# A body of mass 2kg is driven by an engine

Question:

A body of mass $2 \mathrm{~kg}$ is driven by an engine delivering a constant power $1 \mathrm{~J} / \mathrm{s}$. The body starts from rest and moves in a straight line. After 9 seconds, the body has moved a distance (in $\mathrm{m}$ )_______.

Solution:

$P=\operatorname{mav}$

$\mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}} \mathrm{v}=\mathrm{P}$

$\int_{0}^{v} v d v=\frac{P}{m} \int_{0}^{t} d t$

$\frac{\mathrm{v}^{2}}{2}=\frac{\mathrm{Pt}}{\mathrm{m}} \Rightarrow \mathrm{v}=\left(\frac{2 \mathrm{Pt}}{\mathrm{m}}\right)^{1 / 2}$

$\frac{\mathrm{v}^{2}}{2}=\frac{\mathrm{Pt}}{\mathrm{m}} \Rightarrow \mathrm{v}=\left(\frac{2 \mathrm{Pt}}{\mathrm{m}}\right)^{1 / 2}$

$\frac{\mathrm{dx}}{\mathrm{dt}}=\sqrt{\frac{2 \mathrm{P}}{\mathrm{m}}} \mathrm{t}^{1 / 2}$

$x=\sqrt{\frac{2 P}{m}} \frac{t^{3 / 2}}{3 / 2}=\sqrt{\frac{2 P}{m}} \times \frac{2}{3} t^{3 / 2}$

$=\sqrt{\frac{2 \times 1}{2}} \times \frac{2}{3} \times 9^{3 / 2}$

$=\frac{2}{3} \times 27=18$