unknowns involved in the problem. The second step is to solve for the unknown, in this case using Newton’s second law. Finally, we check our answer to ensure it is reasonable. These techniques for integrated concept problems will be useful in applications of physics outside of a physics course, such as in your profession, in other science disciplines, and in everyday life.

Exercise 6.4

The soccer player stops after completing the play described above, but now notices that the ball is in position to be stolen. If she now experiences a force of 126 N to attempt to steal the ball, which is 2.00 m away from her, how long will it take her to get to the ball?

Example 6.7: What Force Acts on a Model Helicopter?

A 1.50-kg model helicopter has a velocity of 5.00 \(\hat{j}\) m/s at t = 0. It is accelerated at a constant rate for two seconds (2.00 s) after which it has a velocity of (6.00 \(\hat{i}\) + 12.00 \(\hat{j}\)) m/s. What is the magnitude of the resultant force acting on the helicopter during this time interval?

We can easily set up a coordinate system in which the x-axis (\(\hat{i}\) direction) is horizontal, and the y-axis (\(\hat{j}\) direction) is vertical. We know that \(\Delta\)t = 2.00s and \(\Delta\)v = (6.00 \(\hat{i}\) + 12.00 \(\hat{j}\) m/s) − (5.00 \(\hat{j}\) m/s). From this, we can calculate the acceleration by the definition; we can then apply Newton’s second law.

\[a = \frac{\Delta v}{\Delta t} = \frac{(6.00 \hat{i} + 12.00 \hat{j}\; m/s) - (5.00 \hat{j}\; m/s)}{2.00\; s} = 3.00 \hat{i} + 3.50 \hat{j}\; m/s^{2}$$ $$\sum \vec{F} = m \vec{a} = (1.50\; kg)(3.00 \hat{i} + 3.50 \hat{j}\; m/s^{2}) = 4.50 \hat{i} + 5.25 \hat{j}\; N \ldotp\]

The magnitude of the force is now easily found:

\[F = \sqrt{(4.50\; N)^{2} + (5.25\; N)^{2}} = 6.91\; N \ldotp\]

The original problem was stated in terms of \(\hat{i}\) − \(\hat{j}\) vector components, so we used vector methods. Compare this example with the previous example.

Exercise 6.5

Find the direction of the resultant for the 1.50-kg model helicopter.

Example 6.8: Baggage Tractor

Figure \(\PageIndex{7}\)(a) shows a baggage tractor pulling luggage carts from an airplane. The tractor has mass 650.0 kg, while cart A has mass 250.0 kg and cart B has mass 150.0 kg. The driving force acting for a brief period of time accelerates the system from rest and acts for 3.00 s. (a) If this driving force is given by F = (820.0t) N, find the speed after 3.00 seconds. (b) What is the horizontal force acting on the connecting cable between the tractor and cart A at this instant?

Figure (a) shows a baggage tractor driving to the left and pulling two luggage carts. The external forces on the system are shown. The forces on the tractor are F sub tractor, horizontally to the left, N sub tractor vertically up, and w sub tractor vertically down. The forces on the cart immediately behind the tractor, cart A, are N sub A vertically up, and w sub A vertically down. The forces on cart B, the one behind cart A, are N sub B vertically up, and w sub B vertically down. Figure (b) shows the free body diagram of the tractor, consisting of F sub tractor, horizontally to the left, N sub tractor vertically up, w sub tractor vertically down, and T horizontally to the right.

A free-body diagram shows the driving force of the tractor, which gives the system its acceleration. We only need to consider motion in the horizontal direction. The vertical forces balance each other and it is not necessary to consider them. For part b, we make use of a free-body diagram of the tractor alone to determine the force between it and cart A. This exposes the coupling force \(\vec{T}\), which is our objective.

  • $$\sum F_{x} = m_{system} a_{x}\; and\; \sum F_{x} = 820.0t,$$so $$820.0t = (650.0 + 250.0 + 150.0)a$$ $$a = 0.7809t \ldotp$$Since acceleration is a function of time, we can determine the velocity of the tractor by using a = \(\frac{dv}{dt}\) with the initial condition that v 0 = 0 at t = 0. We integrate from t = 0 to t = 3: $$\begin{split}dv & = adt \\ \int_{0}^{3} dv & = \int_{0}^{3.00} adt = \int_{0}^{3.00} 0.7809tdt \\ v & = 0.3905t^{2} \big]_{0}^{3.00} = 3.51\; m/s \ldotp \end{split}$$
  • Refer to the free-body diagram in Figure \(\PageIndex{7}\)(b) $$\begin{split} \sum F_{x} & = m_{tractor} a_{x} \\ 820.0t - T & = m_{tractor} (0.7805)t \\ (820.0)(3.00) - T & = (650.0)(0.7805)(3.00) \\ T & = 938\; N \ldotp \end{split}$$

Since the force varies with time, we must use calculus to solve this problem. Notice how the total mass of the system was important in solving Figure \(\PageIndex{7}\)(a), whereas only the mass of the truck (since it supplied the force) was of use in Figure \(\PageIndex{7}\)(b).

