Answer
(a) The car is 100 meters from the intersection when the brakes are applied.
(b) $a = -2.0~m/s^2$
(c) The stopping time is 10.5 seconds.
Work Step by Step
(a) It takes 0.50 seconds to apply the brakes. We can find the distance the car moves in that time.
$d = v~t = (20~m/s)(0.50~s)$
$d = 10~m$
The car is 100 meters from the intersection when the brakes are applied.
(b) $a = \frac{v^2-v_0^2}{2x}$
$a = \frac{0-(20~m/s)^2}{(2)(100~m)}$
$a = -2.0~m/s^2$
(c) We can find the time to stop after the brakes are applied.
$t = \frac{v-v_0}{a} = \frac{0-20~m/s}{-2.0~m/s^2}$
$t = 10~s$
The total time to stop is 10 seconds plus 0.50 seconds for the reaction time. The stopping time is 10.5 seconds.
