Question

1.     The front 1.20 m of a 1,400-kg car is designed as a “crumple zone” that collapses to absorb the shock of a collision. If a car traveling 25.0 m/s stops uniformly in

1.20 m, (a) how long does the collision last, (b) what is the magnitude of the average force on the car, and (c) what is the acceleration of the car? Express the acceleration as a multiple of the acceleration due to gravity.

Answer 1

(a) We can find the duration of the collision as follows:

$$d=\dfrac{1}{2}(v_i+v_f)\Delta t,$$ $$\Delta t=\dfrac{2d}{v}=\dfrac{2\cdot1.20\ m}{25\ \dfrac{m}{s}}=0.096\ s.$$

(b) We can find the magnitude of the average force on the car from the formula:

$$F_{avg}\Delta t=m|(v_f-v_i)|,$$ $$F_{avg}=\dfrac{m|(v_f-v_i)|}{\Delta t},$$ $$F_{avg}=\dfrac{1400\ kg\cdot|(0-25\ \dfrac{m}{s})|}{0.096\ s}=3.64\cdot10^{5}\ N.$$

(c) We can find the acceleration of the car from the Newton's Second Law of Motion:

$$a=\dfrac{F_{avg}}{m}=\dfrac{3.64\cdot10^{5}\ N}{1400\ kg}=260\dfrac{m}{s^2}\cdot\dfrac{1\ g}{9.8\ \dfrac{m}{s^2}}=26.5g.$$