Recall that v = \(\frac{ds}{dt}\) and a = \(\frac{dv}{dt}\). If acceleration is a function of time, we can use the calculus forms developed in Motion Along a Straight Line , as shown in this example. However, sometimes acceleration is a function of displacement. In this case, we can derive an important result from these calculus relations. Solving for dt in each, we have dt = \(\frac{ds}{v}\) and dt = \(\frac{dv}{a}\). Now, equating these expressions, we have \(\frac{ds}{v}\) = \(\frac{dv}{a}\). We can rearrange this to obtain a ds = v dv.

Example 6.9: Motion of a Projectile Fired Vertically

A 10.0-kg mortar shell is fired vertically upward from the ground, with an initial velocity of 50.0 m/s (see Figure \(\PageIndex{8}\)). Determine the maximum height it will travel if atmospheric resistance is measured as F D = (0.0100 v 2 ) N, where v is the speed at any instant.

(a) A photograph of a soldier firing a mortar shell straight up. (b) A free body diagram of the mortar shell shows forces F sub D and w, both pointing vertically down. Force w is larger than force F sub D.

The known force on the mortar shell can be related to its acceleration using the equations of motion. Kinematics can then be used to relate the mortar shell’s acceleration to its position.

Initially, y 0 = 0 and v 0 = 50.0 m/s. At the maximum height y = h, v = 0. The free-body diagram shows F D to act downward, because it slows the upward motion of the mortar shell. Thus, we can write

\[\begin{split} \sum F_{y} & = ma_{y} \\ -F_{D} - w & = ma_{y} \\ -0.0100 v^{2} - 98.0 & = 10.0 a \\ a & = -0.00100 v^{2} - 9.80 \ldotp \end{split}\]

The acceleration depends on v and is therefore variable. Since a = f(v), we can relate a to v using the rearrangement described above,

\[a ds = v dv \ldotp\]

We replace ds with dy because we are dealing with the vertical direction,

\[\begin{split} ady & = vdv \\ (−0.00100v^{2} − 9.80)dy & = vdv \ldotp \end{split}\]

We now separate the variables (v’s and dv’s on one side; dy on the other):

\[\begin{split} \int_{0}^{h} dy & = \int_{50.0}^{0} \frac{vdv}{(-0.00100 v^{2} - 9.80)} \\ & = - \int_{50.0}^{0} \frac{vdv}{(-0.00100 v^{2} + 9.80)} \\ & = (-5 \times 10^{3}) \ln(0.00100v^{2} + 9.80) \Big|_{50.0}^{0} \ldotp \end{split}\]

Thus, h = 114 m.

Notice the need to apply calculus since the force is not constant, which also means that acceleration is not constant. To make matters worse, the force depends on v (not t), and so we must use the trick explained prior to the example. The answer for the height indicates a lower elevation if there were air resistance. We will deal with the effects of air resistance and other drag forces in greater detail in Drag Force and Terminal Speed .

Exercise 6.6

If atmospheric resistance is neglected, find the maximum height for the mortar shell. Is calculus required for this solution?

Explore the forces at work in this simulation when you try to push a filing cabinet. Create an applied force and see the resulting frictional force and total force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. View a free-body diagram of all the forces (including gravitational and normal forces).

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Speed and Velocity

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Algebra is often taught abstractly with little or no emphasis on what algebra is or how it can be used to solve real problems. Just as English can be translated into other languages, word problems can be "translated" into the math language of algebra and easily solved. Real World Algebra explains this process in an easy to understand format using cartoons and drawings. This makes self-learning easy for both the student and any teacher who never did quite understand algebra. Includes chapters on algebra and money, algebra and geometry, algebra and physics, algebra and levers and many more. Designed for children in grades 4-9 with higher math ability and interest but could be used by older students and adults as well. Contains 22 chapters with instruction and problems at three levels of difficulty.

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3.1: Velocity and Acceleration

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  • Page ID 89728

  • Joel Feldman, Andrew Rechnitzer and Elyse Yeager
  • University of British Columbia

If you are moving along the \(x\)–axis and your position at time \(t\) is \(x(t)\text{,}\) then your velocity at time \(t\) is \(v(t)=x'(t)\) and your acceleration at time \(t\) is \(a(t)=v'(t) = x''(t)\text{.}\)

Example 3.1.1 Velocity as derivative of position.

