Question

A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to

Answer 1

Step 1: Given data

A body is moving unidirectionally under the influence of a source of constant power.

Step 2: Formula used

$P=\frac{W}{t}[whereP=power,W=workdone,t=time]$

Step 3: Calculating displacement

Now, the total work done by this force is equal to the product of the magnitude of applied force and the distance travelled by the body.

$W=Fd$

Substituting the value of work done in the power formula

$P=\frac{Fd}{t}$

Now as the velocity is given by $v=\frac{d}{t}[d=dis\tan ce,t=time]$ and the force is given by $F=ma[m=mass,a=acceleration]$

The power can be expressed as $P=mav$

Finally, we substitute $v=at$ to obtain $P=m{a}^{2}t$

$a=\sqrt{\frac{P}{mt}}......\left(i\right)$

Now Newton’s first equation of motion gives the displacement of the body as $d=ut+\frac{1}{2}a{t}^2............\left(ii\right)$

We assume that the body was initially at rest, i.e., $u=0$.
So equation (ii) becomes $d=\frac{1}{2}a{t}^2.......\left(iii\right)$

Substituting equation (i) and (iii) we get, $d=\frac{1}{2}\left(\sqrt{\frac{P}{mt}}\right)t^2$

On simplifying we get,

$$\begin{gathered} d=\frac{1}{2}\sqrt{\frac{P}{m}}\times t^{\left[2-\frac{1}{2}\right]} \\ d=\frac{1}{2}\sqrt{\frac{P}{m}}\times t^{3/2} \\ d\propto t^{3/2} \end{gathered}$$

Hence, its displacement is directly proportional to time $t^{3/2}$.