Answer to Question #171543 in Mechanics | Relativity for John Ivan Cortezano

Question

Karl is driving his car 15 m/s when he noticed that he is approaching a stalled truck that is approximately 35m away. He steps on the breaks and stops 1.0m away from the stalled truck. On what rate dis his car slows down? Assume that Karl reaction time is 2.0s

Answer 1

Let us determine the distance that Karl's car moved after Karl stepped the breaks.

The reaction time is quite big, so the car moved $v\cdot t = 15\cdot 2 = 30$ meters before Karl stepped the breaks. Therefore, the distance from Karl to the truck became $35-30=5$ meters

Also we know that the car stopped at 1 meter from the truck, so the distance of motion after using the breaks is only 4 meters.

We should determine the deceleration

For the motion with constant acceleration the distance is

$s(t) = v_0t + \dfrac{at^2}{2}$

The velocity may be written in form $v(t) = v_0 + at$ , so $t = \dfrac{v(t)-v_0}{a}$

Therefore, $s = \dfrac{v_0(v(t)-v_0)}{a} + \dfrac{(v(t)-v_0)^2}{2a} = \dfrac{v(t)^2-v_0^2}{2a}.$

So $a= \dfrac{v(t)^2-v_0^2}{2s} = \dfrac{0^2-15^2}{2\cdot4}=-28.1\,\mathrm{m/s^2}.$

This is quite large acceleration, due to the small path.