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

d = t (V2+V1) / 2

t = 2d / (V2+V1)

t = 2 (1.2) / (0+25)

t = 0.096S

Explanation:

The stopping distance is known to be 1.20m. the final velocity is 0 meters per second and the starting or initial velocity is 25 meters per second.

Answer: 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