Answer to Question #149043 in Physics for Melanie Belandres

Question

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

By definition, the average deceleration is:

$$a = \dfrac{v_f - v_i}{t} = -\dfrac{ v_i}{t}$$

where $v_f = 0$ is the final speed of the car (stopped to rest), $v_i = 35mi/h\approx 15.6m/s$ is the initial speed of the car, and $t$ is the time needed to stop the car.

The distance travelled by the car is:

$$d = v_it + \dfrac{at^2}{2}$$

If $d = 2.5ft \approx 0.6m$ is the distance, then, substituting the expression for $a$ and expressing the time, obtian:

$$d = v_it - \dfrac{v_it}{2} = \dfrac{v_it}{2}\\ t = 2dv_i\\ t = 2\cdot 0.6\cdot 15.6 = 18.72s$$

The average deceleration is then:

$$a = -\dfrac{ v_i}{t} = -\dfrac{ 15.6}{18.72} \approx 0.83m/s^2$$

Answer. (a) 0.83 m/s^2, (b) 18.72 s.