An advertisement claims that a certain 1200 -kg car can accelerate from rest to a speed of $25 \mathrm{~m} / \mathrm{s}$ in a time of $8.0 \mathrm{~s}$. What average power must the motor develop to produce this acceleration? Give your answer in both watts and horsepower. Ignore friction losses.
The work done in accelerating the car is
$$
\text { Work done }=\text { Change in } \mathrm{KE}=\frac{1}{2} m\left(v_{f}^{2}-v_{i}^{2}\right)=\frac{1}{2} m v_{f}^{2}
$$
The time taken for this work to be performed is $8.0 \mathrm{~s}$. Therefore, to two significant figures,
$$
\text { Power }=\frac{\text { Work }}{\text { Time }}=\frac{\frac{1}{2}(1200 \mathrm{~kg})(25 \mathrm{~m} / \mathrm{s})^{2}}{8.0 \mathrm{~s}}=46875 \mathrm{~W}=47 \mathrm{~kW}
$$
Converting from watts to horsepower, we have
$$
\text { Power }=(46875 \mathrm{~W})\left(\frac{1 \mathrm{hp}}{746 \mathrm{~W}}\right)=63 \mathrm{hp}
$$