Linear Motion - Minimum Retardation to Avoid Crash?

AI Thread Summary
The discussion revolves around calculating the minimum deceleration required for a passenger train traveling at 80 m/s to avoid colliding with a goods train moving at 30 m/s, which is 1500 meters ahead. The poster initially calculated that a collision would occur in 30 seconds and suggested that the passenger train should decelerate to the same speed as the goods train to prevent a crash. They derived a deceleration of -5/3 m/s² but later acknowledged that their method was incorrect and sought clarification on the proper approach. The conversation emphasizes the need for a correct method rather than just the numeric answer. The poster is looking for guidance on how to solve such problems effectively.
project_ILE
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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
 
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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.
 
Spinnor said:
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.
 
The answer might help potential helpers. What was it 8^)
 
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