Question

6. A car traveling at 35 mi/hr is stopped in a distance of 2.5 ft when it hits a tree. (a) Calculate the average deceleration of the car in m/s2. (b) Calculate the time needed to stop the car (assuming constant deceleration).

Answer 1

(a) We can find the deceleration of the car from the kinematic equation:

$$v^2=v_0^2+2ad,$$ $$a=\dfrac{v^2-v_0^2}{2d},$$ $$a=\dfrac{0-(35\ \dfrac{mi}{hr}\cdot \dfrac{1609.34\ m}{1\ mi}\cdot \dfrac{1\ hr}{3600\ s})^2}{2\cdot 2.5\ ft\cdot \dfrac{0.3048\ m}{1\ ft}}=-160.6\ \dfrac{m}{s^2}.$$

The sign minus means that the car decelerates.

b) We can find time needed to stop the car from the another kinematic equation:

$$v=v_0+at,$$ $$t=\dfrac{-v_0}{a}=\dfrac{-35\ \dfrac{mi}{hr}\cdot \dfrac{1609.34\ m}{1\ mi}\cdot \dfrac{1\ hr}{3600\ s}}{-160.6\ \dfrac{m}{s^2}}=0.1\ s.$$

Answer:

(a) $a=-160.6\ \dfrac{m}{s^2},$ car decelerates.

(b) $t=0.1\ s.$