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A train begins at the origin and moves at constant speed v1=11.7 m/s toward another train...

Question

A train begins at the origin and moves at constant speed v1=11.7 m/s toward another train...

A train begins at the origin and moves at constant speed v1=11.7 m/s toward another train which is initially at rest at a distance d=152 m from the first train. If the second train has an acceleration of magnitude a2=0.95 m/s2 toward the first train, how far does the first train move before the trains collide? (Hint : use quadratic equation and choose positive time)

Science Physics
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Answers

Given:

Speed of the first train, v_{1}=11.7m/s

Let d_{1} be the distance covered by the first train before colliding with the second train

Let do be the distance covered by the second train before colliding with the first train

Initial gap between the trains, d= 152m

So d_{1}+d_{2}=d

{\color{Red} d_{1}+d_{2}=152}-------------(1)

Initial speed of the second train, u_{2}=0m/s

Acceleration of the second train, a_{2}=0.95m/s^{2}

Travel time of both the trains is the same. t is the travel time of both the trains.

====================

Consider the first train

Speed = Distance/Time

v_{1}=\frac{d_{1}}{t}

v_{1}*t=d_{1}

11.7m/s*t=d_{1}

{\color{Red} 11.7t=d_{1}}-----------(2)

================

Consider the second train

Use formula s=ut+\frac{1}{2}at^{2}

d_{2}=u_{2}t+\frac{1}{2}a_{2}t^{2}

d_{2}=0m/s*t+0.5*0.95m/s^{2}*t^{2}

d_{2}=0*t+0.5*0.95*t^{2}

{\color{Red} d_{2}=0.475t^{2}}--------(3)

====================

Put equation (2) and equation (3) in equation (1)

d_{1}+d_{2}=152

11.7t+0.475t^{2}=152

0.475t^{2}+11.7t-152=0

Solve the quadratic equation using a calculator

{\color{Red} t=9.4024s}

-----------

Put value of t in (2)

11.7t=d_{1}

11.7*9.4024s=d_{1}

ANSWER: {\color{Red} d_{1}=110m}

===========================

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