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Chapter 8: Rational Expressions
8.8 Rate Word Problems: Speed, Distance and Time
Distance, rate and time problems are a standard application of linear equations. When solving these problems, use the relationship rate (speed or velocity) times time equals distance .
[latex]r\cdot t=d[/latex]
For example, suppose a person were to travel 30 km/h for 4 h. To find the total distance, multiply rate times time or (30km/h)(4h) = 120 km.
The problems to be solved here will have a few more steps than described above. So to keep the information in the problem organized, use a table. An example of the basic structure of the table is below:
The third column, distance, will always be filled in by multiplying the rate and time columns together. If given a total distance of both persons or trips, put this information in the distance column. Now use this table to set up and solve the following examples.
Example 8.8.1
Joey and Natasha start from the same point and walk in opposite directions. Joey walks 2 km/h faster than Natasha. After 3 hours, they are 30 kilometres apart. How fast did each walk?
The distance travelled by both is 30 km. Therefore, the equation to be solved is:
[latex]\begin{array}{rrrrrrl} 3r&+&3(r&+&2)&=&30 \\ 3r&+&3r&+&6&=&30 \\ &&&-&6&&-6 \\ \hline &&&&\dfrac{6r}{6}&=&\dfrac{24}{6} \\ \\ &&&&r&=&4 \text{ km/h} \end{array}[/latex]
This means that Natasha walks at 4 km/h and Joey walks at 6 km/h.
Example 8.8.2
Nick and Chloe left their campsite by canoe and paddled downstream at an average speed of 12 km/h. They turned around and paddled back upstream at an average rate of 4 km/h. The total trip took 1 hour. After how much time did the campers turn around downstream?
The distance travelled downstream is the same distance that they travelled upstream. Therefore, the equation to be solved is:
[latex]\begin{array}{rrlll} 12(t)&=&4(1&-&t) \\ 12t&=&4&-&4t \\ +4t&&&+&4t \\ \hline \dfrac{16t}{16}&=&\dfrac{4}{16}&& \\ \\ t&=&0.25&& \end{array}[/latex]
This means the campers paddled downstream for 0.25 h and spent 0.75 h paddling back.
Example 8.8.3
Terry leaves his house riding a bike at 20 km/h. Sally leaves 6 h later on a scooter to catch up with him travelling at 80 km/h. How long will it take her to catch up with him?
The distance travelled by both is the same. Therefore, the equation to be solved is:
[latex]\begin{array}{rrrrr} 20(t)&=&80(t&-&6) \\ 20t&=&80t&-&480 \\ -80t&&-80t&& \\ \hline \dfrac{-60t}{-60}&=&\dfrac{-480}{-60}&& \\ \\ t&=&8&& \end{array}[/latex]
This means that Terry travels for 8 h and Sally only needs 2 h to catch up to him.
Example 8.8.4
On a 130-kilometre trip, a car travelled at an average speed of 55 km/h and then reduced its speed to 40 km/h for the remainder of the trip. The trip took 2.5 h. For how long did the car travel 40 km/h?
[latex]\begin{array}{rrrrrrr} 55(t)&+&40(2.5&-&t)&=&130 \\ 55t&+&100&-&40t&=&130 \\ &-&100&&&&-100 \\ \hline &&&&\dfrac{15t}{15}&=&\dfrac{30}{15} \\ \\ &&&&t&=&2 \end{array}[/latex]
This means that the time spent travelling at 40 km/h was 0.5 h.
Distance, time and rate problems have a few variations that mix the unknowns between distance, rate and time. They generally involve solving a problem that uses the combined distance travelled to equal some distance or a problem in which the distances travelled by both parties is the same. These distance, rate and time problems will be revisited later on in this textbook where quadratic solutions are required to solve them.
For Questions 1 to 8, find the equations needed to solve the problems. Do not solve.
- A is 60 kilometres from B. An automobile at A starts for B at the rate of 20 km/h at the same time that an automobile at B starts for A at the rate of 25 km/h. How long will it be before the automobiles meet?
- Two automobiles are 276 kilometres apart and start to travel toward each other at the same time. They travel at rates differing by 5 km/h. If they meet after 6 h, find the rate of each.
- Two trains starting at the same station head in opposite directions. They travel at the rates of 25 and 40 km/h, respectively. If they start at the same time, how soon will they be 195 kilometres apart?
- Two bike messengers, Jerry and Susan, ride in opposite directions. If Jerry rides at the rate of 20 km/h, at what rate must Susan ride if they are 150 kilometres apart in 5 hours?
- A passenger and a freight train start toward each other at the same time from two points 300 kilometres apart. If the rate of the passenger train exceeds the rate of the freight train by 15 km/h, and they meet after 4 hours, what must the rate of each be?
- Two automobiles started travelling in opposite directions at the same time from the same point. Their rates were 25 and 35 km/h, respectively. After how many hours were they 180 kilometres apart?
- A man having ten hours at his disposal made an excursion by bike, riding out at the rate of 10 km/h and returning on foot at the rate of 3 km/h. Find the distance he rode.
- A man walks at the rate of 4 km/h. How far can he walk into the country and ride back on a trolley that travels at the rate of 20 km/h, if he must be back home 3 hours from the time he started?
Solve Questions 9 to 22.
- A boy rides away from home in an automobile at the rate of 28 km/h and walks back at the rate of 4 km/h. The round trip requires 2 hours. How far does he ride?
- A motorboat leaves a harbour and travels at an average speed of 15 km/h toward an island. The average speed on the return trip was 10 km/h. How far was the island from the harbour if the trip took a total of 5 hours?
- A family drove to a resort at an average speed of 30 km/h and later returned over the same road at an average speed of 50 km/h. Find the distance to the resort if the total driving time was 8 hours.
- As part of his flight training, a student pilot was required to fly to an airport and then return. The average speed to the airport was 90 km/h, and the average speed returning was 120 km/h. Find the distance between the two airports if the total flying time was 7 hours.
- Sam starts travelling at 4 km/h from a campsite 2 hours ahead of Sue, who travels 6 km/h in the same direction. How many hours will it take for Sue to catch up to Sam?
- A man travels 5 km/h. After travelling for 6 hours, another man starts at the same place as the first man did, following at the rate of 8 km/h. When will the second man overtake the first?
- A motorboat leaves a harbour and travels at an average speed of 8 km/h toward a small island. Two hours later, a cabin cruiser leaves the same harbour and travels at an average speed of 16 km/h toward the same island. How many hours after the cabin cruiser leaves will it be alongside the motorboat?
- A long distance runner started on a course, running at an average speed of 6 km/h. One hour later, a second runner began the same course at an average speed of 8 km/h. How long after the second runner started will they overtake the first runner?
- Two men are travelling in opposite directions at the rate of 20 and 30 km/h at the same time and from the same place. In how many hours will they be 300 kilometres apart?
- Two trains start at the same time from the same place and travel in opposite directions. If the rate of one is 6 km/h more than the rate of the other and they are 168 kilometres apart at the end of 4 hours, what is the rate of each?
- Two cyclists start from the same point and ride in opposite directions. One cyclist rides twice as fast as the other. In three hours, they are 72 kilometres apart. Find the rate of each cyclist.
