## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

(a) The rate of deceleration is $\frac{v_0^2}{2~(d - v_0\cdot t_R)}$ (b) The car will be able to stop in time.
(a) In a time of $t_R$, the car travels a distance of $v_0\cdot t_R$. The remaining distance is $d - v_0\cdot t_R$. We can find the required acceleration. $a = \frac{0-v_0^2}{2~(d - v_0\cdot t_R)}$ The rate of deceleration is $\frac{v_0^2}{2~(d - v_0\cdot t_R)}$ (b) We can use part (a) to find the minimum rate of deceleration required to stop in time. $a = \frac{v_0^2}{2~(d - v_0\cdot t_R)}$ $a = \frac{(21~m/s)^2}{2~[50~m - (21~m/s)\cdot (0.50~s)]}$ $a = 5.6~m/s^2$ Since the car's maximum deceleration of $6.0~m/s^2$ is greater than the minimum required rate of deceleration ($5.6~m/s^2$), the car will be able to stop in time.