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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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Physics
1 year
2021-09-01T21:28:12+00:00
2021-09-01T21:28:12+00:00 2 Answers
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Answers ( )
Answer:
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
Answer:
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.