- Two small planes start from the same point and fly in opposite directions. The first plane is flying 25 km/h slower than the second plane. In two hours, the planes are 430 kilometres apart. Find the rate of each plane.
- On a 130-kilometre trip, a car travelled at an average speed of 55 km/h and then reduced its speed to 40 km/h for the remainder of the trip. The trip took a total of 2.5 hours. For how long did the car travel at 40 km/h?
- Running at an average rate of 8 m/s, a sprinter ran to the end of a track and then jogged back to the starting point at an average of 3 m/s. The sprinter took 55 s to run to the end of the track and jog back. Find the length of the track.
Answer Key 8.8
Intermediate Algebra by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.
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Solving Problems Involving Distance, Rate, and Time
- Pre Algebra & Algebra
- Math Tutorials
- Exponential Decay
- Worksheets By Grade
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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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)
Rearranging the subject of a formula
Substitution into formulae
This topic is relevant for:
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.
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.
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 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 .
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.
- 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.
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.
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
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.
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.
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
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
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
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
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.
- 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.
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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Speed, Distance, and Time
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Classical mechanics.
Hardcore training for the aspiring physicist.
A common set of physics problems ask students to determine either the speed, distance, or travel time of something given the other two variables. These problems are interesting since they describe very basic situations that occur regularly for many people. For example, a problem might say: "Find the distance a car has traveled in fifteen minutes if it travels at a constant speed of \(75 \text {km/hr}\)." Often in these problems, we work with an average velocity or speed, which simplifies the laws of motion used to calculate the desired quantity. Let's see how that works.
Application and Extensions
As long as the speed is constant or average, the relationship between speed , distance , and time is expressed in this equation
\[\mbox{Speed} = \frac{\mbox{Distance}}{\mbox{Time}},\]
which can also be rearranged as
\[\mbox{Time} = \frac{\mbox{Distance}}{\mbox{Speed}}\]
\[\mbox{Distance} = \mbox{Speed} \times \mbox{Time}.\]
Speed, distance, and time problems ask to solve for one of the three variables given certain information. In these problems, objects are moving at either constant speeds or average speeds.
Most problems will give values for two variables and ask for the third.
Bernie boards a train at 1:00 PM and gets off at 5:00 PM. During this trip, the train traveled 360 kilometers. What was the train's average speed in kilometers per hour? In this problem, the total time is 4 hours and the total distance is \(360\text{ km},\) which we can plug into the equation: \[\mbox{Speed} = \frac{\mbox{Distance}}{\mbox{Time}}= \frac{360~\mbox{km}}{4~\mbox{h}} = 90~\mbox{km/h}. \ _\square \]
When working with these problems, always pay attention to the units for speed, distance, and time. Converting units may be necessary to obtaining a correct answer.
A horse is trotting along at a constant speed of 8 miles per hour. How many miles will it travel in 45 minutes? The equation for calculating distance is \[\mbox{Distance} = \mbox{Speed} \times \mbox{Time},\] but we won't arrive at the correct answer if we just multiply 8 and 45 together, as the answer would be in units of \(\mbox{miles} \times \mbox{minute} / \mbox{hour}\). To fix this, we incorporate a unit conversion: \[\mbox{Distance} = \frac{8~\mbox{miles}}{~\mbox{hour}} \times 45~\mbox{minutes} \times \frac{1~\mbox{hour}}{60~\mbox{minutes}} = 6~\mbox{miles}. \ _\square \] Alternatively, we can convert the speed to units of miles per minute and calculate for distance: \[\mbox{Distance} = \frac{2}{15}~\frac{\mbox{miles}}{\mbox{minute}} \times 45~\mbox{minutes} = 6~\mbox{miles},\] or we can convert time to units of hours before calculating: \[\mbox{Distance} = 8~\frac{\mbox{miles}}{\mbox{hour}} \times \frac{3}{4}~\mbox{hours} = 6~\mbox{miles}.\] Any of these methods will give the correct units and answer. \(_\square\)
In more involved problems, it is convenient to use variables such as \(v\), \(d\), and \(t\) for speed, distance, and velocity, respectively.
Alice, Bob, Carly, and Dave are in a flying race!
Alice's plane is twice as fast as Bob's plane. When Alice finishes the race, the distance between her and Carly is \(D.\) When Bob finishes the race, the distance between him and Dave is \(D.\)
If Bob's plane is three times as fast as Carly's plane, then how many times faster is Alice's plane than Dave's plane?
Albert and Danny are running in a long-distance race. Albert runs at 6 miles per hour while Danny runs at 5 miles per hour. You may assume they run at a constant speed throughout the race. When Danny reaches the 25 mile mark, Albert is exactly 40 minutes away from finishing. What is the race's distance in miles? \[\] Let's begin by calculating how long it takes for Danny to run 25 miles: \[\mbox{Time} = \frac{\mbox{Distance}}{\mbox{Speed}}= \frac{25~\mbox{miles}}{5~\mbox{miles/hour}}= 5~\mbox{hours}.\] So, it will take Albert \(5~\mbox{hours} + 40~\mbox{minutes}\), or \(\frac{17}{3}~\mbox{hours}\), to finish the race. Now we can calculate the race's distance: \[\begin{align} \mbox{Distance} &= \mbox{Speed} \times \mbox{Time} \\ &= (6~\mbox{miles/hour}) \times \left(\frac{17}{3}~\mbox{hours}\right) \\ &= 34~\mbox{miles}.\ _\square \end{align}\]
A cheetah spots a gazelle \(300\text{ m}\) away and sprints towards it at \(100\text{ km/h}.\) At the same time, the gazelle runs away from the cheetah at \(80\text{ km/h}.\) How many seconds does it take for the cheetah to catch the gazelle? \[\] Let's set up equations representing the distance the cheetah travels and the distance the gazelle travels. If we set distance \(d\) equal to \(0\) as the cheetah's starting point, we have \[\begin{align} d_\text{cheetah} &= 100t \\ d_\text{gazelle} &= 0.3 + 80t. \end{align}\] Note that time \(t\) here is in units of hours, and \(300\text{ m}\) was converted to \(0.3\text{ km}.\) The cheetah catches the gazelle when \[\begin{align} d_\text{cheetah} &=d_\text{gazelle} \\ 100t &= 0.3 + 80t \\ 20t &= 0.3 \\ t &= 0.015~\mbox{hours}. \end{align}\] Converting that answer to seconds, we find that the cheetah catches the gazelle in \(54~\mbox{seconds}\). \(_\square\)
Two friends are crossing a hundred meter railroad bridge when they suddenly hear a train whistle. Desperate, each friend starts running, one towards the train and one away from the train. The one that ran towards the train gets to safety just before the train passes, and so does the one that ran in the same direction as the train.
If the train is five times faster than each friend, then what is the train-to-friends distance when the train whistled (in meters)?
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Module 9: Multi-Step Linear Equations
Using the distance, rate, and time formula, learning outcomes.
- Use the problem-solving method to solve problems using the distance, rate, and time formula
One formula you’ll use often in algebra and in everyday life is the formula for distance traveled by an object moving at a constant speed. The basic idea is probably already familiar to you. Do you know what distance you traveled if you drove at a steady rate of [latex]60[/latex] miles per hour for [latex]2[/latex] hours? (This might happen if you use your car’s cruise control while driving on the Interstate.) If you said [latex]120[/latex] miles, you already know how to use this formula!
