# Average Speed Question for a Train that Accelerates and then Decelerates

• vryC0nfused
In summary: A4. Using the same methods as before, I determined that the total time taken was 3,360 seconds and the total distance covered was 12,960 meters.(3,360 s) (12,960 m)... = 42,360 metersQ5. The average speed during this time period was 22 m/s.Oh! I'll make sure to provide evidence of my attempts in future questions!
vryC0nfused
TL;DR Summary: A train on a level stretch of tracks, starting from a dead stop, steadily speeds up to 28 m/s in 2.0 minutes, and then slows to 18 m/s with a constant deceleration of 0.011 m/s^2. What is the average speed of the train for this whole thing (speeding up and slowing down)?

I know that if I take the direction of the train as going in the positive direction, then the average speed is the same as the average velocity. The answer is 22 m/s, but I'm not sure how to get to that conclusion since I feel like I'm not picking up on some key information.

Mentor's note: Thread moved to homework forum.

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vryC0nfused said:
TL;DR Summary: A train on a level stretch of tracks, starting from a dead stop, steadily speeds up to 28 m/s in 2.0 minutes, and then slows to 18 m/s with a constant deceleration of 0.011 m/s^2. What is the average speed of the train for this whole thing (speeding up and slowing down)?

I know that if I take the direction of the train as going in the positive direction, then the average speed is the same as the average velocity. The answer is 22 m/s, but I'm not sure how to get to that conclusion since I feel like I'm not picking up on some key information.
Time average or average over distance?

Philip Koeck said:
Time average or average over distance?
I think it's average over distance.

vryC0nfused said:
I think it's average over distance.
The time average would be easier, I think. Maybe that's what you're supposed to work out.
Maybe you could sketch how you would proceed.

Philip Koeck said:
The time average would be easier, I think. Maybe that's what you're supposed to work out.
Maybe you could sketch how you would proceed.
Okay, that sounds like a good idea.

vryC0nfused said:
Okay, that sounds like a good idea.
Try to plot v as a function of t. That should help you.

malawi_glenn
vryC0nfused said:
TL;DR Summary: A train on a level stretch of tracks, starting from a dead stop, steadily speeds up to 28 m/s in 2.0 minutes, and then slows to 18 m/s with a constant deceleration of 0.011 m/s^2. What is the average speed of the train for this whole thing (speeding up and slowing down)?

I know that if I take the direction of the train as going in the positive direction, then the average speed is the same as the average velocity. The answer is 22 m/s, but I'm not sure how to get to that conclusion since I feel like I'm not picking up on some key information.

Mentor's note: Thread moved to homework forum.
hI @vryC0nfused and welcome to PF. The rules here require you to show evidence of your own effort before we offer guidance. But I don't think I'm bending the rules too much with this...

Can you answer each of the following questions? If working through these questions doesn't resolve your problem, post as much of your working as you have been able to do...

Q1. During the first 2 minutes, while speeding-up, what is the total distance covered?

Q2. How long does it take to slow down from 28m/s to 18m/s?

Q3. What distance is covered while slowing down from 28m/s to 18m/s?

Q4. What is the total time taken and the total distance covered?

Q5. What is the average speed?

A good place to start when confused is the definitions. What is the definition of average speed?

nasu
Steve4Physics said:
hI @vryC0nfused and welcome to PF. The rules here require you to show evidence of your own effort before we offer guidance. But I don't think I'm bending the rules too much with this...

Can you answer each of the following questions? If working through these questions doesn't resolve your problem, post as much of your working as you have been able to do...

Q1. During the first 2 minutes, while speeding-up, what is the total distance covered?

Q2. How long does it take to slow down from 28m/s to 18m/s?

Q3. What distance is covered while slowing down from 28m/s to 18m/s?

Q4. What is the total time taken and the total distance covered?

Q5. What is the average speed?
Oh! I'll make sure to provide evidence of my attempts in future questions!

