A person who is properly restrained by an over-the-shoulder seat belt has a good chance of surviving a car collision if the deceleration does not exceed 30 “g’s’’$\left(\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{g}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{80}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}\right)$. Assuming uniform deceleration at 30 g’s, calculate the distance over which the front end of the car must be designed to collapse if a crash brings the car to rest from 95 km/h.

When an object moves with a uniform acceleration, kinematic equations can be used to study its motion. Suppose an object of mass m starts with velocity u and moves with uniform acceleration a. The kinematics equation of motion relating these variables is $v^2-u^2=2as$.

Here, v is the final velocity of the object.

Given,$1.00\mathrm{g}=9.80\mathrm{m}/{\mathrm{s}}^2$

Thus, acceleration of the car is

$\begin{array}{c}a=-30\mathrm{g}\\ =-\left(30\mathrm{g}\right)\times \left(\frac{9.80{\mathrm{ms}}^{-2}}{1.00\mathrm{g}}\right)\\ =-294\mathrm{m}{\mathrm{s}}^{-2}.\end{array}$

The initial velocity of the car before collapsing is

$\begin{array}{c}u=95{\mathrm{kmh}}^{-1}\\ =\left(95{\mathrm{kmh}}^{-1}\right)\times \left(\frac{1000\mathrm{m}}{1\mathrm{km}}\right)\times \left(\frac{1\mathrm{h}}{3600s}\right)\\ =26.39{\mathrm{ms}}^{-1}.\end{array}$

Final velocity of the car after collapsing,$v=0{\mathrm{ms}}^{-1}$

Use the kinematics equation of motion

$\begin{array}{c}{v}^2-u^2=2as.\\ {\left(0\right)}^2-{\left(26.39{\mathrm{ms}}^{-1}\right)}^2=2\left(-294{\mathrm{ms}}^{-2}\right)s\\ s=\frac{{\left(26.39\right)}^2}{2\times 294}\mathrm{m}\\ =1.18\mathrm{m}\end{array}$

After the collision, the car travels a distance of $1.18\mathrm{m}$before coming to rest.

Thus, the front end of the car must be designed at a distance of 1.18 m.