Question

A car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver notices a red light ahead and slows down with constant acceleration −a0. Just as the car comes to a full stop, the light immediately turns green, and the car then accelerates back to its original speed v0 with constant acceleration a0. During the same time interval, the train continues to travel at the constant speed v0.

Answer 1

Answer:

A)The time taken for the car to come to a full stop is:

$t=\displaystyle \frac{v_0}{a_0}$

b) The time taken for the car to accelerate from full stop to  its original cruising speed is:

$t=\displaystyle \frac{v_0}{a_0}$

c) The separation distance between the car and the train is:

$d=v_0^2/a_0$

Completed question:

a) How much time does it take for the car to come to a full stop? Express your answer in terms of v0 and a0.

b) How much time does it take for the car to accelerate from the full stop to its original cruising speed? Express your answer in terms of v0 and a0.

c) The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed again? Express the separation distance in terms of v0 and a0. Your answer should be positive.

Explanation:

a) The car slows down with constant acceleration (-a₀). Therefore, this movement is a linearly accelerated motion:

$v(t)=v_0+t\cdot (-a_0)\\0=v_0+t\cdot (-a_0)\\t=v_0/a_0$

b) for the acceleration process we use the same equation than before:

$v(t)=v_{orig}+t\cdot (a_0)\\v_0=0m/s+t\cdot (a_0)\\t=v_0/a_0$

c) To determine the separation distance between the car and the train we can observe how much distance each of them travels in the time spend for the car to deaccelerate and accelerate again.

For the train:

$x(t_{tot})=x_0+v_0\cdot(t_{tot})\\x(t_{tot})=v_0\cdot(t_{dea}+t_{acc})\\x(t_{tot})=2v_0^2/a_0$

For the car:

$x(t_{tot})=x(t_{dea})+x(t_{acc})\\x(t_{tot})=v_0\cdot(t_{dea})+0.5(-a)(t_{dea})^2+0.5a(t_{acc})^2\\x(t_{tot})=v_0\cdot(t_{dea})\\x(t_{tot})=v_0^2/a_0$

Therefore the separation distance between the car and the train is:

$d=|x_{car}-x_{train}|=v_0^2/a_0$