A1. During the first 2 minutes (120 seconds) the train covered a distance of 3,360 meters.
(28 m/s) (120 s) = 3,360 meters

A2. Using one of of the famous five ( V_x = V_{x0} + a_xt ) and treating the train like it's speeding up from 18 m/s to 28 m/s with an acceleration of 0.011 m/s^2, I got that it took about 910 s.
(28 m/s) = (18 m/s) + (0.011)t ---> t = 909.1 or 910 w/ 3 sig. fig.s

A3. Using the time I got from the previous question, I determined the distance covered during this interval in the same way I did during the first 120 seconds.
(28 m/s)(909.1 s) = 25,454.8 ---> about 25,500 meters

A4. To determine the total time and distance covered, I added the values I got together.
Total time (T) = (120 s) + (909.1 s) ---> T= 1029.1 or 1030 seconds
Total distance (D) = (3,360 m) + (25,454.8 m) ---> D= 28,814.8 or 28,800 meters

A5. The average speed is the total distance (D) divided by the total time (T).
Av. speed = (28,814.8 m) / (1,029.1 s) ---> 28 m/s

I shouldn't be getting 28 m/s as my final answer because the answer key says it's 22 m/s. I think this is because I used 28 m/s in the calculation to find the distance covered during in the 909.1 seconds. I feel like I'm on the right track, but I'm making a mistake somewhere that's making my average speed incorrect.

vryC0nfused said:
I think this is because I used 28 m/s in the calculation to find the distance covered during in the 909.1 seconds. I feel like I'm on the right track, but I'm making a mistake somewhere that's making my average speed incorrect.
Let us assume, for the moment, that your hunch about the source of error is correct. If this 909.1 second interval starts with a speed of 28 m/s and ends with a speed of 18 m/s, what is the average speed over just the 909.1 second interval?

But let us also double check your work. There has to be another error. You cannot average an interval with less then 28 m/s speed and another interval with exactly a 28 m/s speed and end up with a 28 m/s average.

Oh, there it is.
vryC0nfused said:
A1. During the first 2 minutes (120 seconds) the train covered a distance of 3,360 meters.
(28 m/s) (120 s) = 3,360 meters
You've also used the final speed on the first interval as if it were the average speed over that interval.

What is the average speed over just the first 120 second interval?

vryC0nfused said:
A1. During the first 2 minutes (120 seconds) the train covered a distance of 3,360 meters.
(28 m/s) (120 s) = 3,360 meters
Nope. If the train moved at a constant velocity of 28m/s for 120s, you would be correct. But the train isn't moving at a constant velocity!

vryC0nfused said:
A2. Using one of of the famous five ( V_x = V_{x0} + a_xt ) and treating the train like it's speeding up from 18 m/s to 28 m/s with an acceleration of 0.011 m/s^2, I got that it took about 910 s.
(28 m/s) = (18 m/s) + (0.011)t ---> t = 909.1 or 910 w/ 3 sig. fig.s
Ok, but you could use initial velocity = 28m/s, final velocity = 18m/s and a negative value for acceleration. That matches the situation better and gives the same answer.

No need to round the answer as this is an intemediate value used in a subsequent step - as you have correctly done.

vryC0nfused said:
A3. Usin the time I got from the previous question, I determined the distance covered during this interval in the same way I did during the first 120 seconds.
(28 m/s)(909.1 s) = 25,454.8 ---> about 25,500 meters
Nope. If the train moved at 28m/s for 909.1s, you would be correct. But the train isn't moving at 28m/s during that time.

Check your equations and see how you can find displacement (distance here) when the velocity is changing.

vryC0nfused
jbriggs444 said:
Let us assume, for the moment, that your hunch about the source of error is correct. If this 909.1 second interval starts with a speed of 28 m/s and ends with a speed of 18 m/s, what is the average speed over just the 909.1 second interval?

