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Carnegie Mellon University



Problem 37 Medium Difficulty

A train is traveling down a straight track at 20 m/s when the engineer applies the brakes, resulting in an acceleration of 21.0 ${m} / {s}^{2}$ as long as the train is in motion. How far does the train move during a 40-s time interval starting at the instant the brakes are applied?


$S = 200 \mathrm { m }$


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Video Transcript

we have you, we can use the uniformly accelerated motion equation or cinematics weaken. Say doubt The X equals the ex initial t plus 1/2 times the acceleration in the extraction T squared. Now for the full 42nd interval, this would essentially give us 20 meters per second multiplied by 40 seconds plus 1/2 negative 1.0 meters per second squared multiplied by 40 seconds quantity squared and this is equaling zero, which is of course wrong. Now the source of the air is found by computing the time required for the train to come to rest and so the time required t would be equaling velocity final minus Flossie initial. We know that the train is coming to rest so the final velocity will be zero divided by a the acceleration This would be negative 20 meters per second, divided by negative 1.0 meters per second squared and this is giving us 20 seconds. So essentially the train is slowing down for the 1st 20 seconds and then is at rest for the last 20 seconds, so slowing down and then 20 seconds at rest and we can then say to find the stopping distance Delta X. This would be equaling the ex initial T plus 1/2 A T squared. But now here the acceleration. We know the acceleration is not constant during the full 40 seconds. However, it is constant during the 1st 20 seconds as the train slows down to rest. So we're going to be using this cannon back equation. However, the time here would not be 40 seconds. It'll only be 20 seconds. So this would be equaling 20 meters per second, multiplied by 20 seconds plus 1/2 negative, 1.0 meters per second squared, multiplied by 20 seconds. Quantity squared. And we find that doubt tow exes equaling 200 meters. This would be our final answer from the stopping distance. That is the end of the solution. Thank you for watching.

Carnegie Mellon University
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