Question

Power of the only force acting on a particle of mass $m=1 \, kg$ 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 $1 \, m \, s^{- 1}$ , then the displacement of the particle in $ln\left(\right.2\left.\right)$ seconds will be

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

Answer 1

Answer: (C)

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