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

Question

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

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1 year 2021-09-01T21:28:12+00:00 2 Answers 14 views 0

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

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.