Linear Motion - Minimum Retardation to Avoid Crash?

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Homework Help Overview

The discussion revolves around a problem in linear motion, specifically focusing on determining the minimum retardation required for a passenger train to avoid colliding with a goods train. The scenario involves two trains traveling in the same direction at different speeds, with a specified distance between them.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to analyze the situation using kinematic equations and a velocity-time graph to determine the necessary deceleration. They express a desire for guidance on the approach rather than a direct solution.

Discussion Status

Participants are engaged in clarifying the original poster's understanding and approach. There is an indication that the original poster has compared their findings to a known correct answer but acknowledges that their method may not be accurate. The discussion is open, with participants prompting further exploration of the reasoning behind the calculations.

Contextual Notes

The original poster mentions that this question is from an exam paper, which may impose specific constraints or expectations regarding the method of solution. They also indicate that they have a numeric answer but lack clarity on the methodology used to arrive at it.

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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