Question

A 75.0-kg person is riding in a car moving at 20.0 m/s when the car runs into a bridge abutment. (a) calculate the average force on the person if he is stopped by a padded dashboard that compresses an average of 1.00 cm. (b) calculate the average force on the person if he is stopped by an air bag that compresses an average of 15.0 c

Answer 1

(a) $-1.5\cdot 10^6 N$

First of all, we need to calculate the acceleration of the person, by using the following SUVAT equation:

$v^2 -u ^2 = 2ad$

where

v = 0 is the final velocity

u = 20.0 m/s is the initial velocity

a is the acceleration

d = 1.00 cm = 0.01 m is the displacement of the person

Solving for a,

$a=\frac{v^2-u^2}{2d}=\frac{0-20.0^2}{2(0.01)}=-20000 m/s^2$

And the average force on the person is given by

$F=ma$

with m = 75.0 kg being the mass of the person. Substituting,

$F=(75)(-20000)=-1.5\cdot 10^6 N$

where the negative sign means the force is opposite to the direction of motion of the person.

b) $-1.0\cdot 10^5 N$

In this case,

v = 0 is the final velocity

u = 20.0 m/s is the initial velocity

a is the acceleration

d  = 15.00 cm = 0.15 m is the displacement of the person with the air bag

So the acceleration is

$a=\frac{v^2-u^2}{2d}=\frac{0-20.0^2}{2(0.15)}=-1333 m/s^2$

So the average force on the person is

$F=ma=(75)(-1333)=-1.0\cdot 10^5 N$