But let us also double check your work. There has to be another error. You cannot average an interval with less then 28 m/s speed and another interval with exactly a 28 m/s speed and end up with a 28 m/s average.

Oh, there it is.

You've also used the final speed on the first interval as if it were the average speed over that interval.

What is the average speed over just the first 120 second interval?
Wouldn't the average speed for the first 120 second interval also be 28 m/s? You'd divide the total distance (3,360 meters) by the total time taken to cover that distance (120 seconds) and get 28 m/s.

Steve4Physics said:
Nope. If the train moved at a constant velocity of 28m/s for 120s, you would be correct. But the train isn't moving at a constant velocity!Ok, but you could use initial velocity = 28m/s, final velocity = 18m/s and a negative value for acceleration. That matches the situation better and gives the same answer.

No need to round the answer as this is an intemediate value used in a subsequent step - as you have correctly done.Nope. If the train moved at 28m/s for 909.1s, you would be correct. But the train isn't moving at 28m/s during that time.

Check your equations and see how you can find displacement (distance here) when the velocity is changing.
Oh! I think that's where I'm making my mistake, I'm treating the train as if it's moving at a constant velocity! I'll check my notes for the equation I need to use.

Steve4Physics
total Distance over . . . . . is the average speed. { = key info}
total time

- compute d,t
1. distance covered from 0 m/s to 28 m/s in 2.0 minutes: = 1,680 m
2. distance covered from 28 m/s to 18 m/s using a = -0.011 m/s^2 = 20,910 m
3. distance covered steps 1 and 2: = 22,590 m
4. time taken accelerating (2.0 minutes or 120 seconds) and coasting ( = 909 seconds): = 1,029 s
5. average speed = total distance / total time: = ~22 m/s.
key info . don't compute average speed in each zone , just compute distance , time
Don't click spoiler until you try

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vryC0nfused
vryC0nfused said:
Oh! I think that's where I'm making my mistake, I'm treating the train as if it's moving at a constant velocity! I'll check my notes for the equation I need to use.
Ok, since the train is starting from "a dead stop" and I'm assuming that the acceleration is constant, I used the equation
x = x_0 + (1/2)(V_{x0} + V_x)t to find the distance.
x = (0 m) + (1/2)(0 m/s + 28 m/s)(120 s) ---> x = 1,680 meters
I think this is the correct distance covered during this interval and if it is it should help me find my total answer.
The average speed for this interval would then be 14 m/s.
(1,680 m) / (120 s) ---> 14 m/s

vryC0nfused said:
Wouldn't the average speed for the first 120 second interval also be 28 m/s? You'd divide the total distance (3,360 meters) by the total time taken to cover that distance (120 seconds) and get 28 m/s.
That's incorrect logic because the 3,360m was wrong to start with.

You uniformly accelerate in your new Porsche from rest to 100mph in 10s, covering a distance of x metres.

You then drive your Porsche at 100mph for another 10s, covering a distance of y metres.

Without any calculations, which is bigger, x or y? Or are they equal? Can you explain your answer?

Steve4Physics said:
That's incorrect logic because the 3,360m was wrong to start with.

You uniformly accelerate in your new Porsche from rest to 100mph in 10s, covering a distance of x metres.

You then drive your Porsche at 100mph for another 10s, covering a distance of y metres.

Without any calculations, which is bigger, x or y? Or are they equal? Can you explain your answer?
I think the distance covered of y meters would be larger since you're starting from 100 mph instead of from rest or 0 mph. I incorrectly assumed that I was starting with a uniform velocity of 28 m/s when I determined the distance traveled.

