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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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problem solving in speed time and distance

TIME SPEED AND DISTANCE PROBLEMS

Formulas .

problem solving in speed time and distance

To know the shortcuts required to solve problems on time, speed and distance,

please click here

Problem 1 :

If a person drives his car in the speed 50 miles per hour, how far can he cover in 2.5 hours?

Given :  Speed is 50 miles per hour.

So, the distance covered in 1 hour is

Then, the distance covered in 2.5 hours is

= 2.5 ⋅ 50 miles

= 125 miles

So, the person can cover 125 miles of distance in 2.5 hours.

Problem 2 :

If a person travels at a speed of 40 miles per hour. At the same rate, how long will he take to cover 160 miles distance?

Given :  Speed is 40 miles per hour.

The formula to find the time when distance and speed are given is

Time taken to cover the distance of 160 miles is

So, the person will take 4 hours to cover 160 miles distance at the rate of 40 miles per hour.

Problem 3 :

A person travels at a speed of 60 miles per hour. How far will he travel in 4.5 hours?

Given :  Speed is 60 miles per hour.

The distance covered in 1 hour is

Then, the distance covered in 4.5 hours is

= 4.5 ⋅ 60 miles

= 270 miles

So, the person will travel 270 miles distance in 4.5 hours.

Problem 4 :

A person travels at a speed of 60 kms per hour. Then how many meters can he travel in 5 minutes?

Given :  Speed is 60 kms per hour.

The distance covered in 1 hour or 60 minutes is

= 60 ⋅ 1000 meters

= 60000 meters

Then the distance covered in 1 minute is

The distance covered in 5 minutes is

= 5 ⋅ 1000

= 5000 meters

So, the person can cover 5000 meters distance in 5 minutes.

Problem 5 :

A person covers 108 kms in 3 hours. What is his speed in meter per second?

Given :  Distance is 108 kms and time is 3 hours.

The given distance in meters :

= 108 ⋅ 1000

= 108,000 meters

The given time in seconds :

= 3 ⋅ 60 minutes

= 180 minutes

= 180 ⋅ 60 seconds

= 10,800 seconds

The formula to find the speed is

Speed in meter per second is

= ¹⁰⁸⁰⁰⁰⁄₁₀₈₀₀

So, his speed in meter per second is 10.

Problem 6 :

A person covers 90 kms in 2 hours 30 minutes. Find the speed in meter per second.

Given :  Distance is 90 kms and time is 2 hrs 30 min.

= 90 ⋅ 1000

= 90,000 meters

= 2 hrs 30 min

= (120 + 30) min

= 150 minutes

= 150 ⋅ 60 seconds

= 9,000 seconds

= ⁹⁰⁰⁰⁰⁄₉₀₀₀

Problem 7 :

A person travels at the rate of 60 miles per hour and covers 300 miles in 5 hours. If he reduces his speed by 10 miles per hour, how long will he take to cover the same distance?

Original speed is 60 miles per hour.

If the speed is reduced by 10 miles per hour,  then the new speed is

= 50 miles per hour

Distance to be covered is 300 miles.

The formula to find time is

Time taken to cover 300 miles distance at the speed of 50 miles per hour is

So, if the person reduces his speed by 10 miles per hour, he will take 6 hours to cover 300 miles distance.

Problem 8 :

A person travels 50 kms per hour. If he increases his speed by 10 kms per hour, how many minutes will he take to cover 8000 meters?

Original speed is 50 kms per hour.

If the speed is increased by 10 kms per hour,  then the new speed is

= 60 kms per hour

Because, we have to find the time in minutes for the distance given in meters, let us change the speed from kms per hour into meters per minute.

1 hour ----> 60 kms

1 ⋅ 60 minutes ----> 60 ⋅ 1000 meters

60 minutes ----> 60000 meters

1 minute ----> ⁶⁰⁰⁰⁰⁄₆₀ meters

1 minute ----> 1000 meters

So, the speed is 1000 meters/minute.

Time = Distance/Speed

Time taken to cover 8000 meters distance at the speed of 1000 meters per minute is

= ⁸⁰⁰⁰⁄₁₀₀₀

= 8 minutes

So, if the person increases his speed by 10 kms per hour, he will take 8 minutes to cover 8000 meters distance.

Problem 9 :

A person can travel at the speed of 40 miles per hour. If the speed is increased by 50%, how long will it take to cover 330 miles?

Original speed is 40 miles per hour.

If the speed is increased by 50%, then the new speed is

= 150% of 40

= 1.5 ⋅ 40

= 60 miles per hour

Distance to be covered is 330 miles.

Time taken to cover 330 miles distance at the speed of 60 miles per hour is

= 5.5 hours

= 5 hrs 30 minutes

So, if the person is increased by 50%, it will take 5 hrs 30 minutes to cover 330 miles distance.

Problem 10 :

A person speed at a rate of 40 kms per hour. If he increases his speed by 20%, what is his new speed in meter per minute?

Original speed is 40 kms per hour

If the speed is increased by 20%, then the new speed is

= 120 % of 40

= 1.2 ⋅ 40

= 48 kms per hour

Now, let us change the speed from kms per hour into meters per minute.

1 hour ----> 48 kms

1 ⋅ 60 minutes ----> 48 ⋅ 1000 meters

60 minutes ----> 48,000 meters

1 minute ----> ⁴⁸⁰⁰⁰⁄₆₀ meters

1 minute ----> 800 meters

So, the speed is 800 meters/minute.

If the person increases his speed by 20%, his new speed will be 800 meter per minute.

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Solving Problems Involving Distance, Rate, and Time

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In math, distance, rate, and time are three important concepts you can use to solve many problems if you know the formula. Distance is the length of space traveled by a moving object or the length measured between two points. It is usually denoted by d in math problems .

The rate is the speed at which an object or person travels. It is usually denoted by  r  in equations . Time is the measured or measurable period during which an action, process, or condition exists or continues. In distance, rate, and time problems, time is measured as the fraction in which a particular distance is traveled. Time is usually denoted by t in equations. 

Solving for Distance, Rate, or Time

When you are solving problems for distance, rate, and time, you will find it helpful to use diagrams or charts to organize the information and help you solve the problem. You will also apply the formula that solves distance, rate, and time, which is  distance = rate x tim e. It is abbreviated as:

There are many examples where you might use this formula in real life. For example, if you know the time and rate a person is traveling on a train, you can quickly calculate how far he traveled. And if you know the time and distance a passenger traveled on a plane, you could quickly figure the distance she traveled simply by reconfiguring the formula.

Distance, Rate, and Time Example

You'll usually encounter a distance, rate, and time question as a word problem in mathematics. Once you read the problem, simply plug the numbers into the formula.

For example, suppose a train leaves Deb's house and travels at 50 mph. Two hours later, another train leaves from Deb's house on the track beside or parallel to the first train but it travels at 100 mph. How far away from Deb's house will the faster train pass the other train?

To solve the problem, remember that d represents the distance in miles from Deb's house and t  represents the time that the slower train has been traveling. You may wish to draw a diagram to show what is happening. Organize the information you have in a chart format if you haven't solved these types of problems before. Remember the formula:

distance = rate x time

When identifying the parts of the word problem, distance is typically given in units of miles, meters, kilometers, or inches. Time is in units of seconds, minutes, hours, or years. Rate is distance per time, so its units could be mph, meters per second, or inches per year.

Now you can solve the system of equations:

50t = 100(t - 2) (Multiply both values inside the parentheses by 100.) 50t = 100t - 200 200 = 50t (Divide 200 by 50 to solve for t.) t = 4

Substitute t = 4 into train No. 1

d = 50t = 50(4) = 200

Now you can write your statement. "The faster train will pass the slower train 200 miles from Deb's house."

Sample Problems

Try solving similar problems. Remember to use the formula that supports what you're looking for—distance, rate, or time.

d = rt (multiply) r = d/t (divide) t = d/r (divide)

Practice Question 1

A train left Chicago and traveled toward Dallas. Five hours later another train left for Dallas traveling at 40 mph with a goal of catching up with the first train bound for Dallas. The second train finally caught up with the first train after traveling for three hours. How fast was the train that left first going?

Remember to use a diagram to arrange your information. Then write two equations to solve your problem. Start with the second train, since you know the time and rate it traveled:

Second train t x r = d 3 x 40 = 120 miles First train t x r = d 8 hours x r = 120 miles Divide each side by 8 hours to solve for r. 8 hours/8 hours x r = 120 miles/8 hours r = 15 mph

Practice Question 2

One train left the station and traveled toward its destination at 65 mph. Later, another train left the station traveling in the opposite direction of the first train at 75 mph. After the first train had traveled for 14 hours, it was 1,960 miles apart from the second train. How long did the second train travel? First, consider what you know:

First train r = 65 mph, t = 14 hours, d = 65 x 14 miles Second train r = 75 mph, t = x hours, d = 75x miles

Then use the d = rt formula as follows:

d (of train 1) + d (of train 2) = 1,960 miles 75x + 910 = 1,960 75x = 1,050 x = 14 hours (the time the second train traveled)
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Key to Algebra workbook series

Real World Algebra by Edward Zaccaro

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.

Practice makes perfect. Practice math at IXL.com

One to one maths interventions built for KS4 success

Weekly online one to one GCSE maths revision lessons now available

In order to access this I need to be confident with:

Multiplication and division

Converting units of length (distance)

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Substitution into formulae

This topic is relevant for:

GCSE Maths

Speed Distance Time

Speed Distance Time Triangle

Here we will learn about the speed distance time triangle including how they relate to each other, how to calculate each one and how to solve problems involving them.

There are also speed distance time triangle worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

What is speed distance time?

Speed distance time is the formula used to explain the relationship between speed, distance and time. That is speed = distance ÷ time . Or to put it another way distance divided by speed will give you the time. Provided you know two of the inputs you can work out the third.

For example if a car travels for 2 hours and covers 120 miles we can work out speed as 120 ÷ 2 = 60 miles per hour.

The units of the the distance and time tell you the units for the speed.

What is the speed distance time triangle?

The speed distance time triangle is a way to describe the relationship between speed, distance and time as shown by the formula below.

\textbf{Speed } \bf{=} \textbf{ distance } \bf{\div} \textbf{ time}

“Speed equals distance divided by time”

Let’s look at an example to calculate speed.

If a car travels 66km in 1.5 hours then we can use this formula to calculate the speed.