Suppose that you are moving along the \(x\)–axis and that at time \(t\) your position is given by

\begin{align*} x(t)&=t^3-3t+2. \end{align*}

We're going to try and get a good picture of what your motion is like. We can learn quite a bit just by looking at the sign of the velocity \(v(t)=x'(t)\) at each time \(t\text{.}\)

  • If \(x'(t) \gt 0\text{,}\) then at that instant \(x\) is increasing, i.e. you are moving to the right.
  • If \(x'(t)=0\text{,}\) then at that instant you are not moving at all.
  • If \(x'(t) \lt 0\text{,}\) then at that instant \(x\) is decreasing, i.e. you are moving to the left.

From the given formula for \(x(t)\) it is straight forward to work out the velocity

\begin{align*} v(t) = x'(t) &=3t^2-3=3(t^2-1)=3(t+1)(t-1) \end{align*}

This is zero only when \(t=-1\) and when \(t=+1\text{;}\) at no other value 1  of \(t\) can this polynomial be equal zero. Consequently in any time interval that does not include either \(t=-1\) or \(t=+1\text{,}\) \(v(t)\) takes only a single sign 2 . So

  • For all \(t \lt -1\text{,}\) both \((t+1)\) and \((t-1)\) are negative (sub in, for example, \(t=-10\)) so the product \(v(t)=x'(t)=3(t+1)(t-1) \gt 0\text{.}\)
  • For all \(-1 \lt t \lt 1\text{,}\) the factor \((t+1) \gt 0\) and the factor \((t-1) \lt 0\) (sub in, for example, \(t=0\)) so the product \(v(t)=x'(t)=3(t+1)(t-1) \lt 0\text{.}\)
  • For all \(t \gt 1\text{,}\) both \((t+1)\) and \((t-1)\) are positive (sub in, for example, \(t=+10\)) so the product \(v(t)=x'(t)=3(t+1)(t-1) \gt 0\text{.}\)

The figure below gives a summary of the sign information we have about \(t-1\text{,}\) \(t+1\) and \(x'(t)\text{.}\)

It is now easy to put together a mental image of your trajectory.

  • For \(t\) large and negative (i.e. far in the past), \(x(t)\) is large and negative and \(v(t)\) is large and positive. For example 3 , when \(t=-10^6\text{,}\) \(x(t)\approx t^3=- 10^{18}\) and \(v(t)\approx 3t^2 = 3\cdot 10^{12}\text{.}\) So you are moving quickly to the right.
  • For \(t \lt -1\text{,}\) \(v(t)=x'(t) \gt 0\) so that \(x(t)\) is increasing and you are moving to the right.
  • At \(t=-1\text{,}\) \(v(-1)=0\) and you have come to a halt at position \(x(-1)=(-1)^3-3(-1)+2=4\text{.}\)
  • For \(-1 \lt t \lt 1\text{,}\) \(v(t)=x'(t) \lt 0\) so that \(x(t)\) is decreasing and you are moving to the left.
  • At \(t=+1\text{,}\) \(v(1)=0\) and you have again come to a halt, but now at position \(x(1)=1^3-3+2=0\text{.}\)
  • For \(t \gt 1\text{,}\) \(v(t)=x'(t) \gt 0\) so that \(x(t)\) is increasing and you are again moving to the right.
  • For \(t\) large and positive (i.e. in the far future), \(x(t)\) is large and positive and \(v(t)\) is large and positive. For example 4 , when \(t=10^6\text{,}\) \(x(t)\approx t^3= 10^{18}\) and \(v(t)\approx 3t^2 = 3\cdot 10^{12}\text{.}\) So you are moving quickly to the right.

Here is a sketch of the graphs of \(x(t)\) and \(v(t)\text{.}\) The heavy lines in the graphs indicate when you are moving to the right — that is where \(v(t)=x'(t)\) is positive.

And here is a schematic picture of the whole trajectory.

Example 3.1.2 Position and velocity from acceleration.

In this example we are going to figure out how far a body falling from rest will fall in a given time period.

