# Question #020f1

Jan 8, 2017

Velocity of the car just brfore its impact with the stationary truck is ${v}_{i} = 2 \text{m/s}$ The car comes to rest finally, So its final velocity ${v}_{f} = 0 \text{m/s}$.The car travels $5 m$ after its impact.Hence $s = 5 m$

Using equation of kinematics we can write

${v}_{f}^{2} = {v}_{i}^{2} - 2 a s$,where $a$ is the average retardation produced.

${0}^{2} = {20}^{2} - 2 a \times 5 \implies a = 40 \text{m/} {s}^{2}$

So the force exerted by the seat belt on the passenger of mass $m = 90 k g$ will be

$F = m a = 90 \times 40 N = 3600 N$