Answer
a) $\overline P=3.28\times10^4W$
b) $\overline P=5.1\times10^4W$
Work Step by Step
We have $v_0=0$, $v_f=20m/s$, and $t=5.6s$, so $$a=\frac{v_f-v_0}{t}=3.57m/s^2$$ $$s=\Big(\frac{v_0+v_f}{2}\Big)t=56m$$
According to Newton's 2nd law, the force required to accelerate the car is$$F=ma=3.57m$$
The weight of the car does not contribute to the work done to accelerate the car, since it is perpendicular with the car's movement. So we calculate $W$ out of only force $F$.
The average power required to accelerate the car, therefore, is $$\overline P = \frac{W}{t}=\frac{(F\cos\theta)s}{t}$$
Force F is applied along the car's motion, so $\cos\theta=\cos0=1$ $$\overline P =\frac{Fs}{t}=\frac{3.57m\times56}{5.6}=35.7m$$
The final job is to find the mass of the car in each case.
a) $m=\frac{W}{9.8}=\frac{9\times10^3}{9.8}=918.4kg$
$\overline P=3.28\times10^4W$
b) $m=\frac{W}{9.8}=\frac{1.4\times10^4}{9.8}=1428.6kg$
$\overline P=5.1\times10^4W$
