Question

A box is moved along a straight line by a machine delivering constant power. The distance moved by the body in time $t$ is proportional to

• A. $t^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$
• B. $t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$
• C. $t^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}$
• D. $t^3$

Answer 1

The correct option is A

$t^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Step 1. Given data

A box is moved along a straight line by a machine delivering constant power.

Step 2. Finding the distance

Let $S$ be the distance moved by the body

By using the formula of power

Power, $P$ is the product of the force and the velocity.

$P=F.v$ [$eq\left(1\right)$]

By using the formula of force, $F$

Force is equal to the product of the mass, $m$ and the acceleration, $a$.

$F=ma$

By using the formula of velocity,

Velocity, $v$ is the ratio of the distance and the time, $t$

$v=\frac{S}{t}$

By using the formula of acceleration,

$$\begin{gathered} a=\frac{v}{t} \\ a=\frac{S}{t^2} \end{gathered}$$

Putting these value in $eq\left(1\right)$, we get

$P=\frac{mS}{t^2}\left(\frac{S}{t}\right)$

$S^2=\frac{P{t}^3}{m}$

From the above equation, we get

$S\propto t^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Hence, the correct option is $\left(A\right)$