This formula can also be rearranged to calculate distance or calculate time given the other two measures. An easy way to remember the formula and the different rearrangements is to use this speed distance time triangle. 

speed distance time image 1

From this triangle we can work out how to calculate each measure: We can ‘cover up’ what we are trying to find and the formula triangle tells us what calculation to do.

speed distance time image 2

Let’s look at an example to calculate time.

How long does it take for a car to travel 34 miles at a speed of 68 miles per hour?

Let’s look at an example to calculate distance.

What distance does a bike cover if it travels at a speed of 7 metres per second for 50 seconds?

What is the speed distance time triangle?

What is the speed distance time formula?

The speed distance time formula is just another way of referring to the speed distance time triangle or calculation you can use to determine speed, time or distance.

  • speed = distance ÷ time
  • time = distance ÷ speed
  • distance = speed x time

Time problem

We can solve problems involving time by remembering the formula for speed , distance and time . 

problem solving in speed time and distance

Calculate the time if a car travels at 15 miles at a speed of 36 mph.

Time = distance ÷ speed

Time = 15 ÷ 36 = 0.42 hours

0.42 ✕ 60 = 25.2 minutes

A train travels 42km between two stops at an average speed of 36 km/h. 

If the train departs at 4 pm, when does the train arrive?

Time = 42 ÷ 36 = 1.17 hours

1.17 ✕ 60= 70 minutes = 1 hour 10 minutes.

The average speed of a scooter is 18 km/h and the average speed of a cycle is 10 km/h. 

When both have travelled 99 km what is the difference in the time taken?

Time A = 99 ÷ 18 = 5.5 hours

Time B= 99 ÷ 10 = 9.9 hours

Difference in time = 9.9 – 5.5 = 4.4 hours

4.4 hours = 4 hours and 24 minutes

Units of speed, distance and time

  • The speed of an object is the magnitude of its velocity. We measure speed most commonly in metres per second (m/s), miles per hour (mph) and kilometres per hour (km/hr).

The average speed of a small plane is 124mph.

The average walking speed of a person is 1.4m/s.

  • We measure the distance an object has travelled most commonly in millimetres (mm), centimetres (cm), metres (m) and kilometres (km).

The distance from London to Birmingham is 162.54km.

problem solving in speed time and distance

  • We measure time taken in milliseconds, seconds, minutes, hours, days, weeks, months and years. 

The time taken for the Earth to orbit the sun is 1 year or 365 days. We don’t measure this in smaller units like minutes of hours.

A short bus journey however, would be measured in minutes.

Speed, distance and time are proportional.

problem solving in speed time and distance

If we know two of the measurements we can find the other.

A car drives 150 miles in 3 hours.

Calculate the average speed, in mph, of the car.

Distance = 150 miles

Time = 3 hours

Speed = 150÷ 3= 50mph

Speed, distance, time and units of measure

It is very important to be aware of the units being used when calculating speed, distance and time.

  • Examples of units of distance: mm, \ cm, \ m, \ km, \ miles
  • Examples of units of time: seconds (sec), minutes, (mins) hours (hrs), days
  • Examples of units of speed: metres per second (m/s), miles per hour (mph)

Note that speed is a compound measure and therefore involves two units; a combination of a distance in relation to a time.

When you use the speed distance time formula you must check that each measure  is in the appropriate unit before you carry out the calculation. Sometimes you will need to convert a measure into different units. Here are some useful conversions to remember.

Units of length

Units of time

1 minute = 60 seconds

1 hour = 60 minutes

1 day = 24 hours

Let’s look at an example.

What distance does a bike cover if it travels at a speed of 5 metres per second for 3 minutes?

Note here that the speed involves seconds, but the time given is in minutes. So before using the formula you must change 3 minutes into seconds.

3 minutes = 3\times 60 =180 seconds

Note also that sometimes you may need to convert an answer into different units at the end of a calculation.

Constant speed / average speed

For the GCSE course you will be asked to calculate either a constant speed or an average speed . Both of these can be calculated using the same formula as shown above.

However, this terminology is used because in real life speed varies throughout a journey. You should also be familiar with the terms acceleration (getting faster) and deceleration (getting slower).

Constant speed

A part of a journey where the speed stays the same.

Average speed

A journey might involve a variety of different constant speeds and some acceleration and deceleration. We can use the formula for speed to calculate the average speed over the course of the whole journey.

Average Speed Formula

Average speed is the total distance travelled by an object divided by the total time taken. To do this we can use the formula

Average speed =\frac{Total\, distance}{Total\, time}

If we are calculating an average speed in mph or km/h, we will need to ensure we have decimalised the time before we divide.

How to calculate speed distance time

In order to calculate speed, distance or time:

Write down the values of the measures you know with the units.

speed distance time flip card 1 image 1

Check that the units are compatible with each other, converting them if necessary.

Substitute the values into the selected formula and carry out the resulting calculation.

Write your final answer with the required units.

Explain how to calculate speed distance time

Explain how to calculate speed distance time

Speed distance time triangle worksheet

Get your free speed distance time triangle worksheet of 20+ questions and answers. Includes reasoning and applied questions.

Related lessons on   compound measures

Speed distance time  is part of our series of lessons to support revision on  compound measures . You may find it helpful to start with the main compound measures lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

  • Compound measures
  • Mass density volume
  • Pressure force area
  • Formula for speed
  • Average speed formula

Speed distance time triangle examples

Example 1: calculating average speed.

Calculate the average speed of a car which travels 68 miles in 2 hours.

Speed: unknown

Distance: 68 miles

Time: 2 hours

2 Write down the formula you need to use from the speed, distance, time triangle.

speed distance time example 1 image 1

3 Check that the units are compatible with each other, converting them if necessary.

The distance is in miles and the time is in hours. These units are compatible to give the speed in miles per hour.

4 Substitute the values into the formula and carry out the resulting calculation.

5 Write your final answer with the required units.

Example 2: calculating time

A golden eagle can fly at a speed of 55 kilometres per hour. Calculate the time taken for a golden eagle to fly 66 \ km, giving your answer in hours.

Speed: 55 \ km/hour

Distance: 66 \ km

Time: unknown

Write down the formula  you need to use from the speed, distance, time  triangle.

speed distance time flip card2 image 1

T=\frac{D}{S}

Time = distance \div speed

Speed is in km per hour and the distance is in km , so these are compatible to give an answer for time in hours.

Example 3: calculating distance

Calculate the distance covered by a train travelling at a constant speed of 112 miles per hour for 4 hours.

Speed: 112 \ mph

Distance: unknown

Time: 4 hours

speed distance time example 3

D= S \times T

Distance = speed \times time

Speed is in miles per hour. The time is in hours. These units are compatible to find the distance in miles.

Example 4: calculating speed with unit conversion

A car travels for 1 hour and 45 minutes, covering a distance of 63 miles. Calculate the average speed of the car giving your answer in miles per hour (mph) .

Distance: 63 miles

Time: 1 hour and 45 minutes

speed distance time example 4

S = \frac{D}{T}

Speed = distance \div time

The distance is in miles . The time is in hours and minutes. To calculate the speed in miles per hour , the time needs to be converted into hours only.

1 hour 45 minutes = 1\frac{3}{4} hours = 1.75 hours

Example 5: calculating time with unit conversion

A small plane can travel at an average speed of 120 miles per hour. Calculate the time taken for this plane to fly 80 miles giving your answer in minutes.

Speed: 120 \ mph

Distance: 80 \ miles

speed distance time example 5

T = \frac{D}{S}

Speed is in miles per hour and the distance is in miles . These units are compatible to find the time in hours.

\frac{2}{3} hours in minutes

\frac{2}{3} \times 60 = 40

Example 6: calculating distance with unit conversion

A train travels at a constant speed of 96 miles per hour for 135 minutes. Calculate the distance covered giving your answer in miles.

Speed: 96 \ mph

Time: 135 minutes

speed distance time example 6

D = S \times T

The speed is in miles per hour , but the time is in minutes. To make these compatible the time needs changing into hours and then the calculation will give the distance in miles .

135 minutes

135 \div 60 = \frac{9}{4} = 2\frac{1}{4} = 2.25

Common misconceptions

  • Incorrectly rearranging the formula Speed = distance \div time

Make sure you rearrange the formula correctly. One of the simplest ways of doing this is to use the formula triangle. In the triangle you cover up the measure you want to find out and then the triangle shows you what calculation to do with the other two measures.

speed distance time common misconceptions

  • Using incompatible units in a calculation

When using the speed distance time formula you must ensure that the units of the measures are compatible. For example, if a car travels at 80 \ km per hour for 30 minutes and you are asked to calculate the distance, a common error is to substitute the values straight into the formula and do the following calculation. Distance = speed \times time = 80 \times 30 = 2400 \ km The correct way is to notice that the speed uses hours but the time given is in minutes. Therefore you must change 30 minutes into 0.5 hours and substitute these compatible values into the formula and do the following calculation. Distance = speed \times time = 80 \times 0.5 = 40 \ km

Practice speed distance time triangle questions

1. A car drives 120 miles in 3 hours. Calculate its average speed.

GCSE Quiz True

2. A cyclist travels 100 miles at an average speed of 20 \ mph. Calculate how long the journey takes.

3. An eagle flies for 30 minutes at a speed of 66 \ km per hour. Calculate the total distance the bird has flown.

30 minutes = 0.5 hours

4. Calculate the average speed of a lorry travelling 54 miles in 90 minutes. Give your answer in miles per hour (mph).

Firstly convert 90 minutes to hours. 90 minutes = 1.5 hours

5. Calculate the time taken for a plane to fly 90 miles at an average speed of 120 \ mph. Give your answer in minutes.

180 minutes

Convert 0.75 hours to minutes

6. A helicopter flies 18 \ km in 20 minutes. Calculate its average speed in km/h .

Firstly convert 20 minutes to hours. 20 minutes is a third of an hour or \frac{1}{3} hours. \begin{aligned} &Speed = distance \div time \\\\ &Speed =18 \div \frac{1}{3} \\\\ &Speed = 54 \\\\ &54 \ km/h \end{aligned}

Speed distance time triangle GCSE questions

1. A commercial aircraft travels from its origin to its destination in a time of 2 hours and 15 minutes. The journey is 1462.5 \ km.

What is the average speed of the plane in km/hour?

