91Ó°ÊÓ

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

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

The initial velocity of the car before collapsing is

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

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

Use the kinematics equation of motion

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

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.