Question

Two trains face each other on adjacent tracks. They are initially at rest, and their front ends are 49 m apart. The train on the left accelerates rightward at 0.87 m/s². The train on the right accelerates leftward at 0.8 m/s².

Two trains face each other on adjacent tracks. They are initially at rest, and their front ends are 49 m apart. The train on the left accelerates rightward at 0.87 m/s. The train on the right accelerates leftward at 0.8 m/s (a) How far does the train on the left travel before the front ends of the trains pass? x m (b) If the trains are each 150 m in length, how long after the start are they completely past one another, assuming their accelerations are constant? X s Assume that the accelerations of the trains are constant. Let the origin be the initial position of one of the trains. You are interested in the time when both trains are at the same position. Be very careful when assigning signs to the kinematic quantities.

Answer 1

Acceleration of train on left,$a_{L}=0.87m/s^{2}$

acceleration of train on right,$a_{R}=0.8m/s^{2}$

since trains are moving in opposite direction relative acceleration,  $a=a_{L}+a_{R}=0.87m/s^{2}+0.8m/s^{2}$

Relative acceleration ,$a=1.67m/s^{2}$

Relative distance to cover =49m

First find time taken to front end to pass

Use formula $s=ut+1/2at^{2}$

initially both trains are at rest

$s=1/2at^{2}$

$49m=1/2*1.67m/s^{2}*t^{2}$

$49=0.5*1.67*t^{2}$

$49=0.835t^{2}$

$t=7.66s$

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(a)Distance traveled by train on left in t =7.66s

Use formula $s=ut+1/2at^{2}$

initially both trains are at rest

$s=1/2at^{2}$

$s=1/2*0.87m/s^{2}*(7.66s)^{2}$

$s=1/2*0.87*7.66^{2}$

ANSWER: ${\color{Red} s=25.52m}$

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(b)Trains are Completely passed

Relative distance to cover =150m +150m =300m

Relative acceleration ,$a=1.67m/s^{2}$

Use formula $s=ut+1/2at^{2}$

initially both trains are at rest

$s=1/2at^{2}$

$300m=1/2*1.67m/s^{2}*t^{2}$

$300=0.5*1.67*t^{2}$

$300=0.835t^{2}$

ANSWER: ${\color{Red} t=18.96s}$

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