  • time in seconds by \(t\text{,}\)
  • mass in kilograms by \(m\text{,}\)
  • distance fallen (in metres) at time \(t\) by \(s(t)\text{,}\) velocity (in m/sec) by \(v(t)=s'(t)\) and acceleration (in m/sec\(^2\)) by \(a(t)=v'(t)=s''(t)\text{.}\)

It makes sense to choose a coordinate system so that the body starts to fall at \(t=0\text{.}\)

\begin{gather*} \text{the force applied to the body at time } t = m \cdot a(t). \end{gather*}

\begin{align*} \text{the force due to gravity acting on a body of mass } m &= m \cdot g. \end{align*}

\begin{align*} v(0) &= 0. \end{align*}

\begin{align*} m\cdot a(t) &= \text{force due to gravity}\\ m \cdot v'(t) &= m \cdot g & \text{ cancel the } m\\ v'(t) &=g \end{align*}

\begin{gather*} v(t) = gt + c \end{gather*}

for any constant \(c\text{.}\) One can then verify 6  that \(v'(t)=g\text{.}\) Using the fact that \(v(0)=0\) we must then have \(c=0\) and so

\begin{align*} v(t) &= gt. \end{align*}

\begin{align*} s'(t) &= v(t) = g \cdot t. \end{align*}

\begin{align*} s(t) &= \frac{g}{2} t^2 + c \end{align*}

\begin{align*} s(t) &= \frac{g}{2} t^2 & \text{but $g=9.8$, so}\\ &= 4.9 t^2, \end{align*}

Let's now do a similar but more complicated example.

Example 3.1.3 Stoping distance of a braking car.

A car's brakes can decelerate the car at 64000\(\textrm{km/hr}^2\text{.}\) How fast can the car be driven if it must be able to stop within a distance of 50m?

Before getting started, notice that there is a small “trick” in this problem — several quantities are stated but their units are different. The acceleration is stated in kilometres per hour\(^2\text{,}\) but the distance is stated in metres. Whenever we come across a “real world” problem 8  we should be careful of the units used.

  • time (in hours) by \(t\text{,}\)
  • the position of the car (in kilometres) at time \(t\) by \(x(t)\text{,}\) and
  • the velocity (in kilometres per hour) by is \(v(t)\text{.}\)

We can also choose a coordinate system such that \(x(0)=0\) and the car starts braking at time \(t=0\text{.}\)

  • We are told that, at maximum braking, the acceleration \(v'(t)=x''(t)\) of the car is \(-64000\text{.}\)
  • We need to determine the maximum initial velocity \(v(0)\) so that the stopping distance is at most \(50m = 0.05km\) (being careful with our units). Let us call the stopping distance \(x_{stop}\) which is really \(x(t_{stop})\) where \(t_{stop}\) is the stopping time.
  • In order to determine \(x_{stop}\) we first need to determine \(t_{stop}\text{,}\) which we will do by assuming maximum braking from a, yet to be determined, initial velocity of \(v(0)=q\) m/sec.

\begin{align*} v'(t) &= -64000 \end{align*}

This equation is very similar to the ones we had to solve in Example 3.1.2 just above.

As we did there  9  Now is a good time to go back and have a read of that example. , we are going to just guess \(v(t)\text{.}\) First, we just guess one function whose derivative is \(-64000\text{,}\) namely \(-64000 t\text{.}\) Next we observe that, since the derivative of a constant is zero, any function of the form

\begin{gather*} v(t) = -64000\,t + c \end{gather*}

with constant \(c\text{,}\) has the correct derivative. Finally, the requirement that the initial velocity \(v(0)=q\)" forces \(c=q\text{,}\) so

\begin{gather*} v(t) = q - 64000\,t \end{gather*}

\begin{align*} 0 = v(t_{stop}) &= q-64000 \cdot t_{stop} & \text{ and so}\\ t_{stop} &= \frac{q}{64000}. \end{align*}

\begin{align*} x'(t) &= v(t) = q - 64000t. \end{align*}

\begin{align*} x(t) &= qt - 32000t^2 + \text{constant} \end{align*}

\begin{align*} x(t) &= qt - 32000 t^2. \end{align*}

\begin{align*} x_{stop} &= x(t_{stop}) = q t_{stop} - 32000 t_{stop}^2\\ &= \frac{q^2}{64000} - \frac{32000 q^2}{64000^2}\\ &= \frac{q^2}{64000} \left(1 - \frac{1}{2} \right)\\ &= \frac{q^2}{2 \times 64000} \end{align*}

\begin{align*} x_{stop} = \frac{q^2}{2 \times 64000} &\leq \frac{5}{100}\\ q^2 &\leq \frac{2 \times 64000 \times 5}{100} = \frac{64000 \times 10}{100} = 6400 \end{align*}

Exercise \(\PageIndex{1}\)

Suppose you throw a ball straight up in the air, and its height from \(t=0\) to \(t=4\) is given by \(h(t)=-4.9t^2+19.6t\text{.}\) True or false: at time \(t=2\text{,}\) the acceleration of the ball is 0.

