A car weighs 3000 lbm, and is travelling 60 mph when it has to make an emergency stop. The car comes to a stop 5 seconds after the brakes are applied.
A) Assuming the rate of deceleration is constant, what force is required?
B) Assuming the rate of deceleration is constant, how much distance is covered before the car comes to a stop?
A) Force = mass × acceleration
60 \operatorname{mph}\left(\frac{1~ \mathrm{hr}}{3600~ \mathrm{sec}}\right)\left(\frac{5280 ~\mathrm{ft}}{1 ~\mathrm{mile}}\right)=88 \frac{\mathrm{ft}}{\mathrm{sec}}
Acceleration =\frac{\text { velocity }_{\text {final }}-\text { velocity }_{\text {initial }}}{\text { time }}
\begin{aligned}& a=\frac{88 \frac{\mathrm{ft}}{\mathrm{sec}}-0}{5 \,\mathrm{sec}}\\& \mathrm{a}=17.6 \frac{\mathrm{ft}}{\mathrm{sec}^{2}}\\& \rm Force =17.6 \frac{\mathrm{ft}}{\mathrm{sec}^{2}} \times 3000~ \mathrm{lb}_{\mathrm{m}}\\& \rm Force =\bf 52500 \frac{lb_{m} f t}{\bf sec ^{2}}\end{aligned}
B) \begin{aligned}\text{Position}_{\text {final }}=\frac{\text { acceleration } * \text { time }^2}{2}+ \text{velocity} _{\text {inital }} \times \text{time} + \text{position} _{\text {intial }}\end{aligned}
\rm Position_{\mathrm{final}}=\frac{\left(-17.6 \frac{\mathrm{ft}}{\mathrm{sec}^{2}}\right)(5 \mathrm{sec})^{2}}{2}+\left(88 \frac{\mathrm{ft}}{\mathrm{sec}}\right)(5~ \mathrm{sec})+0
\rm Position _{\text {final }}=\bf 220~ \mathbf{\ ft}