Question

A constant power delivering machine has towed a box, which was initially at rest, along a horizontal straight line. The distance moved by the box in time $t$ is proportional to:

• A. $t$
• B. $t^{3/2}$
• C. $t^{1/2}$
• D. $t^{2/3}$

Answer 1

The correct option is B $t^{3/2}$

$\overrightarrow{P}=\overrightarrow{F}.\overrightarrow{v}=Fv\cos \theta$

As force and velocity has same direction the angle between
$\overrightarrow{F}$ and $\overrightarrow{v}$ is $0^{\circ }$. Becuae it starts from rest and moves along a straight line.

$P=Fv=mav$

$P=\frac{mdv}{dt}v$

${\int }_0^{t}Pdt=m{\int }_0^{v}vdv$

$Pt=\frac{m{v}^2}{2}$

$v=\sqrt{\frac{2Pt}{m}}$

Now, $\frac{dx}{dt}=v=\sqrt{\frac{2Pt}{m}}$

$\int dx=\int \sqrt{\frac{2Pt}{m}}dt$

$\Rightarrow x\propto t^{3/2}$