Exercise \(\PageIndex{2}\)

Suppose an object is moving with a constant acceleration. It takes ten seconds to accelerate from \(1\;\frac{\mathrm{m}}{\mathrm{s}}\) to \(2\;\frac{\mathrm{m}}{\mathrm{s}}\text{.}\) How long does it take to accelerate from \(2\;\frac{\mathrm{m}}{\mathrm{s}}\) to \(3\;\frac{\mathrm{m}}{\mathrm{s}}\text{?}\) How long does it take to accelerate from \(3\;\frac{\mathrm{m}}{\mathrm{s}}\) to \(13\;\frac{\mathrm{m}}{\mathrm{s}}\text{?}\)

Exercise \(\PageIndex{3}\)

Let \(s(t)\) be the position of a particle at time \(t\text{.}\) True or false: if \(s''(a) \gt 0\) for some \(a\text{,}\) then the particle's speed is increasing when \(t=a\text{.}\)

Exercise \(\PageIndex{4}\)

Let \(s(t)\) be the position of a particle at time \(t\text{.}\) True or false: if \(s'(a) \gt 0\) and \(s''(a) \gt 0\) for some \(a\text{,}\) then the particle's speed is increasing when \(t=a\text{.}\)

For this section, we will ask you a number of questions that have to do with objects falling on Earth. Unless otherwise stated, you should assume that an object falling through the air has an acceleration due to gravity of 9.8 meters per second per second.

Exercise \(\PageIndex{5}\)

A flower pot rolls out of a window 10m above the ground. How fast is it falling just as it smacks into the ground?

Exercise \(\PageIndex{6}\)

You want to know how deep a well is, so you drop a stone down and count the seconds until you hear it hit bottom.

  • If the stone took \(x\) seconds to hit bottom, how deep is the well?
  • Suppose you think you dropped the stone down the well, but actually you tossed it down, so instead of an initial velocity of 0 metres per second, you accidentally imparted an initial speed of \(1\) metres per second. What is the actual depth of the well, if the stone fell for \(x\) seconds?

Exercise \(\PageIndex{7}\)

You toss a key to your friend, standing two metres away. The keys initially move towards your friend at 2 metres per second, but slow at a rate of 0.25 metres per second per second. How much time does your friend have to react to catch the keys? That is--how long are the keys flying before they reach your friend?

Exercise \(\PageIndex{8}\)

A car is driving at 100 kph, and it brakes with a deceleration of \(50000 \frac{\mathrm{km}}{\mathrm{hr}^2}\text{.}\) How long does the car take to come to a complete stop?

Exercise \(\PageIndex{9}\)

You are driving at 120 kph, and need to stop in 100 metres. How much deceleration do your brakes need to provide? You may assume the brakes cause a constant deceleration.

Exercise \(\PageIndex{10}\)

You are driving at 100 kph, and apply the brakes steadily, so that your car decelerates at a constant rate and comes to a stop in exactly 7 seconds. What was your speed one second before you stopped?

Exercise \(\PageIndex{11}\)

About 8.5 minutes after liftoff, the US space shuttle has reached orbital velocity, 17 500 miles per hour. Assuming its acceleration was constant, how far did it travel in those 8.5 minutes?

Source: http://www.nasa.gov/mission_pages/shuttle/shuttlemissions/sts121/launch/qa-leinbach.html

Exercise \(\PageIndex{12}\)

A pitching machine has a dial to adjust the speed of the pitch. You rotate it so that it pitches the ball straight up in the air. How fast should the ball exit the machine, in order to stay in the air exactly 10 seconds?

You may assume that the ball exits from ground level, and is acted on only by gravity, which causes a constant deceleration of 9.8 metres per second.

Exercise \(\PageIndex{13}\)

A peregrine falcon can dive at a speed of 325 kph. If you were to drop a stone, how high up would you have to be so that the stone reached the same speed in its fall?

Exercise \(\PageIndex{14}\)

You shoot a cannon ball into the air with initial velocity \(v_0\text{,}\) and then gravity brings it back down (neglecting all other forces). If the cannon ball made it to a height of 100m, what was \(v_0\text{?}\)

Exercise \(\PageIndex{15}\)

Suppose you are driving at 120 kph, and you start to brake at a deceleration of \(50 000\) kph per hour. For three seconds you steadily increase your deceleration to \(60 000\) kph per hour. (That is, for three seconds, the rate of change of your deceleration is constant.) How fast are you driving at the end of those three seconds?

Exercise \(\PageIndex{16}\)

You jump up from the side of a trampoline with an initial upward velocity of \(1\) metre per second. While you are in the air, your deceleration is a constant \(9.8\) metres per second per second due to gravity. Once you hit the trampoline, as you fall your speed decreases by \(4.9\) metres per second per second. How many seconds pass between the peak of your jump and the lowest part of your fall on the trampoline?

