Question

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

• A. 18.00
• B. 18
• C. 18.0

Answer 1

Answer: (A), (B), (C)

Given, mass of the body, $m=2\text{kg}$
Power delivered by engine, $P=1\text{J/s}$
Time, $t=9$ seconds
Power, $P=FV$
$\Rightarrow P=mav[\because F=ma]$
$\Rightarrow m\frac{dv}{dt}v=P(\because a=\frac{dv}{dt})$
$\Rightarrow vdv=\frac{P}{m}dt$

Integrating both sides we get

$\Rightarrow {\int }_0^{v}vdv=\frac{P}{m}{\int }_0^{t}dt$
$\Rightarrow \frac{v^2}{2}=\frac{Pt}{m}\Rightarrow v={\left(\frac{2Pt}{m}\right)}^{1/2}$
$\Rightarrow \frac{dx}{dt}=\sqrt{\frac{2P}{m}}{t}^{1/2}(\because v=\frac{dx}{dt})$
$\Rightarrow {\int }_0^{x}dx=\sqrt{\frac{2P}{m}}{\int }_0^t{t}^{1/2}dt$
$\therefore \text{Distance,}x=\sqrt{\frac{2P}{m}}\frac{t^{3/2}}{3/2}=\sqrt{\frac{2P}{m}}\times {\frac{2}{3}}^{3/2}$
$\Rightarrow x=\sqrt{\frac{2\times 1}{2}}\times \frac{2}{3}\times 9^{3/2}=\frac{2}{3}\times 27=18$