Linear Motion - Minimum Retardation to Avoid Crash?

[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 traveling at 80 m s$^{-1}$ is 1500 m behind a goods train which is traveling 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 traveled by the passenger train = distance traveled 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

 

[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.

 

[project_ILE]

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.

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.

 

[Spinnor]

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

 

[asteriatic]

Dear project_ILE,

Thank you for your question. It is important to consider the minimum retardation required to avoid a crash in situations like this. To solve this problem, we can use the equations of motion that you have listed in your post. First, let's consider the situation at the point of collision. At this point, the distance traveled by the passenger train (s) will be equal to the distance traveled by the goods train (s + 1500m). We can set up an equation using the equation s = ut + \frac{1}{2}at^{2} for both trains, where u is the initial velocity, a is the acceleration, and t is the time taken. This will give us two equations:

s = 80t - \frac{1}{2}at^{2}
s + 1500 = 30t - \frac{1}{2}at^{2}

We can solve these equations simultaneously to find the time (t) at which the trains will collide. Substituting the value of t into either equation will give us the distance (s) at the point of collision. Now, to avoid a crash, the passenger train must decelerate to the same speed as the goods train (30 m s^{-1}) before reaching this point. This means that the final velocity (v) of the passenger train will be 30 m s^{-1}. We can use the equation v^{2} = u^{2} + 2as to find the minimum retardation (a) required for the passenger train to reach this final velocity.

30^{2} = 80^{2} + 2a(s + 1500)
a = - \frac{5}{3} m s^{-2}

This is the same answer that you have found, which means your method is correct. However, it is always good to double check your work by using different methods or equations. I hope this helps and good luck with your future problems!