Exercise \(\PageIndex{17}\)

Suppose an object is moving so that its velocity doubles every second. Give an expression for the acceleration of the object.

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Heavy Machinery Meets AI

  • Vijay Govindarajan
  • Venkat Venkatraman

solving speed and velocity problems

Until recently most incumbent industrial companies didn’t use highly advanced software in their products. But now the sector’s leaders have begun applying generative AI and machine learning to all kinds of data—including text, 3D images, video, and sound—to create complex, innovative designs and solve customer problems with unprecedented speed.

Success involves much more than installing computers in products, however. It requires fusion strategies, which join what manufacturers do best—creating physical products—with what digital firms do best: mining giant data sets for critical insights. There are four kinds of fusion strategies: Fusion products, like smart glass, are designed from scratch to collect and leverage information on product use in real time. Fusion services, like Rolls-Royce’s service for increasing the fuel efficiency of aircraft, deliver immediate customized recommendations from AI. Fusion systems, like Honeywell’s for building management, integrate machines from multiple suppliers in ways that enhance them all. And fusion solutions, such as Deere’s for increasing yields for farmers, combine products, services, and systems with partner companies’ innovations in ways that greatly improve customers’ performance.

Combining digital and analog machines will upend industrial companies.

Idea in Brief

The problem.

Until recently most incumbent industrial companies didn’t use the most advanced software in their products. But competitors that can extract complex designs, insights, and trends using generative AI have emerged to challenge them.

The Solution

Industrial companies must develop strategies that fuse what they do best—creating physical products—with what digital companies do best: using data and AI to parse enormous, interconnected data sets and develop innovative insights.

The Changes Required

Companies will have to reimagine analog products and services as digitally enabled offerings, learn to create new value from data generated by the combination of physical and digital assets, and partner with other companies to create ecosystems with an unwavering focus on helping customers solve problems.

For more than 187 years, Deere & Company has simplified farmwork. From the advent of the first self-scouring plow, in 1837, to the launch of its first fully self-driving tractor, in 2022, the company has built advanced industrial technology. The See & Spray is an excellent contemporary example. The automated weed killer features a self-propelled, 120-foot carbon-fiber boom lined with 36 cameras capable of scanning 2,100 square feet per second. Powered by 10 onboard vision-processing units handling almost four gigabytes of data per second, the system uses AI and deep learning to distinguish crops from weeds. Once a weed is identified, a command is sent to spray and kill it. The machine moves through a field at 12 miles per hour without stopping. Manual labor would be more expensive, more time-consuming, and less reliable than the See & Spray. By fusing computer hardware and software with industrial machinery, it has helped farmers decrease their use of herbicide by more than two-thirds and exponentially increase productivity.

  • Vijay Govindarajan is the Coxe Distinguished Professor at Dartmouth College’s Tuck School of Business, an executive fellow at Harvard Business School, and faculty partner at the Silicon Valley incubator Mach 49. He is a New York Times and Wall Street Journal bestselling author. His latest book is Fusion Strategy: How Real-Time Data and AI Will Power the Industrial Future . His Harvard Business Review articles “ Engineering Reverse Innovations ” and “ Stop the Innovation Wars ” won McKinsey Awards for best article published in HBR. His HBR articles “ How GE Is Disrupting Itself ” and “ The CEO’s Role in Business Model Reinvention ” are HBR all-time top-50 bestsellers. Follow him on LinkedIn . vgovindarajan
  • Venkat Venkatraman is the David J. McGrath Professor at Boston University’s Questrom School of Business, where he is a member of both the information systems and strategy and innovation departments. His current research focuses on how companies develop winning digital strategies. His latest book is Fusion Strategy: How Real-Time Data and AI Will Power the Industrial Future.  Follow him on LinkedIn . NVenkatraman

Partner Center

Linear Velocity set to Line looks like teleporting to others

I’ve made a dodging system in my game using LinearVelocity using the Line mode, and it looks normal to the player, but to other players and the server I just seem to phase from 1 place to another.

Code I’m using (although I don’t think the problem comes from the code):

Does anyone know what’s going on?

Not sure, but maybe its not replicated? Meaning the physics of the linear velocity only happen on the client but aren’t replicated to the server, therefore only the position the player ends up in is replicated.

Are you running the provided code on a local script or server script?

It’s running on a server script.

Maybe because the max force property is set to math.huge so it’s moving too fast

the max amount it can move is essentially infinite “ math.huge ”

Doesn’t seem to change anything, player just seems to teleport but just a lower distance

Maybe try using another method? ApplyImpulse might work? I think I made something similar to a dash in the past, might be able to dig through my stuff and find it, but I can’t think of much off the top of my head right now.

