Force exerted by a seatbelt, on driver, after collision

[Iciris.0]

Homework Statement

Suppose that the car is stopped from 60mph within right after the collision and the driver (75kg) didn’t push the brake.

Seatbelt is non-stretchable and produces a constant force to stop the driver in 10cm. Calculate the acceleration of the driver and the force that the seatbelt acts on the driver for above two conditions.

Homework Equations

Time for the driver to hit the seatbelt: t=d/v
Force that the seatbelt exerts on the driver: F=m(vo-vt)/t
Acceleration of the driver: a=F/m

The Attempt at a Solution

60mph=27m/s

Time for the driver to hit the seatbelt:
t=d/v = 0.1m/(27m/s) = 3.7ms

Force that the seatbelt exerts on the driver:
F=m(vo-vt)/t = 75kg(27m/s - 0m/s)/3.7ms = 547,297N

Acceleration of the driver:
a=F/m = 547,297N/75kg = 7,297.3 m/s^2

Is this right, the force seems so high?

 

[CWatters]

You appear to be out by a factor of 2...

I would use the equation of motion..

V²= U²+2as

where
V² = 0
U = 60mph = 26.8m/s
s = 0.1m

Solve for "a"

a = -U²/2s
a = - 26.8²/(2*0.1)
= -3591m/s²

Thats about 360 "g"

Then..

F=m * a
= 75 * 3591
= 269,325 N

The high force is why cars have crumple zones at the front so you don't stop in 10cm.