Need to compute distance during during acceleration of a train

In summary: I hope you have a good rest of the day!In summary, a train traveling at 20 m/s applies brakes resulting in an acceleration of -1.0 m/s2 as long as the train is in motion. The train moves 200 m in a 40-s time interval, but the correct answer is 200 m in a 20 s time interval. The mistake was assuming the train took 40 s to come to a complete stop.
  • #1
mileena
129
0

Homework Statement



A train is traveling down a straight track at 20 m/s when the engineer applies the brakes, resulting in an acceleration of -1.0 m/s2 as long as the train is in motion. How far does the train move in a 40-s time interval starting at the instant the brakes are applied?

Homework Equations



[itex]\Delta[/itex]x = v0t + (1/2)at2

The Attempt at a Solution



Answer from the back of the book: 200 m, but my answer does not agree!

[itex]\Delta[/itex]x = v0t + (1/2)at2

[itex]\Delta[/itex]x = (20 m/s)(40 s) + (1/2)(-1.0 m/s2)(40 s)2

[itex]\Delta[/itex]x = 800 m + (1/2)(-1,600 m)

[itex]\Delta[/itex]x = 800 m - 800 m

[itex]\Delta[/itex]x = 0 mThis has to be wrong! I have gone over the calculations and the equation numerous times, and I can't find anything wrong though. Can anyone help?

Thanks!
 
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  • #2
mileena said:
A train is traveling down a straight track at 20 m/s when the engineer applies the brakes, resulting in an acceleration of -1.0 m/s2 as long as the train is in motion. How far does the train move in a 40-s time interval starting at the instant the brakes are applied?
Note that the problem specifies "as long as the train is in motion". When does the train come to rest?
 
  • #3
The answer at the back of your book doesn't make sense at all. How can distance be measured in ##m/s^2##?

EDIT : I see you change it.
 
  • #4
Ok, thanks for giving me a hint Doc Al. Maybe the train came to a stop in less than 40 s, and they put in "40 s" just to trick us?

This is what is know:

v0 = 20 m/s
v = 0 m/s, assuming the train did stop
a = -1.0 m/s2
t < 40 s, assuming the train stopped before 40 s

If I can compute t from v0 to v, I then can compute [itex]\Delta[/itex]x.

These are the five formulae I have to choose from (all are for constant acceleration):

[itex]\bar{v}[/itex] = (1/2)(v0 + v)

[itex]\Delta[/itex]x = (1/2)(v0 + v)(t)

v = v0 + at

[itex]\Delta[/itex]x = v0t + (1/2)at2

v2 = v02 + 2a[itex]\Delta[/itex]x

Unfortunately, they don't work with the info that I have.
 
  • #5
mileena said:
v = v0 + at
Use this one to figure out how long it takes for the train to come to rest.
 
  • #6
Wow, you post fast!

Ok:

v = v0 + at

t = (1/a)(v - v0)

t = (1/-1.0 m/s2)(0 m/s - 20 m/s)

t = (-1.0 s2/m)(-20 m/s)

t = 20 s

I can see a problem though:

I do not know for a fact that v (or vf) is 0 m/s, because it may have reached that speed after the 40 s. But maybe that formula finds the time from vo to v, and then you have to use that time in my original formula:

[itex]\Delta[/itex]x = v0t + (1/2)at2

Was my mistake assuming the time to stop the train was 40 s?

Also, is there a way to do "strike through" formatting here? I wanted to strike though text I edited.
 
Last edited:
  • #7
mileena said:
Ok:

v = v0 + at

t = (1/a)(v - v0)

t = (1/-1.0 m/s2)(0 m/s - 20 m/s)

t = (-1.0 m/s2)(-20 m/s)

t = 20 m2/s3
You messed up the units. That should come to be seconds.

I can see two problems though:

1. I do not know v (vf) is 0 m/s, because it may have reached that speed after the 40 s.
Try it and see!

2. I did (1/a) in the formula above because I do not know how to write fraction with a horizontal fraction bar on this forum. But when I do that, the answer came out as m2/s3. If I had done the horizontal fraction bar, the units would have been in seconds, which is the correct unit for time. What happened?
You messed up the units, which I'm glad that you realize. Don't blame the notation!

1/a = 1/(-1 m/s^2) = - 1 s^2/m.

You can write fractions using Latex:
[tex]\frac{1}{a}=\frac{1}{-1 m/s^2} = -1 s^2/m[/tex]
 
  • #8
Ok, thank you! I am going to try writing fractions with Latex later on! And I found the "strike through" button above.

Let me try finishing the problem:

[itex]\Delta[/itex]x = v0t + (1/2)at2

[itex]\Delta[/itex]x = (20 m/s)(20 s) + (1/2)(-1.0 m/s2)(20 s)2

[itex]\Delta[/itex]x = 400 m - 200 m

[itex]\Delta[/itex]x = 200 m

Oh my gosh! The answer matched the one in the book!

Thank you.

I guess my mistake was in assuming the train took the 40 s mentioned to come to a complete stop, even though all the variable did fit into the equation in my first post, and appeared to give me an answer.
 
Last edited:
  • #9
mileena said:
Wow, you post fast!

Ok:

v = v0 + at

t = (1/a)(v - v0)

t = (1/-1.0 m/s2)(0 m/s - 20 m/s)

t = (-1.0 s2/m)(-20 m/s)

t = 20 s
OK, much better.

I can see a problem though:

I do not know for a fact that v (or vf) is 0 m/s, because it may have reached that speed after the 40 s.
But you just found that the train came to rest in 20 seconds!

But maybe that formula finds the time from vo to v, and then you have to use that time in my original formula:

[itex]\Delta[/itex]x = v0t + (1/2)at2
Sure.

Was my mistake assuming the time to stop the train was 40 s?
Yes. Something of a trick question.

Also, is there a way to do "strike through" formatting here? I wanted to [STRIKE]strike though[/STRIKE] text I edited.
Sure.
 
  • #10
Thank you so much Doc Al! I am learning so much through the math and physics homework help forums here!
 

1. How is distance calculated during acceleration of a train?

Distance is calculated by multiplying the average velocity by the time during which the train is accelerating. This can be represented by the equation: d = vavg * t.

2. What is the difference between distance and displacement during acceleration of a train?

Distance refers to the total length traveled by the train, while displacement refers to the shortest distance between the starting and ending point of the train's motion. During acceleration, the distance and displacement may be different since the train may change direction multiple times.

3. How does the acceleration of a train affect the distance it travels?

The acceleration of a train can greatly affect the distance it travels. If the train is accelerating at a constant rate, the distance traveled will increase at a faster rate. However, if the train is decelerating, the distance traveled will decrease at a slower rate.

4. Can the distance traveled during acceleration of a train be negative?

Yes, the distance traveled during acceleration can be negative if the train is traveling in the opposite direction of its initial motion. This would result in a negative displacement value.

5. How does the mass of a train affect the distance it travels during acceleration?

The mass of a train does not directly affect the distance it travels during acceleration. However, a heavier train may require more force and time to accelerate, resulting in a longer distance traveled.

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