EDIT: Found another post, solution here was to run the linearVelocity on the client… not sure why its like that, but it might work.

It works now, but how would I remove the force if it’s on the client if handling the roll system is on the server? Since I want the roll’s force to be removed if you dodge an attack with perfect timing.

You could fire an event back to the client to signal them to remove the linear velocity or you could check for the dodge on both the client and server at the same time, on the client you would then handle the damaging depending on dodging, the client would only check if they dodged to remove the linear velocity again, not to detect any damage or anything.

The latter is probably a bit tedious and I’m sure there is a better way, but using a remote event to send the signal back to the client might cause a delay and lag would certainly affect the system in that case.

I would probably make a shared module script in replicated storage that has a function or something to handle the dodge check, then run it on client and server at the same time, that way you don’t don’t have to write the code for checking twice.

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COMMENTS

  1. Velocity and Speed: Problems with Solutions

    Problem 1: A man walks 7 km in 2 hours and 2 km in 1 hour in the same direction. a) What is the man's average speed for the whole journey? b) What is the man's average velocity for the whole journey? Solution to Problem 1 Problem 2: A man walks 7 km East in 2 hours and then 2.5 km West in 1 hour.

  2. Velocity and Speed: Solutions to Problems

    Solution to Problem 1: a) average speed = distance time = 7 km + 2 km 2 hours + 1 hour = 9 km 3 hours = 3 km/h b) average velocity = displacement time = 7 km + 2 km 2 hours + 1 hour = 9 km 3 hours = 3 km/h Problem 2: A man walks 7 km East in 2 hours and then 2.5 km West in 1 hour. a) What is the man's average speed for the whole journey?

  3. 2.2 Speed and Velocity

    distance = v avg × time Suppose, for example, a car travels 150 kilometers in 3.2 hours. Its average speed for the trip is v avg = distance time = 150 km 3.2 h = 47 km/h. A car's speed would likely increase and decrease many times over a 3.2 hour trip. Its speed at a specific instant in time, however, is its instantaneous speed.

  4. Solved Speed, Velocity, and Acceleration Problems

    Solution: First find its total distance traveled ( D D) by summing all distances in each section, which gets D=100+200+50=350\, {\rm m} D = 100 +200+ 50 = 350m. Now, by definition of average speed, divide it by the total time elapsed T=5+7+4=16 T = 5+ 7+4 = 16 minutes.

  5. Speed and velocity questions (practice)

    Choose 1 answer: The average velocity is 40 miles per hour. A The average velocity is 40 miles per hour. The total displacement of the trip is 300 miles. B The total displacement of the trip is 300 miles. The average speed is 37.5 miles per hour. C The average speed is 37.5 miles per hour.

  6. Speed and Velocity

    t means time Example: You run 360 m in 60 seconds. Speed = Δs Δt = 360 m 60 seconds = 6 m 1 second So your speed is 6 meters per second (6 m/s). Units Speed is commonly measured in: meters per second (m/s or m s -1 ), or kilometers per hour (km/h or km h -1)

  7. Speed, Velocity, and Acceleration Problems

    Speed, Velocity, and Acceleration Problems A car is traveling at a velocity of 8ms-1 and experiences an acceleration of 5ms-2. How far does it travel in 4 seconds? Skip to content Home PHYSICS RESOURCES Resources MECHANICS WAVES/ACOUSTICS THERMODYNAMICS ELECTROMAGNETISM OPTICS SOLAR ENERGY MODERN PHYSICS QUESTIONS AND ANSWERS POLICIES

  8. Velocity Problems

    Problem # 1 A car travels at uniform velocity a distance of 100 m in 4 seconds. What is the velocity of the car? (Answer: 25 m/s) Problem # 2 A sailboat is traveling north at 10 km/h, relative to the water. The water is flowing north at 5 km/h. What is the velocity of the boat relative to ground? (Answer: 15 km/h) Problem # 3

  9. 6.1 Solving Problems with Newton's Laws

    3.2 Instantaneous Velocity and Speed; 3.3 Average and Instantaneous Acceleration; 3.4 ... These techniques also reinforce concepts that are useful in many other areas of physics. Many problem-solving strategies are stated outright in the worked examples, so the following techniques should reinforce skills you have already begun to develop ...

  10. Speed and Velocity

    practice problem 1 I went for a walk one day. I walked north 6.0 km at 6.0 km/h and then west 10 km at 5.0 km/hr. Determine the average… speed velocity …for the entire journey. solution Average speed This problem is deceptively easy. Averaging is taught in elementary school, which makes this an elementary problem. Right?

