Finding deceleration of a moving train

[garcia1]

Homework Statement

An engineer in a locomotive sees a car stuck
on the track at a railroad crossing in front of
the train. When the engineer first sees the
car, the locomotive is 250 m from the crossing
and its speed is 12 m/s.
If the engineer’s reaction time is 0.36 s,
what should be the magnitude of the mini-
mum deceleration to avoid an accident?
Answer in units of m/s2.

Homework Equations

I used the kinematics equation x = volt + 1/2at^2 to find my answer.

The Attempt at a Solution

By solving for a, I got the answer 3791.358 m/s^2. This was wrong, and I'm also confused as to whether it should be a positive or negative answer.

 

[ideasrule]

I used the kinematics equation x = volt + 1/2at^2 to find my answer.

What did you use for t? It should be the maximum possible t such that the train doesn't hit the car, but this value is hard to compute.

 

[garcia1]

I used the .36s reaction time, but i see how this is wrong now. Should I be finding the final velocity of the train when the engineer hits the breaks? Is it even necessary to find this time if I can assume that the final velocity of the whole problem would be zero, since the train would stop?

 

[gneill]

I used the .36s reaction time, but i see how this is wrong now. Should I be finding the final velocity of the train when the engineer hits the breaks? Is it even necessary to find this time if I can assume that the final velocity of the whole problem would be zero, since the train would stop?

Find the total distance over which braking can occur (the train keeps moving at its initial velocity during the engineer's reaction period). The train must go from its initial velocity to zero in that distance. You probably know a formula that relates initial and final velocities with acceleration and distance...

 

[garcia1]

So I did do that through this rationale:

x = volt (since a = 0, the rest of this equation cancels out)

Solving for this, x = 12*.36, I got 4.32m.

From here, I subtracted this from the initial distance, 250m, to yield 245.68m.

With the new distance, I calculated the acceleration by using the final velocity as 0m/s, with the following equation:

Vf^2 = Vo^2 + 2a(x - xo)

Vf = 0, Vo = 12, x = 245.68, and xo = 0

I solved for a, getting -.29 m/s^2.
I used the answer +.29 m/s^2 since the problem asks for the magnitude. Does this seem right, or should I be using the negative answer as is for this problem?

 

[gneill]

I solved for a, getting -.29 m/s^2.
I used the answer +.29 m/s^2 since the problem asks for the magnitude. Does this seem right, or should I be using the negative answer as is for this problem?

Magnitudes are always positive. You're fine.

 

[garcia1]

I don't know, this answer still came up wrong on my homework, but it seems like this is the way to do it.

 

[gneill]

I don't know, this answer still came up wrong on my homework, but it seems like this is the way to do it.

0.29 m/s² looks like the correct answer to me. Even the significant figures match those for the given values.