A train \"A\" moving @48 m/s is moving to the right and is 250 metersaway from a

Question

A train "A" moving @48 m/s is moving to the right and is 250 metersaway from another train "B" moving at 52 m/s to the left. The train"A" decelerates at a rate of 8 m/s2 and the train movingto the left also decelerates at a rate of 12 m/s2 . Willthey collide with each other? [Assume that after the trains stop,they don't turn around and accelerate in the backward direction;they remain stopped].
a) if yes, how far from the left train's original position willthey collide? How fast is each one traveling when they collide? Howmany seconds later will they collide?

Explanation / Answer

Suppose the two trains collide at a distance d from initialposition of A. So, s = u*t - 0.5*a*t*t v = u - at v*v = u*u + 2*a*s For A, Time to stop A is :    t = u/a = 48/8 s = 6 s Time to stop B is : t' = 52/12 = 4.33 s In 6s, A travels, s = 48*6 - 0.5*8*6*6 m = 144 m In 4.33s, B travels, s' = 52*4.33 - 0.5*12*4.33*4.33 m = 112.67 m So, s+s' = 144 + 112.67 m = 256.67 m > 250 m Hence the trains would collide. In time t when acceleration of B continues after havingstopped they collide, d = s + s' = 48*t - 0.5*8*t*t + 52*t - 0.5*12*t*t = 250m => 10*t*t - 100*t + 250 = 0 => t*t - 10*t + 25 = 0 => t = 5 s However, since the train B has already stopped aftertravelling 112.67m, and it remains at rest according toquestion, they collide at (250-112.67)m from left train's initialposition = 137.33 m So, 137.33 = s = 48*t - 0.5*8*t*t => t = 4.7087 s (smaller of the two values you haveobtained since it stays at rest after stopping) train A is travelling at v = u + at = 48 - 8*4.7087 = 10.331m/s B is travelling at 0 m/s since it has already stopped. As found before they collide after 4.7087s.

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