The math to calculate the distance might look like this:
[latex]\begin{array}{}\\ \text{distance}=\left(\Large\frac{60\text{ miles}}{1\text{ hour}}\normalsize\right)\left(2\text{ hours}\right)\hfill \\ \text{distance}=120\text{ miles}\hfill \end{array}[/latex]
In general, the formula relating distance, rate, and time is
[latex]\text{distance}\text{=}\text{rate}\cdot \text{time}[/latex]
Distance, Rate, and Time
For an object moving at a uniform (constant) rate, the distance traveled, the elapsed time, and the rate are related by the formula
[latex]d=rt[/latex]
where [latex]d=[/latex] distance, [latex]r=[/latex] rate, and [latex]t=[/latex] time.
Notice that the units we used above for the rate were miles per hour, which we can write as a ratio [latex]\Large\frac{miles}{hour}[/latex]. Then when we multiplied by the time, in hours, the common units “hour” divided out. The answer was in miles.
Jamal rides his bike at a uniform rate of [latex]12[/latex] miles per hour for [latex]3\Large\frac{1}{2}[/latex] hours. How much distance has he traveled?
In the following video we provide another example of how to solve for distance given rate and time.
Rey is planning to drive from his house in San Diego to visit his grandmother in Sacramento, a distance of [latex]520[/latex] miles. If he can drive at a steady rate of [latex]65[/latex] miles per hour, how many hours will the trip take?
Show Solution
In the following video we show another example of how to find rate given distance and time.
- Question ID 145550, 145553,145619,145620. Authored by : Lumen Learning. License : CC BY: Attribution
- Ex: Find the Rate Given Distance and Time. Authored by : James Sousa (Mathispower4u.com). Located at : https://youtu.be/3rYh32ErDaE . License : CC BY: Attribution
- Example: Solve a Problem using Distance = Rate x Time. Authored by : James Sousa (Mathispower4u.com). Located at : https://youtu.be/lMO1L_CvH4Y . License : CC BY: Attribution
- Prealgebra. Provided by : OpenStax. License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]
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
Two Objects Move in Opposite Direction
Train Passes a Moving Object in the Same Direction
Train Passes a Moving Object in the Opposite Direction
Train Passes through a Pole
Train Passes through a Bridge
Two Trains Passes in the Same Direction
Two Trains Passes in the Opposite Direction
8th Grade Math Practice From Problems on Calculating Speed to HOME PAGE
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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.
SPEED DISTANCE TIME PROBLEMS WITH SOLUTIONS
Problem 1 :
A passenger train takes 3 hours less than a slow train for journey of 600 km. If the speed of the slow train is 10 km/hr less than that of the passenger train, find the speed of two trains.
Let x be the speed of the of the passenger train
Speed of the slow train is 10 km/hr less than that of the passenger train
So x-10 be the speed of the slow train
Distance has to be covered = 600 km
Time = Distance/speed
Let T 1 be the time taken by passenger train
Let T 2 be the time taken by the slow train
The differences of time taken by both trains are 3 hours
T 1 = 600/x
T 2 = 600/(x-0)
T 2 –T 1 = 3 hours
(600/(x–10)) – (600/x) = 3
600[(1/(x-10)-(1/x)] = 3
x-(x-10)/(x 2 -10x) = 1/200
(x-x+10)/(x 2 -10x) = 1/200
2000 = x 2 -10x
x 2 -10x-2000 = 0
x 2 –50x+40x-2000 = 0
x(x–50)+40(x–50) = 0
(x+40) (x–50) = 0
By solving, we get x = -40 and x = 50
Therefore the speed of passenger train = 50 km/hr
Speed of slow train = 40 m/hr.
Problem 2 :
The distance between two stations A and B is 192 km. Traveling by a fast train takes 48 minutes less than another train. Calculate the speed of the fast train if the speeds of two trains differ by 20 km/hr
Distance between two stations A and B = 192 km
Fast train takes 48 minutes less then the time taken by the slow train.
Let x be the speed taken by the fast train
Speeds of two trains differ by 20 km/hr
So speed of slow train is x – 20.
Let T 1 be the time taken by the fast train
T 1 = 192/x
T 2 = 192/(x–20)
48/60 = 4/5 hours
T 1 – T 2 = 4/5
[192/(x-20)-192/x] = 4/5
192[(x–x+20)/x(x - 20)] = 4/5
192(20)/x 2 –20 x = 4/5
3840 (5) = 4(x 2 –20 x)
19200 = 4x 2 – 80 x
4800 = x 2 – 20 x
x 2 –20x–4800 = 0
x 2 –60x+40x-4800 = 0
(x–60) (x+40) = 0
x = 60 and x = -40
Speed of fast train is 60 km/hr.
Problem 3 :
A train covers a distance of 300 km at a certain average speed. If its speed was decreased by 10km/hr, the journey would take 1 hour longer. What is the average speed.
Let x be the average speed of the train
So x–10 be the decreased speed
Time = Distance/Speed
T 1 and T 2 be the time taken by the train to cover the distance with speed of x km/hr and (x-10) km/hr respectively.
T 1 = 300/x
T 2 = 300/(x–10)
T 1 – T 2 = 1
[300/x] - [300/(x-10)] = 1
3000/(x 2 – 10x) = 1
3000 = x 2 – 10 x
x 2 – 10x = 3000
x 2 –10x–3000 = 0
x 2 –60x+50x–3000 = 0
x (x – 60) + 50 (x – 60) = 0
(x + 50)(x – 60) = 0
By solving, we get
x = -50 and x = 60
So speed of the 60 km/hr.
Problem 4 :
The time taken by a train to travel a distance of 250 km was reduced by 5/4 hours when average speed was increased by 10km/hr. Calculate the average speed.
Distance to be covered = 250 km
Let x be the required average speed.
If the average speed was increased by 10 km/hr
x+10 be the increased speed
Let T 1 be the time taken to cover the distance in the average speed of x km/hr
Let T 2 be the time taken to cover the distance in the average speed of (x + 10) km/hr
T 1 = 250/x
T 2 = 250/(x+10)
T 1 – T 2 = 5/4
250/x – 250/(x + 10) = 5/4
250 [(x+10–x)/x(x+10)] = 5/4
2500/(x 2 + 10x) = 5/4
2500 (4) = 5(x 2 +10x)
10000 = 5x 2 +10 x
Now we are going to divide the whole equation by 5, so we get
x 2 +10x = 2000
x 2 +10x–2000 = 0
x 2 + 50x-40x-2000 = 0
x(x+50)–40(x+50) = 0
(x–40) (x+50) = 0
x = 40 and x = -50
Therefore the required average speed = 40 km/hr
Increased speed = (40+10)
Problem 5 :
An express train makes run 240 km t a certain speed. Another train whose speed is 12 km/hr less takes an hour longer to make the same trip. Find the speed of the express train.
