Average Speed Question for a Train that Accelerates and then Decelerates

[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.

 

[Philip Koeck]

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?

 

[vryC0nfused]

Time average or average over distance?

I think it's average over distance.

 

[Philip Koeck]

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.

 

[vryC0nfused]

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.

 

[Philip Koeck]

Okay, that sounds like a good idea.

Try to plot v as a function of t. That should help you.

 

[Steve4Physics]

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?

 

[haruspex]

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

 

[vryC0nfused]

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.

 

[jbriggs444]

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.

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?

 

[Steve4Physics]

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!

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.

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]

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.

 

[vryC0nfused]

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.

 

[TonyStewart]

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

 

[vryC0nfused]

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

 

[Steve4Physics]

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.

Think about this...

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?

 

[vryC0nfused]

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

Think about this...

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.

 

[vryC0nfused]

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! :)

 

[jbriggs444]

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.

 

[malawi_glenn]

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.

 

[PeroK]

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.