When the brakes of a car are applied, the car decelerates at a constant rate of $10\;{\rm{m/s}}^2$. If the car is to stop within $20 {\rm m}$ after the brakes are applied, what is the maximum velocity the car could have been traveling? Express the answer in miles per hour.

Solution Let $s(t)$ represent the distance $s$ in meters the car has traveled $t$ seconds after the brakes are applied. Let $v_{0}$ be the velocity of the car at the time the brakes are applied ($t=0$). Since the car decelerates at the rate of $10\;{\rm{m/s}}^2$, its acceleration $a,$ in meters per second squared, is $${a}(t) =\dfrac{{d\kern-1ptv}}{dt}={-10}$$

We solve the differential equation for $v$. $$v(t)=-10t+C_{1}$$

When $t=0$, $v(0)=v_{0}$, the velocity of the car when the brakes are applied, so $C_{1}=v_{0}.$ Then $${v(t)={\frac{ds}{dt}}}={-10t+v_{0}}$$

Now we solve the differential equation ${v(t)={\dfrac{ds}{dt}}}$ for $s$. $$s(t)= -5t^{2}+v _{0}t+C_{2}$$

Since the distance $s$ is measured from the point at which the brakes are applied, the second initial condition is $s( 0) =0$. Then $s( 0) =-5\cdot 0+v_{0}\cdot 0+C_{2}=0$, so $C_{2}=0$. The distance $s$, in meters, the car travels $t$ seconds after applying the brakes is $$s(t)=-5t^{2}+v_{0}t$$

The car stops completely when its velocity equals $0$. That is, when $$\begin{eqnarray*} v(t) ={-10t+v_{0}} &=&0 \\[3pt] t &=&{\frac{v_{0}}{10}} \end{eqnarray*}$$

This is the time it takes the car to come to rest. Substituting $\dfrac{v_{0}}{10}$ for $t$ in $s(t)$, the distance the car has traveled is $$s\left( {\frac{v_{0}}{10}}\right) =-5\left( {\frac{v_{0}}{10}}\right) ^{2}+v_{0}\left( {\frac{v_{0}}{10}}\right) =\frac{v_{0}^{2}}{20}\hbox{ }$$

If the car is to stop within 20 m, then $s\leq 20$; that is, $\dfrac{v_{0}^{2}}{20}\leq 20$ or equivalently $v_{0}^{2}\le 400$. The maximum possible velocity $v_{0}$ for the car is $v_{0}=20 \ {\rm m}/{\rm s}$.

To express this in miles per hour, we proceed as follows: $$\begin{eqnarray*} v_{0}=20\,{\rm{m}/{s}}&=&\left( \frac{20{{\rm m}}}{{{\rm s}}}\right)\, \left( \frac{1{\rm km}}{1000\,{{\rm m}}}\right)\, \left( \frac{3600 {{\rm s}} }{1\,{\rm h}}\right) \\[3pt] &=& 72{\rm km}/{\rm h}\approx \left( \frac{72\,{\rm km} }{{\rm h}}\right)\, \left( \frac{1\,{\rm mi}}{1.6\,{\rm km}}\right) =45 {\rm mi}/{\rm h} \end{eqnarray*}$$

The maximum possible velocity to stop within $20$ m is $45$ mi/h.