Let x be speed of express train
So x–12 be the speed of another train
Distance to be covered = 240 km
Let T 1 be the time taken by the train to cover the distance 240 km at the speed of x km/hr
Let T 2 be the time taken by the train to cover the distance 240 km at the speed of (x + 12) km/hr
Time = Distance /speed
T 1 = 240/x
T 2 = 240/(x - 12)
T 2 - T 1 = 1 hour
[240/(x- 12)] - [240/x] = 1
240[(1/(x -12) - 1/x] = 1
240[(x - x + 12)/x(x - 12)] = 1
240[12/(x 2 - 12 x)] = 1
2880 = (x 2 -12 x)
x 2 -12x-2880 = 0
x 2 +60x-48x-2880 = 0
x(x+60)- 48 (x+60) = 0
(x-48) (x+60) = 0
x = 48 x = -60
Speed of express train = 48 km/hr
Speed of other train = (x - 12)
= (48-12)
= 36 km/hr
Problem 6 :
A plane traveled a distance of 400 km at an average speed of x km/hr. On the return journey, the speed was increased by 40 km/hr. Write down an expression for the time taken for
(i) the onward journey and
(ii) the return journey. If the journey took 30 minutes less than onward journey, write down an equation in x and its value.
Let “x” be average speed of plane
On the return journey, the speed was increased by 40 km/hr
So “x + 40” be the speed of plane
Distance to be covered = 400 km
Let T 1 be the time taken for onward journey in the speed of x km/hr
Let T 2 be the time taken for downward journey to cover the same distance 400 km at the speed of (x + 40) km/hr
T 1 = 400/x
T 2 = 400/(x+40)
T 1 - T 2 = 30 minutes
[400/x]-[400/(x + 40)] = 30/60
400[(1/x) - 1/(x+40)] = 1/2
400[40/(x 2 + 40 x)] = 1/2
16000 (2) = (x 2 +40 x)
x 2 +40x-32000 = 0
x 2 +160x-100x-32000 = 0
(x - 100) (x + 160) = 0
x = 100 x = -160
Speed of the plane = 48 km/hr
Increased speed = (x+40)
= (48+40)
= 88 km/hr
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Rate, Time Distance Problems With Solutions
The relationship between distance, rate (speed) and time
Distance Word Problems
In these lessons, we will learn to solve word problems involving distance, rate (speed) and time.
Related Pages Rate Distance Time Word Problems Distance Problems Average Speed Problems
What Are Distance Word Problems Or Distance Rate Time Problems?
Distance problems are word problems that involve the distance an object will travel at a certain average rate for a given period of time .
The formula for distance problems is: distance = rate × time or d = r × t
Things to watch out for: Make sure that you change the units when necessary. For example, if the rate is given in miles per hour and the time is given in minutes then change the units appropriately.
It would be helpful to use a table to organize the information for distance problems. A table helps you to think about one number at a time instead being confused by the question.
The following diagrams give the steps to solve Distance-Rate-Time Problems. Scroll down the page for examples and solutions.
Distance Problems: Traveling At Different Rates
Example: A bus traveling at an average rate of 50 kilometers per hour made the trip to town in 6 hours. If it had traveled at 45 kilometers per hour, how many more minutes would it have taken to make the trip?
Solution: Step 1: Set up a rtd table.
Step 2: Fill in the table with information given in the question.
A bus traveling at an average rate of 50 kilometers per hour made the trip to town in 6 hours. If it had traveled at 45 kilometers per hour, how many more minutes would it have taken to make the trip?
Let t = time to make the trip in Case 2.
Step 3: Fill in the values for d using the formula d = rt
Step 4: Since the distances traveled in both cases are the same, we get the equation:
Step 5: Beware - the question asked for “how many more minutes would it have taken to make the trip”, so we need to deduct the original 6 hours taken.
Answer: The time taken would have been 40 minutes longer.
How to solve different types of distance, rate, time problems?
The following examples illustrate three types of problems involving distance, rate and time: opposite directions, same direction and roundtrip. Example 1: Lea left home and drove toward the ferry office at an average speed of 22 km.h. Trevon left at the same time and drove in the opposite direction with an average speed of 43 km/h. How long does Trevon need to drive before they are 65 km apart?
Example 2: Maria left home and traveled toward the capital at an average speed of 80 km/h. Some time later, Imani left traveling in the opposite direction with an average speed of 45 km/h. After Maria has traveled for six hours they were 705 km apart. Find the number of hours Imani traveled.
Example 3: A passenger train traveled to the repair yards and back. On the trip there, it traveled 68 mph and on the return trip it went 85 mph. How long did the trip there take if the return trip took four hours?
Example 4: A submarine traveled to St. Vincent and back. It took three hours less time to get there than it did to get back. The average speed on the trip there was 30 km/h. The average speed on the way back was 20 km/h. How many hours did the trip there take?
Example 5: Micaela left the hardware store and traveled toward her friend’s house at an average speed of 25 km/h. Nicole left some time later traveling in the same direction at the average speed of 30 km/h. After traveling for five hours Nicole caught up with Micaela. How long did Micaela travel before Nicole caught up?
Example 6: A passenger plane left New York and flew east. A jet left one hour later flying at 280 km/h in a effort to catch up to the passenger plane. After flying for seven hours the jet finally caught up. Find the passenger plane’s average speed.
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- Speed Distance Time Questions for Class 5
An Introduction to Speed, Distance, and Time
Riya used to come home late after school, cycling at a slower speed, while her friend Diya came home early as she was cycling faster than Riya. Speed refers to how fast something is moving or capable of moving.
In mathematics, speed is measured as the distance travelled/travelled per unit of time . The distance travelled is in kilometres, and the time is in hours, so the unit of measurement for speed is "km" per "hour" or km/h.
Speed in Speedometer
Table of Units in Common Use for Speed
Here, we will see the different terms regarding speed, distance, and time. Also, the abbreviation has been added to make it more readable.
Units used for Speed and their Abbreviations
Solved Example Problems on Speed Distance and Time
Let’s see the speed, distance, and time formula and some solved questions based on it.
Type 1: Formula for Speed = $\dfrac{\text{Distance (in unit of length)}}{\text{Time (in unit of time)}}$
Example 1: A car covers a distance of $224 \mathrm{~km}$ between Goa and Mumbai in 4 hours. Find its average speed.
Ans: Average Speed $=$ $\dfrac{\text{Total distance covered}}{\text{Total time taken }}$
$\dfrac{224\,Km}{4\,hr}=56\,Km/hr$
Therefore, its average speed is 56 km/hr.
Example 2: The train's speed is $108 \mathrm{~km} / \mathrm{h}$. Find its speed in metres per second.
$ 1 \mathrm{~km}=1000m$
$\text { Speed }=108\left(\frac{\mathrm{Km}}{\mathrm{Hr}}\right)$
Also, we know that 1 minute $=60$ seconds
Therefore, that's 1 hour. $=60 \times 60=3600$ seconds
By using the above stats, we get
$=108\left(\dfrac{1000 \mathrm{~m}}{3600 \mathrm{~s}}\right)$
$=30\left(\dfrac{\mathrm{m}}{\mathrm{s}}\right)$
Therefore, its speed is $=30\left(\dfrac{\mathrm{m}}{\mathrm{s}}\right)$.
Example 3: Latvija covers the first 120 km in 2 hours and the next 180 km in 4 hours. What is his average speed over the entire journey in km/h?
