## College Physics (4th Edition)

The engine's average power output is $9.30\times 10^5~Watts$
We can find the drag racer's change in kinetic energy: $KE = \frac{1}{2}mv^2 = \frac{1}{2}(500.0~kg)(125~m/s)^2 = 3.906\times 10^6~J$ The engine does this much work in a time of 4.2 seconds. We can find the power output: $Power = \frac{Energy}{Time} = \frac{3.906\times 10^6~J}{4.2~s} = 9.30\times 10^5~Watts$ The engine's average power output is $9.30\times 10^5~Watts$.