The front 1.20 m of a 1,550-kg car is designed as a "crumple zone" that collapses to absorb the shock of a collision. (a) If a car traveling 24.0 m/s stops uniformly in 1.20 m, how long does the collision last

t=0.1seconds.

Explanation:

The mass of the car m = 1,550 kg

We know the stopping distance, 1.20m,

we know the final velocity, 0m/s (its stopped),

the starting velocity, 24m/s.

d=t (v2+v1) / 2

Rearange to solve for t, and remove the v2^2 as its zero

t=2d/v1

t=2 (1.20m) / (24m/s)

t=0.1seconds.

t = 0.1 s

Explanation:

Given:

- The distance crushed (Stopping-Distance) s = 1.20 m

- The mass of the car m = 1,550 kg

- The initial velocity vi = 24.0 m/s

Find:

- How long does the collision last t?

Solution:

- The stopping distance s is the average velocity v times time t as follows:

s = t * (vf + vi) / 2

Where,

vf = 0 m/s (Stopped)

s = t * (vi) / 2

t = 2*s / vi

t = 2*1.20 / 24

t = 0.1 s