Ans: Total distance travelled = 120 + 180 = 300 km.
Total time taken = 2 + 4 = 6 hours.
Average speed = $\dfrac{\text{Total distance travelled}}{\text{Total time spent}}$
$=\dfrac{300}{6}=50 \mathrm{~km} / \mathrm{h}$
Type 2: Speed and Distance Formula
Average speed = Total distance travelled / Total time spent
Solved Examples based on Distance and Time:
Example 1: Using the average speed formula, find the average speed of Sam, who covers the first 200 kilometres in 4 hours and the next 160 kilometres in another 4 hours.
Ans: To find the average speed, we need the total distance and the total time.
Total distance covered by Sam = 200 Km + 160 km = 360 km
Total time taken by Sam = 4 hour + 4 hour = 8 hour
Average Speed = $\dfrac{\text{Total distance covered}} {\text{Total time taken}}$
Average Speed = $\dfrac{360}{8}$
Therefore, their average speed of Sam is 45lm/hr.
Example 2: A cyclist rides at $10 \mathrm{~km} / \mathrm{h}$ for 2 hours then at $13 \mathrm{~km} / \mathrm{h}$ for 1 hour. Find its average speed.
Distance travelled in the first 2 hours $=10 \times 2=20 \mathrm{~km}$
Distance travelled in the next hour $=13 \times 1=13 \mathrm{~km}$
Total distance travelled $=20+13=33 \mathrm{~km}$
Total time spent $=2+1=3 h r s$
Average speed $=$ $\dfrac{\text{Total distance travelled}}{\text{Total time spent }}$
$=\dfrac{33}{3}=11 \mathrm{~km} / \mathrm{hr}$
Type 3. Average Speed - If the Travel Time is the Same.
Speed Distance Time problems : In this type of problem, the travel time is always the same.
Example 1: A motorist drives for one hour at an average speed of 45 km/h, and for the next hour at an average speed of 65 km/h. What is your average speed?
Ans: $\dfrac{(45 + 65)}{2}=55 \mathrm{~km} / \mathrm{hr}$
The total distance covered by the car driver in these two hours was $65+45=110 \mathrm{~km}$, and it took two hours.
Therefore your average speed $=55 \mathrm{~km} / \mathrm{h}$.
Example 2: Radha can type 960 letters in 20 mins.
Calculate her typing speed in the:
(a) Words per min.
(b) Words per hour.
Words per min. $=\dfrac{960}{20}$
$=48$ words per minute
Words per hour. $=48 \times 60$
$=2880$ words per hour.
Type 4. Average speed: If the distance covered is the same:
Speed Distance Problems: In this type of problem, the distance covered is always the same.
Example: On the way from the hospital to the Pimpri department, I drive at 30 km/h and back at 45 km/h. What is my average travel speed?
Ans: 37.5 km/h is inappropriate as the time travel is different in both cases, and only the distances are the same. Let the distance between the hospital and the Pimpri ward be x km.
The time spent on the outward journey $=\dfrac{x}{30}$ hour and the time spent on the return journey $=\dfrac{x}{45}$
$\therefore$ Total round trip time $=\dfrac{x}{30}+\dfrac{x}{45}$ $=\dfrac{5 x}{90}$ hours.
Total distance travelled in both directions $=2 \times x \mathrm{km}$
$\therefore$ Average speed $\dfrac{2x}{\dfrac{5x}{90}}=36 \mathrm{kmph}$
Practice Questions
Let’s see some of the speed distance time questions for Class 5:
Q 1 An aeroplane covers a certain distance in 5 hours at a speed of 240 km/h. To travel the same distance in 1 hour, at what speed must he travel:
Ans: 720 km/hr.
Q 2. How many minutes for the bus per hour? Excluding stops, the speed of a bus is 54 km/h, and including stops, it is 45 km/h.
Ans: 10 min.
Q 3. Determine the time it takes when the distance is 7150 km, and the speed is 780 km/h.
Ans: 9.16 hr
Speed tells us how fast something or someone is travelling. You can find the average speed of an object if you know the distance travelled and the time it took. The formula for speed is speed = distance ÷ time. To work out what the units are for speed, you need to know the units for distance and time. In mathematics, speed is measured as the distance travelled per unit of time.
The distance travelled is in kilometres, and the time is in hours, so the unit of measurement for speed is "km" per "hour" or km/h. If the distance is in metres and the time is in minutes, then the speed is 'm' for 'minute' or m/min. You can also read this offline. Just download the speed distance time questions for Class 5 PDF from our website.
FAQs on Speed Distance Time Questions for Class 5
Q1. How do we use speed in everyday life?
Ans: The vehicle's speed. How quickly the train is moving. The river is moving at a fluctuating speed. The rate at which water leaves a tap.
Q2. Why are speed, distance, and time important?
Ans: Speed, distance and time refer to the three important variables in physics and maths. Furthermore, these three variables certainly facilitate the solving of several types of problems in Maths.
Q3. Who discovered the formula for speed, namely distance/time?
Ans: Italian physicist Galileo Galilei is credited with being the first to measure speed by taking into account the distance travelled and the time it takes. Galileo defined speed as the distance travelled per unit of time.
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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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Heavy Machinery Meets AI
- Vijay Govindarajan
- Venkat Venkatraman
Until recently most incumbent industrial companies didn’t use highly advanced software in their products. But now the sector’s leaders have begun applying generative AI and machine learning to all kinds of data—including text, 3D images, video, and sound—to create complex, innovative designs and solve customer problems with unprecedented speed.
Success involves much more than installing computers in products, however. It requires fusion strategies, which join what manufacturers do best—creating physical products—with what digital firms do best: mining giant data sets for critical insights. There are four kinds of fusion strategies: Fusion products, like smart glass, are designed from scratch to collect and leverage information on product use in real time. Fusion services, like Rolls-Royce’s service for increasing the fuel efficiency of aircraft, deliver immediate customized recommendations from AI. Fusion systems, like Honeywell’s for building management, integrate machines from multiple suppliers in ways that enhance them all. And fusion solutions, such as Deere’s for increasing yields for farmers, combine products, services, and systems with partner companies’ innovations in ways that greatly improve customers’ performance.
Combining digital and analog machines will upend industrial companies.
Idea in Brief
The problem.
Until recently most incumbent industrial companies didn’t use the most advanced software in their products. But competitors that can extract complex designs, insights, and trends using generative AI have emerged to challenge them.
The Solution
Industrial companies must develop strategies that fuse what they do best—creating physical products—with what digital companies do best: using data and AI to parse enormous, interconnected data sets and develop innovative insights.
The Changes Required
Companies will have to reimagine analog products and services as digitally enabled offerings, learn to create new value from data generated by the combination of physical and digital assets, and partner with other companies to create ecosystems with an unwavering focus on helping customers solve problems.
For more than 187 years, Deere & Company has simplified farmwork. From the advent of the first self-scouring plow, in 1837, to the launch of its first fully self-driving tractor, in 2022, the company has built advanced industrial technology. The See & Spray is an excellent contemporary example. The automated weed killer features a self-propelled, 120-foot carbon-fiber boom lined with 36 cameras capable of scanning 2,100 square feet per second. Powered by 10 onboard vision-processing units handling almost four gigabytes of data per second, the system uses AI and deep learning to distinguish crops from weeds. Once a weed is identified, a command is sent to spray and kill it. The machine moves through a field at 12 miles per hour without stopping. Manual labor would be more expensive, more time-consuming, and less reliable than the See & Spray. By fusing computer hardware and software with industrial machinery, it has helped farmers decrease their use of herbicide by more than two-thirds and exponentially increase productivity.