2 hours 15 minutes = 2\frac{15}{60} = 2\frac{1}{4} = 2.25

2. John travelled 30 \ km in 90 minutes.

Nadine travelled 52.5 \ km in 2.5 hours.

Who had the greater average speed?

You must show your working.

90 minutes = 1.5 hours

John = 30 \div 1.5 = 20 \ km/h

Nadine = 52.5 \div 2.5 = 21 \ km/h

Nadine has the greater average speed.

3. The distance from Birmingham to Rugby is 40 miles.

Omar drives from Rugby to Birmingham at 60 \ mph.

Ayushi drives from Rugby to Birmingham at 50 \ mph.

How much longer was Ayushi’s journey compared to Omar’s journey? Give your answer in minutes.

For calculating time in hours for Omar or Ayushi.

For converting hours into minutes for Omar or Ayushi.

For correct final answer of 8 minutes.

Learning checklist

You have now learned how to:

  • Use compound units such as speed
  • Solve simple kinematic problem involving distance and speed
  • Change freely between related standard units (e.g. time, length) and compound units (e.g. speed) in numerical contexts
  • Work with compound units in a numerical context

The next lessons are

  • Best buy maths
  • Scale maths

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Distance, Time and Speed Word Problems | GMAT GRE Maths

Before you get into distance, time and speed word problems, take a few minutes to read this first and understand: How to build your credit score in USA as an international student .

Problems involving Time, Distance and Speed are solved based on one simple formula.

Distance = Speed * Time

Which implies →

Speed = Distance / Time   and

Time = Distance / Speed

Let us take a look at some simple examples of distance, time and speed problems.   Example 1. A boy walks at a speed of 4 kmph. How much time does he take to walk a distance of 20 km?

Time = Distance / speed = 20/4 = 5 hours.   Example 2. A cyclist covers a distance of 15 miles in 2 hours. Calculate his speed.

Speed = Distance/time = 15/2 = 7.5 miles per hour.   Example 3. A car takes 4 hours to cover a distance, if it travels at a speed of 40 mph. What should be its speed to cover the same distance in 1.5 hours?

Distance covered = 4*40 = 160 miles

Speed required to cover the same distance in 1.5 hours = 160/1.5 = 106.66 mph   Now, take a look at the following example:

Example 4. If a person walks at 4 mph, he covers a certain distance. If he walks at 9 mph, he covers 7.5 miles more. How much distance did he actually cover?

Now we can see that the direct application of our usual formula Distance = Speed * Time or its variations cannot be done in this case and we need to put in extra effort to calculate the given parameters.

Let us see how this question can be solved.

For these kinds of questions, a table like this might make it easier to solve.

  Let the distance covered by that person be ‘d’.

Walking at 4 mph and covering a distance ‘d’ is done in a time of ‘d/4’

IF he walks at 9 mph, he covers 7.5 miles more than the actual distance d, which is ‘d+7.5’.

He does this in a time of (d+7.5)/9.

Since the time is same in both the cases →

d/4 = (d+7.5)/9            →        9d = 4(d+7.5)   →        9d=4d+30        →        d = 6.

So, he covered a distance of 6 miles in 1.5 hours.   Example 5. A train is going at 1/3 of its usual speed and it takes an extra 30 minutes to reach its destination. Find its usual time to cover the same distance.

Here, we see that the distance is same.

Let us assume that its usual speed is ‘s’ and time is ‘t’, then

  s*t = (1/3)s*(t+30)      →        t = t/3 + 10      →        t = 15.

So the actual time taken to cover the distance is 15 minutes.

Note: Note the time is expressed in terms of ‘minutes’. When we express distance in terms of miles or kilometers, time is expressed in terms of hours and has to be converted into appropriate units of measurement.

Solved Questions on Trains

Example 1. X and Y are two stations which are 320 miles apart. A train starts at a certain time from X and travels towards Y at 70 mph. After 2 hours, another train starts from Y and travels towards X at 20 mph. At what time do they meet?

Let the time after which they meet be ‘t’ hours.

Then the time travelled by second train becomes ‘t-2’.

Distance covered by first train+Distance covered by second train = 320 miles

70t+20(t-2) = 320

Solving this gives t = 4.

So the two trains meet after 4 hours.   Example 2. A train leaves from a station and moves at a certain speed. After 2 hours, another train leaves from the same station and moves in the same direction at a speed of 60 mph. If it catches up with the first train in 4 hours, what is the speed of the first train?

Let the speed of the first train be ‘s’.

Distance covered by the first train in (2+4) hours = Distance covered by second train in 4 hours

Therefore, 6s = 60*4

Solving which gives s=40.

So the slower train is moving at the rate of 40 mph.  

Questions on Boats/Airplanes

For problems with boats and streams,

Speed of the boat upstream (against the current) = Speed of the boat in still water – speed of the stream

[As the stream obstructs the speed of the boat in still water, its speed has to be subtracted from the usual speed of the boat]

Speed of the boat downstream (along with the current) = Speed of the boat in still water + speed of the stream

[As the stream pushes the boat and makes it easier for the boat to reach the destination faster, speed of the stream has to be added]

Similarly, for airplanes travelling with/against the wind,

Speed of the plane with the wind = speed of the plane + speed of the wind

Speed of the plane against the wind = speed of the plane – speed of the wind

Let us look at some examples.

Example 1. A man travels at 3 mph in still water. If the current’s velocity is 1 mph, it takes 3 hours to row to a place and come back. How far is the place?

Let the distance be ‘d’ miles.

Time taken to cover the distance upstream + Time taken to cover the distance downstream = 3

Speed upstream = 3-1 = 2 mph

Speed downstream = 3+1 = 4 mph

So, our equation would be d/2 + d/4 = 3 → solving which, we get d = 4 miles.   Example 2. With the wind, an airplane covers a distance of 2400 kms in 4 hours and against the wind in 6 hours. What is the speed of the plane and that of the wind?

Let the speed of the plane be ‘a’ and that of the wind be ‘w’.

Our table looks like this:  

  4(a+w) = 2400 and 6(a-w) = 2400

Expressing one unknown variable in terms of the other makes it easier to solve, which means

a+w = 600 → w=600-a

Substituting the value of w in the second equation,

a-(600-a) = 400 → a = 500

The speed of the plane is 500 kmph and that of the wind is 100 kmph.  

More solved examples on Speed, Distance and Time

Example 1. A boy travelled by train which moved at the speed of 30 mph. He then boarded a bus which moved at the speed of 40 mph and reached his destination. The entire distance covered was 100 miles and the entire duration of the journey was 3 hours. Find the distance he travelled by bus.

Let the time taken by the train be ‘t’. Then that of bus is ‘3-t’.

The entire distance covered was 100 miles

So, 30t + 40(3-t) = 100

Solving which gives t=2.

Substituting the value of t in 40(3-t), we get the distance travelled by bus is 40 miles.

Alternatively, we can add the time and equate it to 3 hours, which directly gives the distance.

d/30 + (100-d)/40 = 3

Solving which gives d = 60, which is the distance travelled by train. 100-60 = 40 miles is the distance travelled by bus.   Example 2. A plane covered a distance of 630 miles in 6 hours. For the first part of the trip, the average speed was 100 mph and for the second part of the trip, the average speed was 110 mph. what is the time it flew at each speed?

Our table looks like this.

Assuming the distance covered in the 1 st part of journey to be ‘d’, the distance covered in the second half becomes ‘630-d’.

Assuming the time taken for the first part of the journey to be ‘t’, the time taken for the second half becomes ‘6-t’.

From the first equation, d=100t

The second equation is 630-d = 110(6-t).

Substituting the value of d from the first equation, we get

630-100t = 110(6-t)

Solving this gives t=3.

So the plane flew the first part of the journey in 3 hours and the second part in 3 hours.   Example 2. Two persons are walking towards each other on a walking path that is 20 miles long. One is walking at the rate of 3 mph and the other at 4 mph. After how much time will they meet each other?

  Assuming the distance travelled by the first person to be ‘d’, the distance travelled by the second person is ’20-d’.

The time is ‘t’ for both of them because when they meet, they would have walked for the same time.

Since time is same, we can equate as

d/3 = (20-d)/4

Solving this gives d=60/7 miles (8.5 miles approximately)

Then t = 20/7 hours

So the two persons meet after 2 6/7 hours.  

Practice Questions for you to solve

Problem 1: Click here

A boat covers a certain distance in 2 hours, while it comes back in 3 hours. If the speed of the stream is 4 kmph, what is the boat’s speed in still water?

A) 30 kmph B) 20 kmph C) 15 kmph D) 40 kmph

Answer 1: Click here

Explanation

Let the speed of the boat be ‘s’ kmph.

Then, 2(s+4) = 3(s-4) → s = 20

Problem 2: Click here

A cyclist travels for 3 hours, travelling for the first half of the journey at 12 mph and the second half at 15 mph. Find the total distance he covered.

A) 30 miles B) 35 miles C) 40 miles D) 180 miles

Answer 2: Click here

Since it is mentioned, that the first ‘half’ of the journey is covered in 12 mph and the second in 15, the equation looks like

(d/2)/12 + (d/2)/15 = 3

Solving this gives d = 40 miles

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17 thoughts on “Distance, Time and Speed Word Problems | GMAT GRE Maths”

Meera walked to school at a speed of 3 miles per hour. Once she reached the school, she realized that she forgot to bring her books, so rushed back home at a speed of 6 miles per hour. She then walked back to school at a speed of 4 miles per hour. All the times, she walked in the same route. please explain above problem

When she walks faster the time she takes to reach her home and school is lower. There is nothing wrong with the statement. They never mentioned how long she took every time.

a man covers a distance on a toy train.if the train moved 4km/hr faster,it would take 30 min. less. if it moved 2km/hr slower, it would have taken 20 min. more .find the distance.

Let the speed be x. and time be y. A.T.Q, (x+4)(y-1/2)=d and (x-2)(y+1/3)=d. Equate these two and get the answer

Could you explain how ? you have two equations and there are 3 variables.

The 3rd equation is d=xy. Now, you have 3 equations with 3 unknowns. The variables x and y represent the usual speed and usual time to travel distance d.

Speed comes out to be 20 km/hr and the time taken is 3 hrs. The distance traveled is 60 km.

