Question

Power of the only force acting on a particle of mass $m=1kg$ moving in a straight line depends on its velocity as $P=v^2$ where $v$ is in $m{s}^{-1}$ and $P$ is in $watt$ . If the initial velocity of the particle is $1m{s}^{-1}$ , then the displacement of the particle in $ln(2)$ seconds will be

• A. $\left(ln2-1\right)m$
• B. ${\left(ln2\right)}^{2}m$
• C. $1m$
• D. $2m$

Answer 1

Answer: (C)

$P=Fv$
$v^2=Fv$ ( since $P=v^2$ )
therefore $F=v$
$ma=v$
$1\times v\frac{dv}{dx}=v$
${\int }_1^{v}dv={\int }_0^{x}dx$
$v-1=x$
$v=x+1$
$\frac{dx}{dt}=x+1$
${\int }_0^x\frac{dx}{x+1}={\int }_0^{t}dt$
$ln\left(x+1\right)-ln\left(0+1\right)=t$
$x+1=e^t$
$x=e^t-1$
$x=e^t-1$
At $t=ln2$
$x=2-1=1m$