- Vijay Govindarajan is the Coxe Distinguished Professor at Dartmouth College’s Tuck School of Business, an executive fellow at Harvard Business School, and faculty partner at the Silicon Valley incubator Mach 49. He is a New York Times and Wall Street Journal bestselling author. His latest book is Fusion Strategy: How Real-Time Data and AI Will Power the Industrial Future . His Harvard Business Review articles “ Engineering Reverse Innovations ” and “ Stop the Innovation Wars ” won McKinsey Awards for best article published in HBR. His HBR articles “ How GE Is Disrupting Itself ” and “ The CEO’s Role in Business Model Reinvention ” are HBR all-time top-50 bestsellers. Follow him on LinkedIn . vgovindarajan
- Venkat Venkatraman is the David J. McGrath Professor at Boston University’s Questrom School of Business, where he is a member of both the information systems and strategy and innovation departments. His current research focuses on how companies develop winning digital strategies. His latest book is Fusion Strategy: How Real-Time Data and AI Will Power the Industrial Future. Follow him on LinkedIn . NVenkatraman
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About iOS 17 Updates
iOS 17 brings big updates to Phone, Messages, and FaceTime that give you new ways to express yourself as you communicate. StandBy delivers a new full-screen experience with glanceable information designed to view from a distance when you turn iPhone on its side while charging. AirDrop makes it easier to share and connect with those around you and adds NameDrop for contact sharing. Enhancements to the keyboard make entering text faster and easier than ever before. iOS 17 also includes updates to Widgets, Safari, Music, AirPlay, and more.
For information on the security content of Apple software updates, please visit this website: https://support.apple.com/kb/HT201222
This update provides bug fixes for your iPhone including:
Text may unexpectedly duplicate or overlap while typing
This update introduces additional security measures with Stolen Device Protection. This release also includes a new Unity wallpaper to honor Black history and culture in celebration of Black History Month, as well as other features, bug fixes, and security updates for your iPhone.
Stolen Device Protection
Stolen Device Protection increases security of iPhone and Apple ID by requiring Face ID or Touch ID with no passcode fallback to perform certain actions
Security Delay requires Face ID or Touch ID, an hour wait, and then an additional successful biometric authentication before sensitive operations like changing device passcode or Apple ID password can be performed
Lock Screen
New Unity wallpaper honors Black history and culture in celebration of Black History Month
Collaborate on playlists allows you to invite friends to join your playlist and everyone can add, reorder, and remove songs
Emoji reactions can be added to any track in a collaborative playlist
This update also includes the following improvements:
AirPlay hotel support lets you stream content directly to the TV in your room in select hotels
AppleCare & Warranty in Settings shows your coverage for all devices signed in with your Apple ID
Crash detection optimizations (all iPhone 14 and iPhone 15 models)
Some features may not be available for all regions or on all Apple devices. For information on the security content of Apple software updates, please visit this website:
https://support.apple.com/kb/HT201222
This update provides important bug fixes and is recommended for all users.
For information on the security content of Apple software updates, please visit this website:
This update introduces Journal, an all-new way to reflect on life’s moments and preserve your memories. This release also includes Action button and Camera enhancements, as well as other features, bug fixes, and security updates for your iPhone.
Journal is a new app that lets you write about the small moments and big events in your life so you can practice gratitude and improve your wellbeing
Journaling suggestions make it easy to remember your experiences by intelligently grouping your outings, photos, workouts, and more into moments you can add to your journal
Filters let you quickly find bookmarked entries or show entries with attachments so you can revisit and reflect on key moments in your life
Scheduled notifications help you keep a consistent journaling practice by reminding you to write on the days and time you choose
Option to lock your journal using Touch ID or Face ID
iCloud sync keeps your journal entries safe and encrypted on iCloud
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Translate option for the Action button on iPhone 15 Pro and iPhone 15 Pro Max to quickly translate phrases or have a conversation with someone in another language
Spatial video lets you capture video on iPhone 15 Pro and iPhone 15 Pro Max so you can relive your memories in three dimensions on Apple Vision Pro
Improved Telephoto camera focusing speed when capturing small faraway objects on iPhone 15 Pro and iPhone 15 Pro Max
Catch-up arrow lets you easily jump to your first unread message in a conversation by tapping the arrow visible in the top-right corner
Add sticker option in the context menu lets you add a sticker directly to a bubble
Memoji updates include the ability to adjust the body shape of any Memoji
Contact Key Verification provides automatic alerts and Contact Verification Codes to help verify people facing extraordinary digital threats are messaging only with the people they intend
Precipitation amounts help you stay on top of rain and snow conditions for a given day over the next 10 days
New widgets let you choose from next-hour precipitation, daily forecast, sunrise and sunset times, and current conditions such as Air Quality, Feels Like, and wind speed
Wind map snapshot helps you quickly assess wind patterns and access the animated wind map overlay to prepare for forecasted wind conditions for the next 24 hours
Interactive moon calendar lets you easily visualize the phase of the moon on any day for the next month
This update also includes the following improvements and bug fixes:
Siri support for privately accessing and logging Health app data using your voice
AirDrop improvements including expanded contact sharing options and the ability to share boarding passes, movie tickets, and other eligible passes by bringing two iPhones together
Favorite Songs Playlist in Apple Music lets you quickly get back to the songs you mark as favorites
Use Listening History in Apple Music can be disabled in a Focus so music you listen to does not appear in Recently Played or influence your recommendations
A new Digital Clock Widget lets you quickly catch a glimpse of the time on your Home Screen and while in StandBy
Enhanced AutoFill identifies fields in PDFs and other forms enabling you to populate them with information such as names and addresses from your contacts
New keyboard layouts provide support for 8 Sámi languages
Sensitive Content Warning for stickers in Messages prevents you from being unexpectedly shown a sticker containing nudity
Qi2 charger support for all iPhone 13 models and iPhone 14 models
Fixes an issue that may prevent wireless charging in certain vehicles
This update provides important security fixes and is recommended for all users.
In rare circumstances, Apple Pay and other NFC features may become unavailable on iPhone 15 models after wireless charging in certain cars
Weather Lock Screen widget may not correctly display snow
This update introduces the ability for AirDrop transfers to continue over the internet when you step out of AirDrop range. This release also includes enhancements to StandBy and Apple Music, as well as other features, bug fixes, and security updates for your iPhone.