(s + 4) (t – 1/2)= st 1…new equotion = -1/2s + 4t = 2

(s – 2) (t + 1/3)= st 2…new equotion = 1/3s – 2t = 2/3

Multiply all by 6 1… -3s + 24t = 12 2… 2s – 12t = 4 Next, use elimination t= 3 Find s: -3s + 24t = 12 -3s + 24(3) = 12 -3s = -60 s= 20

st or distance = 3 x 20 = 60 km/h

It’s probably the average speed that we are looking for here. Ave. Speed= total distance/ total time. Since it’s harder to look for one variable since both are absent, you can use, 3d/ d( V2V3 + V1V3 + V1V2/ V1V2V3)

2 girls meenu and priya start at the same time to ride from madurai to manamadurai, 60 km away.meenu travels 4kmph slower than priya. priya reaches manamadurai and at turns back meeting meenu 12km from manamaduai. find meenu’s speed?

Hi, when the two girls meet, they have taken equal time to travel their respective distance. So, we just need to equate their time equations

Distance travelled by Meenu = 60 -12 = 48 Distance travelled by Priya = 60 + 12 = 72 Let ‘s’ be the speed of Meenu

Time taken by Meenu => t1 = 48/s Time taken by Priya => t2 = 72/(s+4)

t1=t2 Thus, 48/s = 72/(s+4) => 24s = 192 => s = 8Km/hr

A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 KMS away from A at the same time. On the way, however the train lost about 12.5 minutes while stopping at the stations. The speed of the car is:

Let speed of the CAR BE x kmph.

Then, speed of the train = 3/2(x) .’. 75/x – 75/(3/2)x= 125/(10*60) — subtracting the times travelled by two them hence trains wastage time

therefore x= 120 kmph

A cyclist completes a distance of 60 km at the same speed throughout. She travels 10 km in one hour. She stops every 20 km for one hour to have a break. What are the two variables involved in this situation?

For the answer, not variables: 60km divided by 10km/h=6 hours 60 divided by 20= 3 hours 3 hours+6 hours= 9 hours Answer: 9 hours

Let the length of the train to prod past a point be the intrinsic distance (D) of the train and its speed be S. Its speed, S in passing the electric pole of negligible length is = D/12. The length of the platform added to the intrinsic length of the train. So, the total distance = D + 200. The time = 20 secs. The Speed, S = (D + 200)/20 At constant speed, D/12 = (D + 200)/20 Cross-multiplying, 20D = 12D + 200*12 20D – 12D = 200*12; 8D = 200*12 D = 200*12/8 = 300m. 4th Aug, 2018

Can anyone solve this? Nathan and Philip agree to meet up at the park at 5:00 pm. Nathan lives 300 m due north of the park, and Philip lives 500 m due west of the park. Philip leaves his house at 4:54 pm and walks towards the park at a pace of 1.5 m/s, but Nathan loses track of time and doesn’t leave until 4:59 pm. Trying to avoid being too late, he jogs towards the park at 2.5 m/s. At what rate is the distance between the two friends changing 30 seconds after Nathan has departed?

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Algebra Topics  - Distance Word Problems

Algebra topics  -, distance word problems, algebra topics distance word problems.

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Algebra Topics: Distance Word Problems

Lesson 10: distance word problems.

/en/algebra-topics/introduction-to-word-problems/content/

What are distance word problems?

Distance word problems are a common type of algebra word problems. They involve a scenario in which you need to figure out how fast , how far , or how long one or more objects have traveled. These are often called train problems because one of the most famous types of distance problems involves finding out when two trains heading toward each other cross paths.

In this lesson, you'll learn how to solve train problems and a few other common types of distance problems. But first, let's look at some basic principles that apply to any distance problem.

The basics of distance problems

There are three basic aspects to movement and travel: distance , rate , and time . To understand the difference among these, think about the last time you drove somewhere.

problem solving in speed time and distance

The distance is how far you traveled. The rate is how fast you traveled. The time is how long the trip took.

The relationship among these things can be described by this formula:

distance = rate x time d = rt

In other words, the distance you drove is equal to the rate at which you drove times the amount of time you drove. For an example of how this would work in real life, just imagine your last trip was like this:

  • You drove 25 miles—that's the distance .
  • You drove an average of 50 mph—that's the rate .
  • The drive took you 30 minutes, or 0 .5 hours—that's the time .

According to the formula, if we multiply the rate and time , the product should be our distance.

problem solving in speed time and distance

And it is! We drove 50 mph for 0.5 hours—and 50 ⋅ 0.5 equals 25 , which is our distance.

What if we drove 60 mph instead of 50? How far could we drive in 30 minutes? We could use the same formula to figure this out.

60 ⋅ 0.5 is 30, so our distance would be 30 miles.

Solving distance problems

When you solve any distance problem, you'll have to do what we just did—use the formula to find distance , rate , or time . Let's try another simple problem.

On his day off, Lee took a trip to the zoo. He drove an average speed of 65 mph, and it took him two-and-a-half hours to get from his house to the zoo. How far is the zoo from his house?

First, we should identify the information we know. Remember, we're looking for any information about distance, rate, or time. According to the problem:

  • The rate is 65 mph.
  • The time is two-and-a-half hours, or 2.5 hours.
  • The distance is unknown—it's what we're trying to find.

You could picture Lee's trip with a diagram like this:

problem solving in speed time and distance

This diagram is a start to understanding this problem, but we still have to figure out what to do with the numbers for distance , rate , and time . To keep track of the information in the problem, we'll set up a table. (This might seem excessive now, but it's a good habit for even simple problems and can make solving complicated problems much easier.) Here's what our table looks like:

We can put this information into our formula: distance = rate ⋅ time .

We can use the distance = rate ⋅ time formula to find the distance Lee traveled.

The formula d = rt looks like this when we plug in the numbers from the problem. The unknown distance is represented with the variable d .

d = 65 ⋅ 2.5

To find d , all we have to do is multiply 65 and 2.5. 65 ⋅ 2.5 equals 162.5 .

We have an answer to our problem: d = 162.5. In other words, the distance Lee drove from his house to the zoo is 162.5 miles.

Be careful to use the same units of measurement for rate and time. It's possible to multiply 65 miles per hour by 2.5 hours because they use the same unit: an hour . However, what if the time had been written in a different unit, like in minutes ? In that case, you'd have to convert the time into hours so it would use the same unit as the rate.

Solving for rate and time

In the problem we just solved we calculated for distance , but you can use the d = rt formula to solve for rate and time too. For example, take a look at this problem:

After work, Janae walked in her neighborhood for a half hour. She walked a mile-and-a-half total. What was her average speed in miles per hour?

We can picture Janae's walk as something like this:

problem solving in speed time and distance

And we can set up the information from the problem we know like this:

The table is repeating the facts we already know from the problem. Janae walked one-and-a-half miles or 1.5 miles in a half hour, or 0.5 hours.

As always, we start with our formula. Next, we'll fill in the formula with the information from our table.

The rate is represented by r because we don't yet know how fast Janae was walking. Since we're solving for r , we'll have to get it alone on one side of the equation.

1.5 = r ⋅ 0.5

Our equation calls for r to be multiplied by 0.5, so we can get r alone on one side of the equation by dividing both sides by 0.5: 1.5 / 0.5 = 3 .

r = 3 , so 3 is the answer to our problem. Janae walked 3 miles per hour.

In the problems on this page, we solved for distance and rate of travel, but you can also use the travel equation to solve for time . You can even use it to solve certain problems where you're trying to figure out the distance, rate, or time of two or more moving objects. We'll look at problems like this on the next few pages.

Two-part and round-trip problems

Do you know how to solve this problem?

Bill took a trip to see a friend. His friend lives 225 miles away. He drove in town at an average of 30 mph, then he drove on the interstate at an average of 70 mph. The trip took three-and-a-half hours total. How far did Bill drive on the interstate?

This problem is a classic two-part trip problem because it's asking you to find information about one part of a two-part trip. This problem might seem complicated, but don't be intimidated!

problem solving in speed time and distance

You can solve it using the same tools we used to solve the simpler problems on the first page:

  • The travel equation d = rt
  • A table to keep track of important information

Let's start with the table . Take another look at the problem. This time, the information relating to distance , rate , and time has been underlined.

Bill took a trip to see a friend. His friend lives 225 miles away. He drove in town at an average of 30 mph , then he drove on the interstate at an average of 70 mph . The trip took three-and-a-half hours total. How far did Bill drive on the interstate?

If you tried to fill in the table the way we did on the last page, you might have noticed a problem: There's too much information. For instance, the problem contains two rates— 30 mph and 70 mph . To include all of this information, let's create a table with an extra row. The top row of numbers and variables will be labeled in town , and the bottom row will be labeled interstate .

We filled in the rates, but what about the distance and time ? If you look back at the problem, you'll see that these are the total figures, meaning they include both the time in town and on the interstate. So the total distance is 225 . This means this is true:

Interstate distance + in-town distance = Total distance

Together, the interstate distance and in-town distance are equal to the total distance. See?

problem solving in speed time and distance

In any case, we're trying to find out how far Bill drove on the interstate , so let's represent this number with d . If the interstate distance is d , it means the in-town distance is a number that equals the total, 225 , when added to d . In other words, it's equal to 225 - d .

problem solving in speed time and distance

We can fill in our chart like this:

We can use the same technique to fill in the time column. The total time is 3.5 hours . If we say the time on the interstate is t , then the remaining time in town is equal to 3.5 - t . We can fill in the rest of our chart.

Now we can work on solving the problem. The main difference between the problems on the first page and this problem is that this problem involves two equations. Here's the one for in-town travel :

225 - d = 30 ⋅ (3.5 - t)

And here's the one for interstate travel :

If you tried to solve either of these on its own, you might have found it impossible: since each equation contains two unknown variables, they can't be solved on their own. Try for yourself. If you work either equation on its own, you won't be able to find a numerical value for d . In order to find the value of d , we'll also have to know the value of t .

We can find the value of t in both problems by combining them. Let's take another look at our travel equation for interstate travel.

While we don't know the numerical value of d , this equation does tell us that d is equal to 70 t .

Since 70 t and d are equal , we can replace d with 70 t . Substituting 70 t for d in our equation for interstate travel won't help us find the value of t —all it tells us is that 70 t is equal to itself, which we already knew.

But what about our other equation, the one for in-town travel?

When we replace the d in that equation with 70 t , the equation suddenly gets much easier to solve.

225 - 70t = 30 ⋅ (3.5 - t)

Our new equation might look more complicated, but it's actually something we can solve. This is because it only has one variable: t . Once we find t , we can use it to calculate the value of d —and find the answer to our problem.

