problem solving involving average speed

Average Speed Problems

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

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

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

How to calculate Average Speed?

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

Examples Of Average Speed Problems

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

Solution: Step 1: The formula for distance is

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

Step 2: Total time = 3 + 2 = 5

Step 3: Using the formula:

Answer: The average speed is 54 miles per hour.

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

How To Solve The Average Speed Problem?

How to calculate the average speed?

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

How To Find The Average Speed For A Round Trip?

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

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

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

How To Relate Speed To Distance And Time?

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

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

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

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

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

How To Calculate Average Speed In Word Problems?

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

How To Use Average Speed To Calculate The Distance Traveled?

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

How To Use Average Speed To Calculate The Time Taken?

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

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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

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AVERAGE SPEED PROBLEMS

Problem 1 :

A person travels from New York to Washington at the rate of 45 miles per hour and comes backs to the New York at the rate of 55 miles per hour. What is his average speed for the whole journey?

Here, both the ways, he covers the same distance.

Then, the formula to find average speed is

= 2xy/(x + y)

x ----> Rate at which he travels from New York to Washington

y ----> Rate at which he travels from New York to Washington

So, the average speed is

= (2 ⋅ 45 ⋅ 55)/(45 + 55)

So, the average speed for the whole journey is 45 miles per hour.

Problem 2 :

A man takes 10 hours to go to a place and come back by walking both the ways. He could have gained 2 hours by riding both the ways. The distance covered in the whole journey is 18 miles. Find the average speed for the whole journey if he goes by walking and comes back by riding.

Given : A man takes 10 hours to go to a place and come back by walking both the ways.

walking + walking = 10 hours

2 ⋅ w alking = 10 hours

walking = 5 hours

Given : He could have gained 2 hours by riding both the ways.

riding + riding = 8 hours

2 ⋅ riding = 8 hours

riding = 4 hours

If he goes by walking and comes back by riding, time taken by him :

walking + riding = 5 + 4

walking + riding = 9 hours

total time taken = 9 hours

total distance covered = 18 miles

= total distance/total time

So, the required average speed is 2 miles per hour.

Problem 3 :

Lily takes 3 hours to travel from place A to place B at the rate of 60 miles per hour. She takes 2 hours to travel from place B to C with 50% increased speed. Find the average speed from place A to C.

speed (from A to B) = 60 miles/hour

speed (from B to C) = 90 miles/hour (50% increased)

Formula to find distance is

= rate ⋅ time

Distance from A to B is

= 60 ⋅ 3

= 180 miles

Distance from B to C :

= 90 ⋅ 2

Total distance traveled from A to B is

= 180 + 180

= 360 miles

Total time taken from A to B is

Formula to find average speed is

So, the average speed from place A to B is 72 miles/hour.

Problem 4 :

Distance from A to B = 200 miles

Distance from B to C = 300 miles

Distance from C to D = 540 miles

The speed from B to C is 50% more than A to B. The speed from C to D is 50% more than B to C. If the speed from A to B is 40 miles per hour, find the average speed from A to D.

speed (from A to B) = 40 miles/hour

speed (from B to C) = 60 miles/hour (50% more)

speed (from C to D) = 90 miles/hour (50% more)

Formula to find time is

= distance/time

time (A to B) = 200/40 = 5 hours

time (B to C) = 300/60 = 5 hours

time (C to D) = 540/90 = 6 hours

Total time taken from A to D is

= 5 + 5 + 6

Total distance from A to D is

= 200 + 300 + 540

= 1040 miles

So, the average speed from A to D is 65 miles per hour.

Problem 5 :

Time (A to B) = 3 hours

Time (B to C) = 5 hours

Time (C to D) = 6 hours

If the distances from A to B, B to C and C to D are equal and the speed from A to B is 70 miles per hour, find the average speed from A to D

= 70 ⋅ 3

= 210 miles

Given : Distance from A to B, B to C and C to D are equal.

= 210 + 210 + 210

= 630 miles

Total time taken A to D is

= 3 + 5 + 6

So, the average speed from A to D is 45 miles per hour.

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Average speed and average velocity – problems and solutions

Solved Problems in Linear Motion – Average speed and average velocity

1. A car travels along a straight road to the east for 100 meters in 4 seconds, then go the west for 50 meters in 1 second. Determine average speed and average velocity.

Distance = 100 meters + 50 meters = 150 meters

Displacement = 100 meters – 50 meters = 50 meters, to east.

Time elapsed = 4 seconds + 1 second = 5 seconds .

Average speed = Distance / time elapsed = 150 meters / 5 seconds = 30 meters/second.

Average velocity = Displacement / time elapsed = 50 meters / 5 seconds = 10 meters/second .

2. A person walks 4 meters east in 1 second, then walks 3 meters north in 1 second. Determine average speed and average velocity.

Average speed and average velocity - problems and solutions 1

Time elapsed = 1 second + 1 second = 2 seconds.

Average speed = distance / time elapsed = 7 meters / 2 seconds = 3.5 meters/second

Average velocity = displacement / time elapsed = 5 meters / 2 seconds = 2.5 meters/second

3. A runner travels around rectangle track with length = 50 meters and width = 20 meters. After travels around rectangle track two times, runner back to starting point. If time elapsed = 100 seconds, determine average speed and average velocity.

Circumference of rectangle = 2(50 meters) + 2(20 meters) = 100 meters + 40 meters = 140 meters.

Travels around rectangle 2 times = 2(140 meters) = 280 meters.

Distance = 280 meter .

Displacement = 0 meter . ( runner back to start point)

Average speed = distance / time elapsed = 280 meters / 100 seconds = 2.8 meters/second.

Average velocity = displacement / time elapsed = 0 / 100 seconds = 0.

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Mission KC6: Average Speed Computations

Mission KC6 pertains to the relationship between the distance traveled, the time of travel, and the average speed. Questions involve calculations. The mission consists of 14 questions organized into 7 Question Groups. You must answer one question from each Question Group to complete the mission. The learning outcomes for this mission are ...

Learning Outcomes

The student should be able to use the average speed equation to perform simple computations and solve word problems.

Launch Mission KC6

Getting help.

If you are not familiar with this topic, then you should first learn about the topic using our written Tutorial or our Video Tutorial: The Physics Classroom, One-Dimensional Kinematics Unit, Lesson 1, Part d

How can the average speed equation be used to solve problems involving distance, time and speed?

Video Tutorial: Speed vs. Velocity

Question-Specific Help

Each Question Group has its own Help page with information specific to the question. You can access the Help page from within the mission by tapping on the Help Me! icon (textbook). For your convenience, links to those pages are provided below: