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

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A

$t^{3/4}$

B

$t^{3/2}$

C

$t^{1/4}$

D

$t^{1/2}$

Solution:

Power $= \frac{\text{Work}}{ Time } = \frac{\text{Force} \times \text{distance}}{\text{Time}} = \text{Force} \times \text{velocity}$
$\therefore \text{Force}\times \text{velocity}=\text{constant }\left(\right.K\left.\right)$
or $\left(\right.m a\left.\right)\left(\right.at⁡\left.\right)=K⁡$
or $a=\left(\right.\frac{K }{m ⁡ t ⁡}\left(\left.\right)^{1 / 2}$
$\because s =\frac{1}{2}at⁡^{2}$
$\therefore s =\frac{1}{2}\left(\right.\frac{K ⁡}{m ⁡ t ⁡}\left(\left.\right)^{1 / 2}t⁡^{2}=\frac{1}{2}\left(\right.\frac{K ⁡}{m ⁡}\left(\left.\right)^{1 / 2}t^{3 / 2}$
or $s$ is proportional to $t^{3/2}$ .

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