# Linear Motion - Minimum Retardation to Avoid Crash?

1. Jan 30, 2012

### project_ILE

First post! I would be grateful if anyone could give me any advice on this particular type of problem (i.e min retardation to avoid a crash). I'm not necessarily looking for the answer to this specific question, I would rather if someone could point me in the right direction as to how to go about these questions. Thanks!

1. A passenger train, which is travelling at 80 m s$^{-1}$ is 1500 m behind a goods train which is travelling at 30 m s$^{-1}$ in the same direction on the same track. At what rate must the passenger train decelerate to avoid a crash? (Ignore the lengths of the trains)

2. s = ut + $\frac{1}{2}$at$^{2}$
v = u + at
s = ($\frac{u + v}{2}$)t
v$^{2}$ = u$^{2}$ + 2as

3. I worked out that at current speeds, the trains would collide after 30 seconds, using a velocity-time graph, letting the distance travelled by the passenger train = distance travelled by the goods train + 1500. (30T + 1500 = 80V). From there I summized that for the trains to never crash, the passenger train should decelerate to at least the same speed as the goods train ( 30 m s$^{-1}$). Then I used v = u + at for the passenger train, and had 30 = 80 + a(30), where a comes out at - $\frac{5}{3}$ m s$^{-2}$. I've compared this answer to the correct one.

Thanks,
project_ILE

2. Jan 30, 2012

### Spinnor

You wrote,

"I've compared this answer to the correct one."

%^) So did you get the right answer? If you did fine, if not then someone might care to find the error.

3. Jan 30, 2012

### project_ILE

Sorry, I thought I had finished my post! This question is from an exam paper, I have the numeric answer, but not the method. My own answer (above) is not correct.

4. Jan 31, 2012

### Spinnor

The answer might help potential helpers. What was it 8^)