To simplify this equation and find the value of t , we'll have to get the t alone on one side of the equals sign. We'll also have to simplify the right side as much as possible.

Let's start with the right side: 30 times (3.5 - t ) is 105 - 30 t .

225 - 70t = 105 - 30t

Next, let's cancel out the 225 next to 70 t . To do this, we'll subtract 225 from both sides. On the right side, it means subtracting 225 from 105 . 105 - 225 is -120 .

- 70t = -120 - 30t

Our next step is to group like terms—remember, our eventual goal is to have t on the left side of the equals sign and a number on the right . We'll cancel out the -30 t on the right side by adding 30 t to both sides. On the right side, we'll add it to -70 t . -70 t + 30 t is -40 t .

- 40t = -120

Finally, to get t on its own, we'll divide each side by its coefficient: -40. -120 / - 40 is 3 .

So t is equal to 3 . In other words, the time Bill traveled on the interstate is equal to 3 hours . Remember, we're ultimately trying to find the distanc e Bill traveled on the interstate. Let's look at the interstate row of our chart again and see if we have enough information to find out.

It looks like we do. Now that we're only missing one variable, we should be able to find its value pretty quickly.

To find the distance, we'll use the travel formula distance = rate ⋅ time .

We now know that Bill traveled on the interstate for 3 hours at 70 mph , so we can fill in this information.

d = 3 ⋅ 70

Finally, we finished simplifying the right side of the equation. 3 ⋅ 70 is 210 .

So d = 210 . We have the answer to our problem! The distance is 210 . In other words, Bill drove 210 miles on the interstate.

problem solving in speed time and distance

Solving a round-trip problem

It might have seemed like it took a long time to solve the first problem. The more practice you get with these problems, the quicker they'll go. Let's try a similar problem. This one is called a round-trip problem because it describes a round trip—a trip that includes a return journey. Even though the trip described in this problem is slightly different from the one in our first problem, you should be able to solve it the same way. Let's take a look:

Eva drove to work at an average speed of 36 mph. On the way home, she hit traffic and only drove an average of 27 mph. Her total time in the car was 1 hour and 45 minutes, or 1.75 hours. How far does Eva live from work?

If you're having trouble understanding this problem, you might want to visualize Eva's commute like this:

problem solving in speed time and distance

As always, let's start by filling in a table with the important information. We'll make a row with information about her trip to work and from work .

1.75 - t to describe the trip from work. (Remember, the total travel time is 1.75 hours , so the time to work and from work should equal 1.75 .)

From our table, we can write two equations:

  • The trip to work can be represented as d = 36 t .
  • The trip from work can be represented as d = 27 (1.75 - t ) .

In both equations, d represents the total distance. From the diagram, you can see that these two equations are equal to each other—after all, Eva drives the same distance to and from work .

problem solving in speed time and distance

Just like with the last problem we solved, we can solve this one by combining the two equations.

We'll start with our equation for the trip from work .

d = 27 (1.75 - t)

Next, we'll substitute in the value of d from our to work equation, d = 36 t . Since the value of d is 36 t , we can replace any occurrence of d with 36 t .

36t = 27 (1.75 - t)

Now, let's simplify the right side. 27 ⋅(1.75 - t ) is 47.25 .

36t = 47.25 - 27t

Next, we'll cancel out -27 t by adding 27 t to both sides of the equation. 36 t + 27 t is 63 t .

63t = 47.25

Finally, we can get t on its own by dividing both sides by its coefficient: 63 . 47.25 / 63 is .75 .

t is equal to .75 . In other words, the time it took Eva to drive to work is .75 hours . Now that we know the value of t , we'll be able to can find the distance to Eva's work.

If you guessed that we were going to use the travel equation again, you were right. We now know the value of two out of the three variables, which means we know enough to solve our problem.

First, let's fill in the values we know. We'll work with the numbers for the trip to work . We already knew the rate : 36 . And we just learned the time : .75 .

d = 36 ⋅ .75

Now all we have to do is simplify the equation: 36 ⋅ .75 = 27 .

d is equal to 27 . In other words, the distance to Eva's work is 27 miles . Our problem is solved.

problem solving in speed time and distance

Intersecting distance problems

An intersecting distance problem is one where two things are moving toward each other. Here's a typical problem:

Pawnee and Springfield are 420 miles apart. A train leaves Pawnee heading to Springfield at the same time a train leaves Springfield heading to Pawnee. One train is moving at a speed of 45 mph, and the other is moving 60 mph. How long will they travel before they meet?

This problem is asking you to calculate how long it will take these two trains moving toward each other to cross paths. This might seem confusing at first. Even though it's a real-world situation, it can be difficult to imagine distance and motion abstractly. This diagram might help you get a sense of what this situation looks like:

problem solving in speed time and distance

If you're still confused, don't worry! You can solve this problem the same way you solved the two-part problems on the last page. You'll just need a chart and the travel formula .

Pawnee and Springfield are 420 miles apart. A train leaves Pawnee heading toward Springfield at the same time a train leaves Springfield heading toward Pawnee. One train is moving at a speed of 45 mph , and the other is moving 60 mph . How long will they travel before they meet?

Let's start by filling in our chart. Here's the problem again, this time with the important information underlined. We can start by filling in the most obvious information: rate . The problem gives us the speed of each train. We'll label them fast train and slow train . The fast train goes 60 mph . The slow train goes only 45 mph .

problem solving in speed time and distance

We can also put this information into a table:

We don't know the distance each train travels to meet the other yet—we just know the total distance. In order to meet, the trains will cover a combined distance equal to the total distance. As you can see in this diagram, this is true no matter how far each train travels.

problem solving in speed time and distance

This means that—just like last time—we'll represent the distance of one with d and the distance of the other with the total minus d. So the distance for the fast train will be d , and the distance for the slow train will be 420 - d .

Because we're looking for the time both trains travel before they meet, the time will be the same for both trains. We can represent it with t .

The table gives us two equations: d = 60 t and 420 - d = 45 t . Just like we did with the two-part problems, we can combine these two equations.

The equation for the fast train isn't solvable on its own, but it does tell us that d is equal to 60 t .

The other equation, which describes the slow train, can't be solved alone either. However, we can replace the d with its value from the first equation.

420 - d = 45t

Because we know that d is equal to 60 t , we can replace the d in this equation with 60 t . Now we have an equation we can solve.

420 - 60t = 45t

To solve this equation, we'll need to get t and its coefficients on one side of the equals sign and any other numbers on the other. We can start by canceling out the -60 t on the left by adding 60 t to both sides. 45 t + 60 t is 105 t .

Now we just need to get rid of the coefficient next to t . We can do this by dividing both sides by 105 . 420 / 105 is 4 .

t = 4 . In other words, the time it takes the trains to meet is 4 hours . Our problem is solved!

If you want to be sure of your answer, you can check it by using the distance equation with t equal to 4 . For our fast train, the equation would be d = 60 ⋅ 4 . 60 ⋅ 4 is 240 , so the distance our fast train traveled would be 240 miles. For our slow train, the equation would be d = 45 ⋅ 4 . 45 ⋅ 4 is 180 , so the distance traveled by the slow train is 180 miles . Remember how we said the distance the slow train and fast train travel should equal the total distance? 240 miles + 180 miles equals 420 miles , which is the total distance from our problem. Our answer is correct.

Practice problem 1

Here's another intersecting distance problem. It's similar to the one we just solved. See if you can solve it on your own. When you're finished, scroll down to see the answer and an explanation.

Jon and Dani live 270 miles apart. One day, they decided to drive toward each other and hang out wherever they met. Jon drove an average of 65 mph, and Dani drove an average of 70 mph. How long did they drive before they met up?

Problem 1 answer

Here's practice problem 1:

Jon and Dani live 270 miles apart. One day, they decided to drive toward each other and hang out wherever they met. Jon drove an average of 65 mph, and Dani drove 70 mph. How long did they drive before they met up?

Answer: 2 hours .

Let's solve this problem like we solved the others. First, try making the chart. It should look like this:

Here's how we filled in the chart:

  • Distance: Together, Dani and Jon will have covered the total distance between them by the time they meet up. That's 270 . Jon's distance is represented by d , so Dani's distance is 270 - d .
  • Rate: The problem tells us Dani and Jon's speeds. Dani drives 65 mph , and Jon drives 70 mph .
  • Time: Because Jon and Dani drive the same amount of time before they meet up, both of their travel times are represented by t .

Now we have two equations. The equation for Jon's travel is d = 65 t . The equation for Dani's travel is 270 - d = 70 t . To solve this problem, we'll need to combine them.

The equation for Jon tells us that d is equal to 65 t . This means we can combine the two equations by replacing the d in Dani's equation with 65 t .

270 - 65t = 70t

Let's get t on one side of the equation and a number on the other. The first step to doing this is to get rid of -65 t on the left side. We'll cancel it out by adding 65 t to both sides: 70 t + 65 t is 135 t .

All that's left to do is to get rid of the 135 next to the t . We can do this by dividing both sides by 135 : 270 / 135 is 2 .

That's it. t is equal to 2 . We have the answer to our problem: Dani and Jon drove 2 hours before they met up.

Overtaking distance problems

The final type of distance problem we'll discuss in this lesson is a problem in which one moving object overtakes —or passes —another. Here's a typical overtaking problem:

The Hill family and the Platter family are going on a road trip. The Hills left 3 hours before the Platters, but the Platters drive an average of 15 mph faster. If it takes the Platter family 13 hours to catch up with the Hill family, how fast are the Hills driving?

You can picture the moment the Platter family left for the road trip a little like this:

problem solving in speed time and distance

The problem tells us that the Platter family will catch up with the Hill family in 13 hours and asks us to use this information to find the Hill family's rate . Like some of the other problems we've solved in this lesson, it might not seem like we have enough information to solve this problem—but we do. Let's start making our chart. The distance can be d for both the Hills and the Platters—when the Platters catch up with the Hills, both families will have driven the exact same distance.

Filling in the rate and time will require a little more thought. We don't know the rate for either family—remember, that's what we're trying to find out. However, we do know that the Platters drove 15 mph faster than the Hills. This means if the Hill family's rate is r , the Platter family's rate would be r + 15 .

