The front 1.20 m of a 1,350-kg car is designed as a "crumple zone" that collapses to absorb the shock of a collision. (a) If a car traveling 30.0 m/s stops uniformly in 1.20 m, how long does the collision last

0.08s

Explanation:

From the question:

Initial velocity (u) = 30m/s

Final velocity (v) = 0

Distance (s) = 1.2m

Calculating the acceleration of the car using the third equation of motion ;

v^2 = u^2 + 2as

a = acceleration, s= distance

0 = 30^2 + (2 × a × 1.2)

0 = 900 + (2.4 × a)

a = (- 900) ÷ 2.4

a = - 375 m/s^2

Therefore, time taken

v = u + at

Where, t = time taken

0 = 30 + (-375 × t)

0 = 30 - 375 × t

375 × t = 30

t = 30 ÷ 375

t = 0.08 seconds

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