Need to compute distance during during acceleration of a train

[mileena]

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

\Deltax = v₀t + (1/2)at²

The Attempt at a Solution

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

\Deltax = v₀t + (1/2)at²

\Deltax = (20 m/s)(40 s) + (1/2)(-1.0 m/s²)(40 s)²

\Deltax = 800 m + (1/2)(-1,600 m)

\Deltax = 800 m - 800 m

\Deltax = 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!

 

[Doc Al]

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?

 

[STEMucator]

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.

 

[mileena]

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 \Deltax.

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

\bar{v} = (1/2)(v₀ + v)

\Deltax = (1/2)(v₀ + v)(t)

v = v₀ + at

\Deltax = v₀t + (1/2)at²

v² = v₀² + 2a\Deltax

Unfortunately, they don't work with the info that I have.

 

[Doc Al]

v = v₀ + at

Use this one to figure out how long it takes for the train to come to rest.

 

[mileena]

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 v_f) is 0 m/s, because it may have reached that speed after the 40 s. But maybe that formula finds the time from v_o to v, and then you have to use that time in my original formula:

\Deltax = 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.

 

[Doc Al]

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 (v_f) 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:
\frac{1}{a}=\frac{1}{-1 m/s^2} = -1 s^2/m

 

[mileena]

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:

\Deltax = v₀t + (1/2)at²

\Deltax = (20 m/s)(20 s) + (1/2)(-1.0 m/s²)(20 s)²

\Deltax = 400 m - 200 m

\Deltax = 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.

 

[Doc Al]

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 v_f) 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 v_o to v, and then you have to use that time in my original formula:

\Deltax = 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.

 

[mileena]

Thank you so much Doc Al! I am learning so much through the math and physics homework help forums here!