Now all that's left is the time. We know it took the Platters 13 hours to catch up with the Hills. However, remember that the Hills left 3 hours earlier than the Platters—which means when the Platters caught up, they'd been driving 3 hours more than the Platters. 13 + 3 is 16 , so we know the Hills had been driving 16 hours by the time the Platters caught up with them.

Our chart gives us two equations. The Hill family's trip can be described by d = r ⋅ 16 . The equation for the Platter family's trip is d = ( r + 15) ⋅ 13 . Just like with our other problems, we can combine these equations by replacing a variable in one of them.

The Hill family equation already has the value of d equal to r ⋅ 16. So we'll replace the d in the Platter equation with r ⋅ 16 . This way, it will be an equation we can solve.

r ⋅ 16 = (r + 15) ⋅ 13

First, let's simplify the right side: r ⋅ 16 is 16 r .

16r = (r + 15) ⋅ 13

Next, we'll simplify the right side and multiply ( r + 15) by 13 .

16r = 13r + 195

We can get both r and their coefficients on the left side by subtracting 13 r from 16 r : 16 r - 13 r is 3 r .

Now all that's left to do is get rid of the 3 next to the r . To do this, we'll divide both sides by 3: 195 / 3 is 65 .

So there's our answer: r = 65. The Hill family drove an average of 65 mph .

You can solve any overtaking problem the same way we solved this one. Just remember to pay special attention when you're setting up your chart. Just like the Hill family did in this problem, the person or vehicle who started moving first will always have a greater travel time.

Practice problem 2

Try solving this problem. It's similar to the problem we just solved. When you're finished, scroll down to see the answer and an explanation.

A train moving 60 mph leaves the station at noon. An hour later, a train moving 80 mph leaves heading the same direction on a parallel track. What time does the second train catch up to the first?

Problem 2 answer

Here's practice problem 2:

Answer: 4 p.m.

To solve this problem, start by making a chart. Here's how it should look:

Here's an explanation of the chart:

  • Distance: Both trains will have traveled the same distance by the time the fast train catches up with the slow one, so the distance for both is d .
  • Rate: The problem tells us how fast each train was going. The fast train has a rate of 80 mph , and the slow train has a rate of 60 mph .
  • Time: We'll use t to represent the fast train's travel time before it catches up. Because the slow train started an hour before the fast one, it will have been traveling one hour more by the time the fast train catches up. It's t + 1 .

Now we have two equations. The equation for the fast train is d = 80 t . The equation for the slow train is d = 60 ( t + 1) . To solve this problem, we'll need to combine the equations.

The equation for the fast train says d is equal to 80 t . This means we can combine the two equations by replacing the d in the slow train's equation with 80 t .

80t = 60 (t + 1)

First, let's simplify the right side of the equation: 60 ⋅ ( t + 1) is 60 t + 60 .

80t = 60t + 60

To solve the equation, we'll have to get t on one side of the equals sign and a number on the other. We can get rid of 60 t on the right side by subtracting 60 t from both sides: 80 t - 60 t is 20 t .

Finally, we can get rid of the 20 next to t by dividing both sides by 20 . 60 divided by 20 is 3 .

So t is equal to 3 . The fast train traveled for 3 hours . However, it's not the answer to our problem. Let's look at the original problem again. Pay attention to the last sentence, which is the question we're trying to answer.

Our problem doesn't ask how long either of the trains traveled. It asks what time the second train catches up with the first.

The problem tells us that the slow train left at noon and the fast one left an hour later. This means the fast train left at 1 p.m . From our equations, we know the fast train traveled 3 hours . 1 + 3 is 4 , so the fast train caught up with the slow one at 4 p.m . The answer to the problem is 4 p.m .

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How to Solve Speed Distance and Time Questions Quickly

September 9, 2023

Solve Speed, Distance and Time Question Quickly

Speed Distance and Time is a part of our study since beginning but students generally fail to focus on this topic . This page will provide you with  the most  simple approach to  Solve Speed Distance and Time questions quickly along with the help of some  Examples

Speed distance and time quickly

How to Solve Speed Distance and Time Questions Quickly:-

Before solving the question, we need to know about the Formulas of Speed Time and Distance.

Basic Formulas

  • Distance (D1) = Speed × Time
  • Time = Distance/Speed
  • Speed = Distance / Time
  •  To Convert Km/h to M/s =Km/h * 5/18 = Values in  m/s or m/sec *18/5 =  Values in km/h
  • Calculate Average Speed = 2ab/a + b
  • X and Y may walk in same direction or in opposite direction.

How to Solve Problems Quickly

Question 1 : Yamini runs from one side of a road to the other to save herself from a dog who was chasing her . The road was 100meters across. In order to save herself from the dog Yamini  takes 25 seconds to cross the road. What is the speed of Yamini ?

A. 2 meters/second

B. 3.5 meters/second

C. 5.1 meters/second

D. 4 meters/second

Solution:    S = \frac{d}{t} \

S = \frac{100.0}{25.0} \

⇒ 4 meters/second

Correct Answer: D

Question 2 : Two person X and a Y are working together in for a task  and stay together in a nearby apartment. X takes 30 minutes and  Y takes 20 minutes to walk from apartment to office. If one day X started at 10:00 AM and the Y at 10:05 AM from the apartment to office, when will they meet?

A. 10: 40 AM

B. 10: 15 AM

C. 10:00 AM

D. 10:05 am

Solution:    Ratio of old man speed to young man speed = 3:2

The distance covered by the old man in 5 min = 10units

The 10 unit is covered with relative speed= \frac{10}{(3-2)} \ =10 min

so, they will meet at 10:15 am.

Correct Answer: B

Question 3 : A freight train takes 5 hours to travel a certain distance. If it reduces its speed by 10 km/h, it would take 6 hours to cover the same distance. Find the original speed of the freight train and the distance it traveled.

                   Options

                     A. 12Km/h, 60 km

                    B. 23 Km/h, 1 km

                    C.45Km/h, 30 km

                    D.18km/h, 1 km

Solution: Let the original speed of the freight train be S km/h, and let the distance it traveled

be D kilometers.

According to the problem, time taken to travel the distance at the original speed is 5 hours, and time taken to travel the same distance at a reduced speed (S – 10) km/h is 6 hours.

We can set up the equations: D / S = 5 D / (S – 10) = 6

Now, we can solve for S and D. From the first equation, we get: S = D / 5 Substitute the value of S into the second equation: D / (D / 5 – 10) = 6 D / (D – 50) = 6 D = 6D – 300 5D = 300 D = 60 km

Now, substitute the value of D back into the first equation to find S: S = D / 5 S = 60 km / 5 S = 12 km/h

Therefore, the original speed of the freight train is 12 km/h, and it traveled a distance of 60 kilometers.

Correct Answer: A

Question 4 Walking \frac{6}{7^{th}} \ of his usual speed, a man is 12 minutes late. What is the usual time taken by him to cover the distance?

B. 1 hr 30 min

C. 1 hr 12 min

Solution:    New speed of man = \frac{6}{7} \ of usual speed

As we know speed & time are inversely proportional

Hence, new time= \frac{7}{6} \ of usual time

Hence \frac{7}{6} \ of usual time- usual time= 12 minutes ⇒ \frac{1}{6} \ of usual time= 12 minutes

Therefore, usual time = 12×6 = 72 minutes

It means 1 hr 12 minutes

Correct Answer: C

Question 5:  Walking barefoot shanti takes 6 hours for walking to a certain place and coming back. She would have taken 2 hours less by riding both ways. What will be the time required by her  to walk both ways?

Solution:    Time taken by the kid in walking to a certain place & riding back is 6 hrs

Time walk+ Time ride = 6

2 hrs of the time is reduced if he rides both the ways.

Time ride+ Time ride = 4

ride = \frac{4}{2} \

Now by putting the value of ride, we get

time walk + 2= 6

timewalk = 6-2 =4

Here we are finding the time taken by the kid if he walk both the ways

Therefore, he will take 8 hrs to walk both the ways.

Correct Answer: D 

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Problems on Calculating Speed

Here we will learn to solve different types of problems on calculating speed.

We know, the speed of a moving body is the distance traveled by it in unit time.             

Formula to find out speed = distance/time

Word problems on calculating speed:

1.  A man walks 20 km in 4 hours. Find his speed.

Solution:            

Distance covered = 20 km

Time taken = 4 hours

We know, speed = distance/time            

                       = 20/4 km/hr

Therefore, speed = 5 km/hr

2. A car covers a distance of 450 m in 1 minute whereas a train covers 69 km in 45 minutes. Find the ratio of their speeds.

Speed of car = Distance covered/Time taken = 450/60 m/sec = 15/2

                                                            = 15/2 × 18/5 km/hr

                                                            = 27 km/hr

Distance covered by train = 69 km

Time taken = 45 min = 45/60 hr = 3/4 hr

Therefore, speed of trains = 69/(3/4) km/hr

                                    = 69/1 × 4/3 km/hr

                                    = 92 km/hr

Therefore, ratio of their speed i.e., speed of car/speed of train = 27/92 = 27 : 92

3. Kate travels a distance of 9 km from her house to the school by auto-rickshaw at 18 km/hr and returns on rickshaw at 15 km/hr. Find the average speed for the whole journey.

Time taken by Kate to reach school = distance/speed = 9/18 hr = 1/2 hr

Time taken by Kate to reach house to school = 9/15 = 3/5 hr

Total time of journey = (1/2 + 3/5) hr

Total time of journey = (5 + 6)/10 = 11/10 hr

Total distance covered = (9 + 9) km = 18 km

Therefore, average speed for the whole journey = distance/speed = 18/(11/10) km/hr

= 18/1 × 10/11 = (18 × 10)/(1 × 11) km/hr

                      = 180/11 km/hr

                      = 16.3 km/hr (approximately)

Speed of Train

Relationship between Speed, Distance and Time

Conversion of Units of Speed

Problems on Calculating Distance

Problems on Calculating Time

Two Objects Move in Same Direction

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Train Passes a Moving Object in the Opposite Direction

Train Passes through a Pole

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8th Grade Math Practice From Problems on Calculating Speed to HOME PAGE

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Chapter 6.8: 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 .

\[r\cdot t=d\]

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. 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.

