I need help for this exercise:
Crumple zones in cars increase the time and the distance over the which a car stop during a collision. The length of the crumple zone is 1.6m.
a) Find the maximum speed that the car could have when hitting a wall head-on without subjecting the passager to an acceleration of more than 46.2g (i.e., 46.2 times the acceleration due the gravity)
b) How long would that collision last?

By increasing the time and the distance over which a car stops during a collision, the change in the velocity (acceleration) is reduced, reducing, in consequence, the impact.

The equation to determine the change in velocity for a uniformly accelerated motion is given by the relation between the speed, the distance, and the acceleration:

v² - u² = 2×acceleration × distance

The acceleration must be less than or equal to 46.2g, which is 46.2 × 9.8m/s² = 452.76m/s². Thus, the condition is acceleration ≤ 452.76m/s²

Here, using the maximum acceleration:

u = final velocity = 0

acceleration = 457.76m/s²

distance = 1.6 m

Thus:

v² = 2 (457.76m/s²) × 1.6m

v = 38.06m/s

v = 38 m/s

To determine the time use the impulse notion.

Impulse = Force × Δt

Change in momentum = mass × Δv

Impulse = Change in momentum

Force × Δt = mass × Δv

Force / mass = Δv / Δt

acceleration = Δv / Δt

Since the final velocity is 0 (the car stops), Δv is the speed of the car: Δv = v.

Substitute:

v/Δt = 457.76 m/s²

Δt = v / 457.76 m/s²

Δt = (38 m/s) / (457.76m/s²) = 0.083s