Solving a Rectilinear Motion Problem

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.