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:

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

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

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:

\begin{array}{rrlll} 12(t)&=&4(1&-&t) \\ 12t&=&4&-&4t \\ +4t&&&+&4t \\ \midrule \dfrac{16t}{16}&=&\dfrac{4}{16}&& \\ \\ t&=&0.25&& \end{array}

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

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:

\begin{array}{rrrrr} 20(t)&=&80(t&-&6) \\ 20t&=&80t&-&480 \\ -80t&&-80t&& \\ \midrule \dfrac{-60t}{-60}&=&\dfrac{-480}{-60}&& \\ \\ t&=&8&& \end{array}

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

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?

\begin{array}{rrrrrrr} 55(t)&+&40(2.5&-&t)&=&130 \\ 55t&+&100&-&40t&=&130 \\ &-&100&&&&-100 \\ \midrule &&&&\dfrac{15t}{15}&=&\dfrac{30}{15} \\ \\ &&&&t&=&2 \end{array}

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.

Answers to odd questions

20t+25t=60

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How to Calculate Speed, Distance & Time

Working out speed, distance and time is an important part of many roles, including those in the armed forces or transport industries.

If you are applying for a role in these industries, you may be tested on this as part of your recruitment process. The questions will allow your employer to test your applied mathematical abilities.

How to calculate speed, distance & time: the formula triangle you’ll need?

To work out speed, divide the distance of the journey by the time it took to travel, so speed = distance divided by time. To calculate time, divide the distance by speed. To get the distance, multiply the speed by time.

You may see these equations simplified as s=d/t, where s is speed, d is distance, and t is time.

speed distance time triangle

This formula can be arranged into the triangle above. In the triangle, speed and time form the base, as they are what is multiplied together to work out the distance.

The triangle is an easy way to remember the formula and can save you time when working out your exam questions. The triangle will help you remember the three formulas:

  • The formula of speed is Speed = Distance ÷ Time
  • The formula of time is Time = Distance ÷ Speed
  • The formula of distance is Distance = Speed x Time

The triangle shows you what calculation you should use. As distance is at the top of the triangle, to work it out, you need to multiply speed with time.

As speed and time are at the bottom of the triangle, you need to divide this number from the figure for distance, to work out the correct answer.

When beginning to study for your practice test, make sure you write out the triangle on your paper. This will help you to memorize it.

How to calculate speed

speed distance time triangle

To calculate speed, you need to divide the distance by the time. You can work this out by using the triangle. If you cover up speed, you are left with distance over time.

Here is an example:

If a driver has travelled 180 miles and it took them 3 hours to make that distance, then to work out their speed you would take:

180 miles / 3 hours -\> 180 / 3 = 60

So the driver’s speed would be 60mph.

How to calculate distance

speed distance time triangle

To work out the distance travelled, you will need to multiply speed and time.

If a driver travelled at 100mph for 4 hours, then to work out the distance you will need to multiply the speed with the time.

100mph x 4 hours -\> 100 x 4 = 400

The distance is 400 miles.

How to calculate the time

speed distance time triangle

To work out the time that the journey took, you will need to know the speed of the journey and the distance that was travelled.

If the driver travelled 50 miles at 5mph, then to work out the time taken you would divide:

50miles / 5 miles per hour -\> 50 / 5 = 10

The time taken to travel this distance is 10 miles per hour.

Three example speed/distance/time questions

  • Andy drives his lorry for 400 miles, which takes him 8 hours. Harry drives 200 miles, which takes him 4.5 hours. Who is travelling faster?

Answer: Andy is driving at 50mph, and Harry is driving at 44.44mph. So, Andy is driving faster.

  • Tessa runs a 5km race with her running club every Saturday. She runs this in 40 minutes. If she maintains the same speed, how long would it take her to run the 8km race?

Answer: Her speed is 7.5kph. If she runs 10km at 7.5kph, to work out the time, you need to divide 8 by 7.5, which equals 1.066. If we convert this into hours and minutes, it will take her 1 hour and 4 minutes to run 8km.

  • Hannah goes on a cycling trip. In the first half of the journey, she travels at 10mph for 2 hours. In the second half, she travels at 20mph for 90 minutes. How far does she travel in total?

Answer: In the first half of her journey she travels 20 miles (10 x 2 = 20). In the second half, she travels 30 miles, (20 x 1.5 = 30). 20+30 =50, so she travels 50 miles in total.

Methods to get better at these questions

To improve your skill in answering speed distance time questions, there are two main things you can do.

First, make sure that you are familiar with the formula triangle. The key to answering these questions is knowing the formula inside out, so that you always know what equation to use, whether the exam question is asking you to work out speed or distance.

By making sure you have memorized the formula in all its variations, you will save yourself time when answering questions.

Secondly, you should focus on improving your general mathematical skills. You may get a compound question, which asks you to use the formula in conjunction with other maths skills to work out the answer to the problem.

You can try a few maths practice tests, such as these , which will help you to practise your basic numeracy skills.

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problem solving in speed time and distance

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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Problem on Time Speed and Distance

Question 1: A racing car covers a certain distance at a speed of 320 km/hr in 5 hours. To cover the same distance in 5/3 hours it must travel at a speed of:  Solution: Given Distance is constant.  So, Speed is inversely proportional to time.   

1 unit -> 320 km/hr  3 unit -> 320 x 3 = 960 km/hr is required speed  Question 2: A train running at a speed of 36 km/hr and 100 meter long. Find the time in which it passes a man standing near the railway line is :  Solution: Speed = 36 km/hr  Change in m/s  So, speed = 36 * 5/18 = 10 m/s  Time required = Distance/speed  = 100/10  = 10 second   Question 3: If an employee walks 10 km at a speed of 3 km/hr, he will be late by 20 minutes. If he walks at 4 km/hr, how early from the fixed time he will reach ?  Solution: Time taken at 3 km/hr = Distance/speed  = 10/3  Actual time is obtained by subtracting the late time  So, Actual time = 10/3 – 1/3 = 9/3 = 3 hour  Time taken at 4 km/hr = 10/4 hr  Time difference = Actual time – time taken at 4 km/hr  = 3 – 10/4  = 1/2 hour  Hence, he will be early by 30 minutes .  Question 4: The diameter of each wheel of a truck is 140 cm, If each wheel rotates 300 times per minute then the speed of the truck (in km/hr) (take pi=22/7)  Solution: Circumference of the wheel= 2 * 22/7 * r  = 2 * 22/7 * 140/2  = 440 cm  Speed of the car = (440 * 300 * 60)/(1000 * 100 )  = 79.2 km/hr   Question 5: A man drives at the rate of 18 km/hr, but stops at red light for 6 minutes at the end of every 7 km. The time that he will take to cover a distance of 90 km is  Solution: Total Red light at the end of 90 km = 90/7 = 12 Red light + 6 km  Time taken in 12 stops= 12 x 6 = 72 minutes  Time taken by the man to cover the 90 km with 18 km/hr without stops = 90/18 = 5 hours  Total time to cover total distance = 5 hour + 1 hour 12 minute  = 6 hour 12 minute   Question 6: Two jeep start from a police station with a speed of 20 km/hr at intervals of 10 minutes.A man coming from opposite direction towards the police station meets the jeep at an interval of 8 minutes.Find the speed of the man.  Solution:   

Here, 4 units -> 20 km/hr  1 unit -> 5 km/hr  Speed of the man = 1 unit = 1 x 5 = 5 km/hr   Question 7: Two city A and B are 27 km away. Two buses start from A and B in the same direction with speed 24 km/hr and 18 km/hr respectively. Both meet at point C beyond B. Find the distance BC.  Solution: Relative speed = 24 – 18  = 6 km/hr  Time required by faster bus to overtake the slower bus = Distance/time  =27/6 hr  Distance between B and C= 18*(27/6)= 81 km   Question 8: A man travels 800 km by train at 80 km/hr, 420 km by car at 60 km/hr and 200 km by cycle at 20 km/hr. What is the average speed of the journey?  Solution: Avg. Speed = Total distance/time taken  (800 + 420 + 200) / [(800/80) + (420/60) + (200/20)]  =>1420 / (10 + 7 + 10)  =>1420/27  => 1420/27 km/hr   Question 9: Ram and Shyam start at the same with speed 10 km/hr and 11 km/hr respectively.  If Ram takes 48 minutes longer than Shyam in covering the journey, then find the total distance of the journey.  Solution: 

Ram takes 1 hour means 60 minutes more than Shyam.  But actual more time = 48 minute.  60 unit -> 48 min  1 unit -> 4/5  Distance travelled by them= Speed x time  = 11 x 10 = 110 unit  Actual distance travelled = 110 x 4/5  = 88 km   Question 10: A person covered a certain distance at some speed. Had he moved 4 km/hr faster, he would have taken 30 minutes less. If he had moved 3 km/hr slower, he would have taken 30 minutes more. Find the distance (in km)  Solution: Distance = [S 1 S 2 / (S 1 – S 2 )] x T   S 1 = initial speed  S 2 = new speed  Distance travelled by both are same so put equal  [S (S + 4) / 4 ] * (30/60) =[ S (S – 3)/ 3 ]* (30/60)  S = 24  Put in 1st  Distance=(24 * 28) / 4 * (30/60) = 84 km   Question 11: Ram and Shyam start from the same place P at same time towards Q, which are 60km apart. Ram’s speed is 4 km/hr more than that of Shyam.Ram turns back after reaching Q and meet Shyam at 12 km distance from Q.Find the speed of Shyam.  Solution: Let the speed of the Shyam = x km/hr  Then Ram speed will be = (x + 4) km/hr  Total distance covered by Ram = 60 + 12 = 72 km  Total distance covered by Shyam = 60 – 12 = 48 km  Acc. to question, their run time are same.  72/ (x + 4) = 48/ x  72x = 48x + 192  24x= 192  x= 8  Shyam speed is 8 km/hr   Question 12: A and B run a kilometre and A wins by 20 second. A and C run a kilometre and A wins by 250 m. When B and C run the same distance, B wins by 25 second. The time taken by A to run a kilometre is  Solution: Let the time taken by A to cover 1 km = x sec  Time taken by B and C to cover the same distance are x + 20 and x + 45 respectively  Given A travels 1000 then C covers only 750.   

A/C = 3/4 = x/(x+45)  3x + 135 = 4x  x =135  Time taken by A is 2 min 15 second  

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Average Speed Problems

Related Pages Rate, Time, Distance Solving Speed, Time, Distance Problems Using Algebra More Algebra Lessons

In these lessons, we will learn how to solve word problems involving average speed.

There are three main types of average problems commonly encountered in school algebra: Average (Arithmetic Mean) Weighted Average and Average Speed.

How to calculate Average Speed?