Content continues to transfer over the internet when you step out of AirDrop range
New options to control when the display turns off (iPhone 14 Pro, iPhone 14 Pro Max, iPhone 15 Pro, and iPhone 15 Pro Max)
Favorites expanded to include songs, albums, and playlists, and you can filter to display your favorites in the library
New cover art collection offers designs that change colors to reflect the music in your playlist
Song suggestions appear at the bottom of every playlist, making it easy to add music that matches the vibe of your playlist
Option to choose a specific album to use with Photo Shuffle on the Lock Screen
Home key support for Matter locks
Improved reliability of Screen Time settings syncing across devices
Fixes an issue that may cause the Significant Location privacy setting to reset when transferring an Apple Watch or pairing it for the first time
Resolves an issue where the names of incoming callers may not appear when you are on another call
Addresses an issue where custom and purchased ringtones may not appear as options for your text tone
Fixes an issue that may cause the keyboard to be less responsive
Fixes an issue that may cause display image persistence
https://support.apple.com/HT201222
This update provides important bug fixes, security updates, and addresses an issue that may cause iPhone to run warmer than expected.
This update provides important bug fixes, security updates, and fixes an issue that may prevent transferring data directly from another iPhone during setup.
This update provides important bug fixes and security updates and is recommended for all users.
Contact Posters let you customize how you appear on other people’s devices when you call them with a customized poster
Live Voicemail displays a live transcription as someone leaves a message and allows you to pick up the call
Stickers iMessage app brings all your stickers into one place including Live Stickers, Memoji, Animoji, emoji stickers, and your third party sticker packs
Live Stickers can be created by lifting the subject from photos or videos and stylizing them with effects like Shiny, Puffy, Comic, and Outline
Check In automatically notifies a family member or friend when you arrive at a destination safely and can share helpful information with them in case of a delay
Audio message transcription is available for audio messages you receive so you can read them in the moment and listen later
Search improvements help you find messages faster by allowing you to combine search filters such as people, keywords, and content types like photos or links to find exactly what you are looking for
Swipe to reply to a message inline by swiping to the right on any bubble
One-time verification code cleanup automatically deletes verification codes from the Messages app after using them with AutoFill in other apps
Leave a video or audio message to capture exactly what you want to say when someone does not pick up your FaceTime call
Enjoy FaceTime calls on Apple TV by using your iPhone as a camera (Apple TV 4K 2nd generation and later)
Reactions layer 3D effects like hearts, balloons, confetti, and more around you in video calls and can be triggered with gestures
Video effects allow you to adjust the intensity of Studio Lighting and Portrait mode
Full-screen experience with glanceable information like clocks, photos, and widgets designed to view from a distance when iPhone is on its side and charging in places such as your nightstand, kitchen counter, or desk
Clocks are available in a variety of styles including Digital, Analog, Solar, Float, and World Clock, with elements you can personalize like the accent color
Photos automatically shuffle through your best shots or showcase a specific album you choose
Widgets give you access to information at a distance and appear in Smart Stacks that deliver the right information at the right time
Night Mode lets clocks, photos, and widgets take on a red tone in low light
Preferred view per MagSafe charger remembers your preference for each place you charge with MagSafe, whether that’s a clock, photos, or widgets
Interactive widgets let you take actions, like mark a reminder as complete, directly from the widget by tapping it on the Home Screen, Lock Screen, or in StandBy
iPhone widgets on Mac enable you to add widgets from your iPhone to your Mac desktop
NameDrop lets you exchange contact information with someone new by bringing your iPhones close together
New way to initiate AirDrop allows you to share content or start a SharePlay session over AirDrop by bringing your iPhones close together
Improved autocorrect accuracy makes typing even easier by leveraging a powerful transformer-based language model (iPhone 12 and later)
Easier autocorrect editing temporarily underlines corrected words and lets you revert back to what you originally typed with just a tap
Enhanced sentence corrections can correct more types of grammatical mistakes when you finish sentences (iPhone 12 and later)
Inline predictive text shows single and multi-word predictions as you type that can be added by tapping space bar (iPhone 12 and later)
Safari and Passwords
Profiles keep your browsing separate for topics like work and personal, separating your history, cookies, extensions, Tab Groups, and favorites
Private Browsing enhancements include locking your private browsing windows when you’re not using them, blocking known trackers from loading, and removing identifying tracking from URLs
Password and passkey sharing lets you create a group of passwords to share with trusted contacts that stays up to date as members of the group make changes
One-time verification code AutoFill from Mail autofill in Safari so you can log in without leaving the browser
SharePlay makes it easy for everyone to control and play Apple Music in the car
Crossfade smoothly transitions between songs by fading out the currently playing song while fading in the next so the music never stops
Intelligent AirPlay device list makes finding the right AirPlay-compatible TV or speaker even easier by showing your devices in order of relevance, based on your preferences
Suggested AirPlay device connections are proactively shown to you as a notification to make it even more seamless to connect to your preferred AirPlay devices
Automatic AirPlay device connections are made between your iPhone and the most relevant AirPlay-compatible device so all you have to do is tap “Play” to begin enjoying your content
Adaptive Audio delivers a new listening mode that dynamically blends Active Noise Cancellation and Transparency to tailor the noise control experience based on the conditions of your environment (AirPods Pro (2nd generation) with firmware version 6A300 or later)
Personalized Volume adjusts the volume of your media in response to your environment and listening preferences over time (AirPods Pro (2nd generation) with firmware version 6A300 or later)
Conversation Awareness lowers your media volume and enhances the voices of the people in front of the user, all while reducing background noise (AirPods Pro (2nd generation) with firmware version 6A300 or later)
Press to mute and unmute your microphone by pressing the AirPods stem or the Digital Crown on AirPods Max when on a call (AirPods (3rd generation), AirPods Pro (1st and 2nd generation), or AirPods Max with firmware version 6A300 or later)
Offline Maps allow you to select an area you want to access, search, and explore rich information for places to download for use when your iPhone doesn’t have a Wi-Fi or cellular signal
EV routing improvements give you routes based on real-time EV charger availability for supported chargers
Option to say “Siri” in addition to “Hey Siri” for an even more natural way to make requests
Back-to-back requests can be issued without needing to reactivate Siri in between commands (iPhone 11 and later)
Visual Look Up
Expanded domains in Visual Look Up help you discover similar recipes from photos of food, Maps information from photos of storefronts, and the meaning of signs and symbols on things like laundry tags
Multiple or single subjects can be lifted from the background of photos and videos and placed into apps like Messages
Visual Look Up in Video helps you learn about objects that appear in paused video frames
Visual Look Up for subjects in photos enables you to look up information about objects you lift from photos directly from the callout bar
State of Mind reflection allows you to log your momentary emotion and daily mood, choose what factors are having the biggest impact on you, and describe your feelings
Interactive charts give you insights into your state of mind, how it has changed over time, and what factors may have influence such as exercise, sleep, and mindful minutes
Mental health assessments help you understand your current risk for depression and anxiety and if you might benefit from getting support
Screen Distance leverages the TrueDepth camera that powers Face ID to encourage you to increase the distance you view your device to reduce digital eye strain and can help reduce the risk of myopia in children