Steve4Physics
TonyStewart said:
total Distance over . . . . . is the average speed. { = key info}
total time

- compute d,t
1. distance covered from 0 m/s to 28 m/s in 2.0 minutes: = 1,680 m
2. distance covered from 28 m/s to 18 m/s using a = -0.011 m/s^2 = 20,910 m
3. distance covered steps 1 and 2: = 22,590 m
4. time taken accelerating (2.0 minutes or 120 seconds) and coasting ( = 909 seconds): = 1,029 s
5. average speed = total distance / total time: = ~22 m/s.
key info . don't compute average speed in each zone , just compute distance , time
Don't click spoiler until you try
Oh! Okay, I was thinking that I was supposed to calculate the individual average speed for each of the intervals and then somehow add them up. Thank you so much, I see why that wouldn't make any sense now! :)

TonyStewart
vryC0nfused said:
Oh! Okay, I was thinking that I was supposed to calculate the individual average speed for each of the intervals and then somehow add them up. Thank you so much, I see why that wouldn't make any sense now! :)
Computing the average speed for each of the intervals and then averaging them is actually a correct approach.

However, it is a bit tricky to do it right. You want a time-weighted average. So you would need to weight each of the two averages by the time spent during that interval.

Total distance divided by total time yields the same result with a bit less complicaation.

The area in the v-t diagram is the distance traveled (more generally, the total displacement)
Complete this diagram and you are basically done with your problem.

vryC0nfused, MatinSAR, Steve4Physics and 2 others
Note that the average speed during a period of constant acceleration is halfway between the initial and final speeds. Assuming no changes in direction.

The average speed during each phase is, therefore, ##14 \ m/s## and ##23\ m/s##.

To find the overall average speed you must, as @jbriggs444 Indicated, take a weighted average of these. In other words, if the train moves at an average speed of ##v_1## for time ##t_1## and at an average speed of ##v_2## for time ##t_2##, then the overall average speed is
$$v_{avg} = \frac{v_1t_1 + v_2t_2}{t_1+t_2}$$That represents an alternative way to solve this problem.

MatinSAR and jbriggs444

## 1. How do you calculate the average speed of a train that accelerates and then decelerates?

To calculate the average speed of a train that accelerates and then decelerates, you need to know the total distance traveled and the total time taken. The average speed is given by the formula: Average Speed = Total Distance / Total Time. If the motion involves different phases with different speeds, you must first calculate the time and distance for each phase and then sum them up to find the total distance and total time.

## 2. What information is needed to solve an average speed problem for a train with varying speeds?

To solve an average speed problem for a train that accelerates and then decelerates, you need the following information: the acceleration rate, the deceleration rate, the maximum speed reached, and the distances or times for each phase of the journey (acceleration, constant speed, and deceleration). With this information, you can determine the total distance and total time to calculate the average speed.

## 3. Can you provide an example problem and solution for calculating the average speed of a train that accelerates and decelerates?

Sure! Let's say a train accelerates from rest at 2 m/s² for 10 seconds, then travels at a constant speed for 20 seconds, and finally decelerates at 1 m/s² until it stops. First, calculate the distance and time for each phase:1. Acceleration: Distance = 0.5 * 2 * (10)² = 100 meters, Time = 10 seconds2. Constant speed: Speed = 2 * 10 = 20 m/s, Distance = 20 * 20 = 400 meters, Time = 20 seconds3. Deceleration: Time = 20 / 1 = 20 seconds, Distance = 0.5 * 20 * 20 = 200 metersTotal Distance = 100 + 400 + 200 = 700 metersTotal Time = 10 + 20 + 20 = 50 secondsAverage Speed = 700 / 50 = 14 m/s

## 4. How does the acceleration and deceleration phase affect the average speed?

The acceleration and deceleration phases affect the average speed by changing the overall distance traveled and the time taken. During acceleration, the train speeds up, covering more distance over time, whereas during deceleration, it slows down, covering less distance over time. These phases must be accounted for to accurately calculate the average speed, as they influence the total distance and total time of the journey.

## 5. What are common mistakes to avoid when calculating average speed in these scenarios?

Common mistakes include not correctly calculating the distances and times for each phase

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