# Force exerted by a seatbelt, on driver, after collision

## 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
Homework Helper
Gold Member
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