The following diagram shows the formula for average speed. Scroll down the page for more examples and solutions on calculating the average speed.

Average Speed

Examples Of Average Speed Problems

Example: John drove for 3 hours at a rate of 50 miles per hour and for 2 hours at 60 miles per hour. What was his average speed for the whole journey?

Solution: Step 1: The formula for distance is

Distance = Rate × Time Total distance = 50 × 3 + 60 × 2 = 270

Step 2: Total time = 3 + 2 = 5

Step 3: Using the formula:

Answer: The average speed is 54 miles per hour.

Be careful! You will get the wrong answer if you add the two speeds and divide the answer by two.

How To Solve The Average Speed Problem?

How to calculate the average speed?

Example: The speed paradox: If I drive from Oxford to Cambridge at 40 miles per hour and then from Cambridge to Oxford at 60 miles per hour, what is my average speed for the whole journey?

How To Find The Average Speed For A Round Trip?

Example: On Alberto’s drive to his aunt’s house, the traffic was light, and he drove the 45-mile trip in one hour. However, the return trip took his two hours. What was his average trip for the round trip?

How To Find The Average Speed Of An Airplane With Good And Bad Weather?

Example: Mae took a non-stop flight to visit her grandmother. The 750-mile trip took three hours and 45 minutes. Because of the bad weather, the return trip took four hours and 45 minutes. What was her average speed for the round trip?

How To Relate Speed To Distance And Time?

If you are traveling in a car that travels 80km along a road in one hour, we say that you are traveling at an average of 80kn/h.

Average speed is the total distance divided by the total time for the trip. Therefore, speed is distance divided by time.

Instantaneous speed is the speed at which an object is traveling at any particular instant.

If the instantaneous speed of a car remains the same over a period of time, then we say that the car is traveling with constant speed.

The average speed of an object is the same as its instantaneous speed if that object is traveling at a constant speed.

How To Calculate Average Speed In Word Problems?

Example: Keri rollerblades to school, a total distance of 4.5km. She has to slow down twice to cross busy streets, but overall the journey takes her 0.65h. What is Keri’s average speed during the trip?

How To Use Average Speed To Calculate The Distance Traveled?

Example: Elle drives 169 miles from Sheffield to London. Her average speed is 65 mph. She leaves Sheffield at 6:30 a.m. Does she arrive in London by 9:00 a.m.?

How To Use Average Speed To Calculate The Time Taken?

Example: Marie Ann is trying to predict the time required to ride her bike to the nearby beach. She knows that the distance is 45 km and, from other trips, that she can usually average about 20 km/h. Predict how long the trip will take.

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Aptitude - Time and Distance

Why should i learn to solve aptitude questions and answers section on "time and distance".

Learn and practise solving Aptitude questions and answers section on "Time and Distance" to enhance your skills so that you can clear interviews, competitive examinations, and various entrance tests (CAT, GATE, GRE, MAT, bank exams, railway exams, etc.) with full confidence.

Where can I get the Aptitude questions and answers section on "Time and Distance"?

IndiaBIX provides you with numerous Aptitude questions and answers based on "Time and Distance" along with fully solved examples and detailed explanations that will be easy to understand.

Where can I get the Aptitude section on "Time and Distance" MCQ-type interview questions and answers (objective type, multiple choice)?

Here you can find multiple-choice Aptitude questions and answers based on "Time and Distance" for your placement interviews and competitive exams. Objective-type and true-or-false-type questions are given too.

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You can download the Aptitude quiz questions and answers section on "Time and Distance" as PDF files or eBooks.

How do I solve Aptitude quiz problems based on "Time and Distance"?

You can easily solve Aptitude quiz problems based on "Time and Distance" by practising the given exercises, including shortcuts and tricks.

  • Time and Distance - Formulas
  • Time and Distance - General Questions
  • Time and Distance - Data Sufficiency 1

   = 2 m/sec.

Converting m/sec to km/hr (see important formulas section)

   = 7.2 km/hr.

Distance = (240 x 5) = 1200 km.

Speed = Distance/Time

Let the actual distance travelled be x km.

Let speed of the car be x kmph.

Due to stoppages, it covers 9 km less.

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  5. Distance, speed and time word problem

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  6. How to Solve a Velocity, Distance, and Time Problem (Easy)

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VIDEO

  1. Time , Speed and Distance

  2. Speed-Time-and-Distance Part-1

  3. SPEED, TIME & DISTANCE MATHS CLASSES

  4. Speed, Distance Time Formula Trick

  5. SPEED TIME & DISTANCE L2 👉important method #maths #youtubeshorts #mathsbyvishwambarsir #youtube

  6. SPEED TIME & DISTANCE L4 👉AVERAGE SPEED #maths #mathsbyvishwambarsir #youtubeshorts #youtube

COMMENTS

  1. 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.

  2. TIME SPEED AND DISTANCE PROBLEMS

    Solution : Given : Speed is 60 miles per hour. The distance covered in 1 hour is = 60 miles Then, the distance covered in 4.5 hours is = 4.5 ⋅ 60 miles = 270 miles So, the person will travel 270 miles distance in 4.5 hours. Problem 4 : A person travels at a speed of 60 kms per hour. Then how many meters can he travel in 5 minutes? Solution :

  3. Solving Problems With a Distance-Rate-Time Formula

    Solving Problems Involving Distance, Rate, and Time Paul Taylor/Getty Images By Deb Russell Updated on July 12, 2019 In math, distance, rate, and time are three important concepts you can use to solve many problems if you know the formula. Distance is the length of space traveled by a moving object or the length measured between two points.

  4. Speed, time, and distance problems worksheets

    There are SEVEN different types of word problems to choose from, ranging from easy to advanced, so you can create a great variety of 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.

  5. Speed Distance Time

    i.e. speed = distance ÷ time S = D/T time = distance ÷ speed T = D/S distance = speed x time D = ST Time problem

  6. Speed Distance Time Calculator

    To solve for distance use the formula for distance d = st, or distance equals speed times time. distance = speed x time Rate and speed are similar since they both represent some distance per unit time like miles per hour or kilometers per hour. If rate r is the same as speed s, r = s = d/t.

  7. Distance, Time and Speed Word Problems

    Problems involving Time, Distance and Speed are solved based on one simple formula. Distance = Speed * Time Which implies → Speed = Distance / Time and Time = Distance / Speed Let us take a look at some simple examples of distance, time and speed problems. Example 1. A boy walks at a speed of 4 kmph.

  8. 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.

  9. Algebra Topics: Distance Word Problems

    Solving distance problems When you solve any distance problem, you'll have to do what we just did—use the formula to find distance, rate, or time. Let's try another simple problem. On his day off, Lee took a trip to the zoo. He drove an average speed of 65 mph, and it took him two-and-a-half hours to get from his house to the zoo.

  10. How to Solve Speed Distance and Time Questions

    How to Solve Speed Distance and Time Questions Quickly:- Before solving the question, we need to know about the Formulas of Speed Time and Distance. Basic Formulas Distance (D1) = Speed × Time Time = Distance/Speed Speed = Distance / Time To Convert Km/h to M/s =Km/h * 5/18 = Values in m/s or m/sec *18/5 = Values in km/h

  11. Problems on Calculating Speed

    Solution: Distance covered = 20 km Time taken = 4 hours We know, speed = distance/time = 20/4 km/hr Therefore, speed = 5 km/hr 2. A car covers a distance of 450 m in 1 minute whereas a train covers 69 km in 45 minutes. Find the ratio of their speeds. Solution: Speed of car = Distance covered/Time taken = 450/60 m/sec = 15/2 = 15/2 × 18/5 km/hr

  12. Speed, Distance, Time Textbook Exercise

    The Corbettmaths Textbook Exercise on Speed, Distance, Time. Previous: Simultaneous Equations: Graphical Textbook Exercise

  13. The Speed, Distance and Time trick [No Ads]

    Xcelerate Math resources https://xceleratemath.com/number/speedTime stamps⏰00:00 Introduction00:20 DST triangle01:19 Question 1: Find the distance (fast car)...

  14. Chapter 6.8: 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. 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.

  15. How To Calculate Speed, Distance And Time Using A Triangle

    To calculate time, divide the distance by speed. To get the distance, multiply the speed by time. You may see these equations simplified as s=d/t, where s is speed, d is distance, and t is time. This formula can be arranged into the triangle above. In the triangle, speed and time form the base, as they are what is multiplied together to work ...

  16. 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?

  17. Speed, Time and Distance

    Understanding the relationship between speed, time, and distance is essential to solve problems. Speed, Time, and Distance Speed = Distance/Time The speed of an object describes how fast or slow it moves and is calculated as distance divided by time. Speed is directly proportional to distance and inversely proportional to time.

  18. Calculating Speed, Distance and Time and Solving Problems ...

    13 1.1K views 1 year ago Educational Videos 2 In this lesson you will learn how to calculate speed, distance, and time, and to solve problems involving average rate and speed with the use...

  19. Problem on Time Speed and Distance

    Problem on Time Speed and Distance Read Courses Question 1: A racing car covers a certain distance at a speed of 320 km/hr in 5 hours. To cover the same distance in 5/3 hours it must travel at a speed of: Solution: Given Distance is constant. So, Speed is inversely proportional to time.

  20. Problems on Speed, Distance & Time: Formulas, Unit

    Speed = distance time If the distance is in kilometers and the time is in hours, the speed is in kilometers per hour. The speed is m / s e c if the distance is measured in metres and the time is measured in seconds. D i s t a n c e = s p e e d × t i m e

  21. Solving Word Problems SPEED, DISTANCE and TIME

    Solving Word Problems SPEED, DISTANCE and TIME | LET and Civil Service Exam Reviewer #speeddistancetime #letreviewer #civilserviceexam #mathteachergon

  22. Average Speed Problems (video lessons, examples and solutions)

    Solution: Step 1: The formula for distance is. Distance = Rate × Time. Total distance = 50 × 3 + 60 × 2 = 270. Step 2: Total time = 3 + 2 = 5. Step 3: Using the formula: Answer: The average speed is 54 miles per hour. Be careful! You will get the wrong answer if you add the two speeds and divide the answer by two.

  23. Time and Distance

    Time and Distance - Data Sufficiency 1 1. A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour? 3.6 7.2 8.4 10 2. An aeroplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in 1 hours, it must travel at a speed of: 300 kmph 360 kmph 600 kmph 720 kmph 3.