Explanation:
The initial kinetic energy [tex]KE_0[/tex] is
[tex]KE_0 = \frac{1}{2}mv_0^2 = \frac{1}{2}(400\:\text{kg})(10\:\text{m/s})^2[/tex]
[tex]\:\:\:\:\:\;\:= 2×10^4\:\text{J} = 20\:\text{kJ}[/tex]
The final kinetic energy [tex]KE[/tex] is
[tex]KE = \frac{1}{2}mv^2 = \frac{1}{2}(400\:\text{kg})(30\:\text{m/s})^2[/tex]
[tex]\:\:\:\:\:\:\:= 1.8×10^5\:\text{J} = 180\:\text{kJ}[/tex]
The work done W on the car is
[tex]W = \Delta{KE} = KE - KE_0[/tex]
[tex]\:\:\:\:\:\:\:= 180\:\text{kJ} - 20\:\text{kJ} = 1.6×10^5\:\text{J}[/tex]
The power expended P is
[tex]P = \dfrac{W}{t} = \dfrac{1.6×10^5\:\text{J}}{15\:\text{s}} = 10667\:\text{Watts}[/tex]
[tex]\:\:\:\:\:= 10.7\:\text{kW}[/tex]