Sensitive Content Warnings can be enabled to prevent users from unexpectedly being shown images containing nudity in Messages, AirDrop, Contact Posters in the Phone app, and FaceTime messages
Expanded Communication Safety protections for children now detect videos containing nudity in addition to photos that children may receive or attempt to send in Messages, AirDrop, Contact Posters in the Phone app, FaceTime messages, and the system Photo picker
Improved sharing permissions give you even more control over what you share with apps, with an embedded photo picker and an add-only Calendar permission
Link tracking protection removes extra information from links shared in Messages, Mail, and Safari Private Browsing that some websites use in their URLs to track you across other websites, and links still work as expected
Accessibility
Assistive Access distills apps and experiences to their essential features in Phone and FaceTime, Messages, Camera, Photos, and Music, including large text, visual alternatives, and focused choices to lighten cognitive load
Live Speech lets you type what you want to say and have it be spoken out loud in phone calls, FaceTime calls, and for in-person conversations
Personal Voice enables users who are at risk of losing their voice to privately and securely create a voice that sounds like them on iPhone, and use it with Live Speech in phone and FaceTime calls
Point and Speak in Magnifier Detection Mode uses iPhone to read text out loud on physical objects with small text labels, such as keypads on doors and buttons on appliances
This release also includes other features and improvements:
Roadside Assistance via satellite lets you contact AAA to help you with vehicle issues when out of Wi-Fi or cellular range (iPhone 14, iPhone 14 Plus, iPhone 14 Pro, iPhone 14 Pro Max)
Pets in the People album in Photos surfaces individual pets in the album just like friends or family members
Photos Album widget lets you select a specific album from the Photos app to appear in the widget
Item sharing in Find My allows you to share an AirTag or Find My network accessory with up to five other people
Activity History in Home displays a recent history of events for door locks, garage doors, security systems, and contact sensors
Grid Forecast in Home shows when your electrical grid has cleaner energy sources available (Contiguous US only)
Grocery Lists in Reminders automatically group related items into sections as you add them
Inline PDFs and document scans in Notes are presented full-width, making them easy to view and mark them up
New Memoji stickers in Keyboard include Halo, Smirk, and Peekaboo
App Shortcuts in Spotlight Top Hit offer you app shortcuts to your next action when you search for an app
Redesigned Sharing tab in Fitness provides highlights of your friends’ activity like workout streaks and awards
Email or phone number sign-in lets you sign into your iPhone with any email address or phone number listed in your Apple ID account
New drawing tools in Freeform include a fountain pen, watercolor brush, ruler and more to create expressive boards
Crash Detection optimizations (iPhone 14, iPhone 14 Plus, iPhone 14 Pro, iPhone 14 Pro Max)
Some features may not be available for all regions or on all Apple devices. For more information, please visit this website:
https://www.apple.com/ios/ios-17
Some features may not be available for all regions or on all iPhone models. For information on the security content of Apple software updates, please visit this website:
Explore Apple Support Community
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IMAGES
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COMMENTS
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.
Solution : Given : Speed is 50 miles per hour. So, the distance covered in 1 hour is = 50 miles 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.
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. It is usually denoted by d in math problems .
To solve for speed or rate use the formula for speed, s = d/t which means speed equals distance divided by time. speed = distance/time To solve for time use the formula for time, t = d/s which means time equals distance divided by speed. time = distance/speed Time Entry Formats hh:mm:ss
i.e. speed = distance ÷ time S = D/T time = distance ÷ speed T = D/S distance = speed x time D = ST Time problem
Solution Time = Distance / speed = 20/4 = 5 hours. Example 2. A cyclist covers a distance of 15 miles in 2 hours. Calculate his speed. Solution 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?
In this problem, the total time is 4 hours and the total distance is \ (360\text { km},\) which we can plug into the equation: \ [\mbox {Speed} = \frac {\mbox {Distance}} {\mbox {Time}}= \frac {360~\mbox {km}} {4~\mbox {h}} = 90~\mbox {km/h}. \ _\square \]
Use the problem-solving method to solve problems using the distance, rate, and time formula; One formula you'll use often in algebra and in everyday life is the formula for distance traveled by an object moving at a constant speed. The basic idea is probably already familiar to you. Do you know what distance you traveled if you drove at a ...
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 years ago I don't understand how displacement can be negative, after all it is the shortest distance between the initial and final point ? • 2 comments ( 45 votes)
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
The seven types of problems are explained in detail in the actual generator below. All worksheets include an answer key on the 2nd page of the file. Please use the quick links below to generate some common types of worksheets. Easy speed, time, and distance worksheet 1: How far can it go or how long does the trip take - using whole or half hours
A video revising the techniques and strategies for solving problems with speed without a calculator (Higher & Foundation)This video is part of the Compound M...
...more Xcelerate Math resources https://xceleratemath.com/number/speedTime stamps⏰00:00 Introduction00:20 DST triangle01:19 Question 1: Find the distance (fast car)...
Click here for Answers speed, distance, time Practice Questions Previous: Pressure Practice Questions Next: Congruent Triangles Practice Questions The Corbettmaths Practice Questions on Speed, Distance, Time
Solution : Let x be speed of express train. So x-12 be the speed of another train. Distance to be covered = 240 km. Let T 1 be the time taken by the train to cover the distance 240 km at the speed of x km/hr. Let T 2 be the time taken by the train to cover the distance 240 km at the speed of (x + 12) km/hr.
Speed Distance Time Word Problems With Solutions. Q1. A train is travelling at a speed of 160 km/hour. It takes 15 hours to cover the distance from city A to city B. Find the distance between the two cities. Solution: Speed= 160 km/hr. Time= 15 hours. Using the formula, $\text {distance=speed }\!\!\times\!\!\text { time}$. Distance= $160\times ...
The Corbettmaths Textbook Exercise on Speed, Distance, Time. Previous: Simultaneous Equations: Graphical Textbook Exercise
Solution to Problem 1: After t hours the distances D1 and D2, in miles per hour, travelled by the two cars are given by D1 = 40 t and D2 = 50 t After t hours the distance D separating the two cars is given by D = D1 + D2 = 40 t + 50 t = 90 t Distance D will be equal to 450 miles when D = 90 t = 450 miles
Distance Problems traveling at different rates, word problems involving distance, rate (speed) and time, How to solve distance, rate and time problems: opposite directions, same direction and round trip, with video lessons, examples and step-by-step solutions.
Speed Distance Time problems: In this type of problem, the travel time is always the same. Example 1: A motorist drives for one hour at an average speed of 45 km/h, and for the next hour at an average speed of 65 km/h. ... Furthermore, these three variables certainly facilitate the solving of several types of problems in Maths. ...
Last Modified 22-06-2023 Real-life Problems Based on Speed, Time and Distance: Formulas, Examples When children or regular people apply mathematics to address Real-life Problems Based on Speed, Time and Distance, they learn that math is more than a task to accomplish for the purpose of the teacher.
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
Britta rides a bicycle from her home to school at a constant speed. If she increases her speed by 3 m/s , she will arrive at school three times as fast. How ...
The automated weed killer features a self-propelled, 120-foot carbon-fiber boom lined with 36 cameras capable of scanning 2,100 square feet per second. Powered by 10 onboard vision-processing ...
iOS 17 brings big updates to Phone, Messages, and FaceTime that give you new ways to express yourself as you communicate. StandBy delivers a new full-screen experience with glanceable information designed to view from a distance when you turn iPhone on its side while charging. AirDrop makes it easier to share and connect with those around you and adds NameDrop for contact sharing. Enhancements ...
118 likes, 2 comments - goldensparrow.assistancedog on February 12, 2024: " ️ "SPARROW, REMOTE" Lesson learned: don't underestimate a dog's ability to ...