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

$t^{3/2}$

Explanation for the correct answer:

Option (B): $t^{3/2}$

Step 1. Given data:

Power delivered by machine is , $P$

Force acting on the machine, $F$

As we know,

$Power=Force\times Velocity$

The above equation can be written as,

$\begin{array}{l}\overrightarrow{P}=\overrightarrow{F}.\overrightarrow{V}\\ \Rightarrow \overrightarrow{P}=\left|\overrightarrow{F}\right|\left|\overrightarrow{V}\right|cos\theta \end{array}$

As force and velocity have the same direction the angle between,

$\begin{array}{l}\overrightarrow{F}\end{array}$ and $\begin{array}{l}\overrightarrow{V}\end{array}$ is $0^{\circ }$

Now, $\overrightarrow{P}=m\times a\times v\cos 0^{\circ }$

$\begin{array}{l}\Rightarrow \overrightarrow{P}=\frac{mVdV}{dt}\end{array}$ $\left[a=\frac{dv}{dt}\&\cos 0°=1\right]$

Step 2. Integrating both sides, we get,

$\begin{array}{l}{\int }_0^{t}Pdt=m{\int }_0^{V}VdV\end{array}$

$\Rightarrow \begin{array}{l}Pt=\frac{m{V}^2}{2}\end{array}$

$\Rightarrow \begin{array}{l}V=\sqrt{\frac{2Pt}{m}}\end{array}$

$\Rightarrow \begin{array}{l}\frac{dx}{dt}=\sqrt{\frac{2Pt}{m}}\end{array}$

Again integrating both sides, we get

$\Rightarrow \begin{array}{l}\int dx=\int \sqrt{\frac{2Pt}{m}}dt\end{array}$

$\Rightarrow x=\sqrt{\frac{2P}{m}}\times t^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+C$

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

Hence, the correct option is (B).