Answer
(a) $7.8s$
(b) less than
(c) $3.9s$
Work Step by Step
(a) We can find the required time as follows:
$P=\frac{\Delta K.E}{t}$
$\implies P=\frac{\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2}{t}=\frac{\frac{1}{2}m(v_f^2-v_i^2)}{t}$
This can be rearranged:
$t=\frac{\frac{1}{2}m(v_f^2-v_i^2)}{P}$
We plug in the known values to obtain:
$t=\frac{\frac{1}{2}[(25)^2-(15)^2]}{48hp\times 746W/hp}$
$t=7.8s$
(b) We know that $t\propto v_f^2-v_i^2$. Thus, the time required to increase the speed from $5.0m/s$ to $15.0m/s$ will be less than than that found in part (a).
(c) We can find the required time as follows:
$t=\frac{\frac{1}{2}m(v_f^2-v_i^2)}{P}$
We plug in the known values to obtain:
$t=\frac{\frac{1}{2}[(15)^2-(5)^2]}{48hp\times 746W/hp}$
$t=3.9s$