  11. 8.8 Rate Word Problems: Speed, Distance and Time

    When solving these problems, use the relationship rate (speed or velocity) times time equals distance. r⋅t = d r ⋅ t = d For example, suppose a person were to travel 30 km/h for 4 h. To find the total distance, multiply rate times time or (30km/h) (4h) = 120 km. The problems to be solved here will have a few more steps than described above.

  12. Kinematic Equations: Sample Problems and Solutions

    A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance traveled. See Answer See solution below. A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is 1.67 m/s 2.

  13. Calculating average velocity or speed (video)

    What is displacement? Calculating average velocity or speed Solving for time Displacement from time and velocity example Instantaneous speed and velocity What is velocity? Position vs. time graphs What are position vs. time graphs? Average velocity and average speed from graphs Instantaneous velocity and instantaneous speed from graphs Science >

  14. 6.3: Solving Problems with Newton's Laws (Part 2)

    t = 2.50 s. The unknown is acceleration, which can be found from its definition: a = Δv Δt. a = Δ v Δ t. (6.3.1) Substituting the known values yields a = 8.00m / s 2.50 s = 3.20 m / s2. a = 8.00 m / s 2.50 s = 3.20 m / s 2. (6.3.2) Here we are asked to find the average force the ground exerts on the runner to produce this acceleration ...

  15. Calculating average velocity or speed

    Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now: https://www.khanacademy.org/science/physics/one-dimensional-mot...

  16. Solving for time (video)

    Solving for time. Rate of change in position, or speed, is equal to distance traveled divided by time. To solve for time, divide the distance traveled by the rate. For example, if Cole drives his car 45 km per hour and travels a total of 225 km, then he traveled for 225/45 = 5 hours. Created by Sal Khan.

  17. Speed and Velocity

    Distance and Displacement, Speed and Velocity Examples: 1. A girl walks 210m in two minutes. Calculate her average speed. 2. A cyclist travels at an average speed of 6m/s for 2 hours. Calculate the distance she covers in this time. 3. The maximum speed limit on UK roads is 70 miles per hour.

  18. Speed, time, and distance problems worksheets

    The seven types of problems are explained in detail in the actual generator below. All worksheets include an answer key on the 2nd page of the file. Please use the quick links below to generate some common types of worksheets. Easy speed, time, and distance worksheet 1: How far can it go or how long does the trip take - using whole or half hours

  19. 3.1: Velocity and Acceleration

    Before getting started, notice that there is a small "trick" in this problem — several quantities are stated but their units are different. The acceleration is stated in kilometres per hour\(^2\text{,}\) but the distance is stated in metres. Whenever we come across a "real world" problem 8 we should be careful of the units used.

  20. What are the kinematic formulas? (article)

    1. v = v 0 + a t. 2. Δ x = ( v + v 0 2) t. 3. Δ x = v 0 t + 1 2 a t 2. 4. v 2 = v 0 2 + 2 a Δ x. Since the kinematic formulas are only accurate if the acceleration is constant during the time interval considered, we have to be careful to not use them when the acceleration is changing.

  21. Average Velocity and Average Speed Problems and Solutions

    Answer: Given: s1 = 10 km, t1 = 4 hours and s2 = 4 km, t2 = 1 hours, and travel Fig.2 of girls are shown below Fig.2 (a) average speed = distance/time = (10 km + 4 km)/ (4 h + 1 h) = 2.8 km/h (b) average velocity = displacement/time = (10 km - 4 km)/ (4 h + 1 h) = 1.2 km/h Problem #4 A car takes 20 s to move around a round about of radius 14 m.

  22. Velocity

    Number of problems found: 966 Speed problem Convert 540 meter/minutes to Kilometer/hour. Speed of light Calculate the speed of light if a distance of 3000 km travels in 0.01 s. Fan The fan has a speed of 179 RPM. Calculate for the time of one fan period. Magnitude of angle

  23. Heavy Machinery Meets AI

    The automated weed killer features a self-propelled, 120-foot carbon-fiber boom lined with 36 cameras capable of scanning 2,100 square feet per second. Powered by 10 onboard vision-processing ...

  24. Average velocity and speed in one direction: word problems

    Average velocity and speed in one direction: word problems. Grant sprints 50 m to the right with an average velocity of 3.0 m s .

  25. Linear Velocity set to Line looks like teleporting to others

    I've made a dodging system in my game using LinearVelocity using the Line mode, and it looks normal to the player, but to other players and the server I just seem to phase from 1 place to another. Code I'm using (although I don't think the problem comes from the code): local RollForce = Instance.new("LinearVelocity",HumRP) RollForce.Attachment0 = Attachment RollForce.Name = "RollForce ...