Question

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

How much time does it take for the car to come to a full stop? t1 = ?
How much time does it take for the car to accelerate from the full stop to its original speed? t2 = ?
The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v(initial) again?
delta d = ?

Answer 1

Answer:

$t_1 = \frac{v_i}{a_i}$

$t_2 = \frac{v_i}{a_i}$

Δd = $v_it_1 = v_i^2/a_i$

Explanation:

As $v(t) = v_i + at$, when the car is making full stop, $v(t_1) = 0$ . $a = -a_i$ . Therefore,

$0 = v_i - a_it_1\\v_i = a_it_1\\t_1 = \frac{v_i}{a_i}$

Apply the same formula above, with $v(t_2) = v_i$ and $a = a_i$, and the car is starting from 0 speed,  we have

$v_i = 0 + a_it_2\\t_2 = \frac{v_i}{a_i}$

As $s(t) = vt + \frac{at^2}{2}$. After $t = t_1 + t_2$, the car would have traveled a distance of

$s(t) = s(t_1) + s(t_2)\\s(t_1) = (v_it_1 - \frac{a_it_1^2}{2})\\ s(t_2) = \frac{a_it_2^2}{2}$

Hence $s(t) = (v_it_1 - \frac{a_it_1^2}{2}) + \frac{a_it_2^2}{2}$

As $t_1 = t_2$ we can simplify $s(t) = v_it_1$

After t time, the train would have traveled a distance of $s(t) = v_i(t_1 + t_2) = 2v_it_1$

Therefore, Δd would be $2v_it_1 - v_it_1 = v_it_1 = v_i^2/a_i$