JEE MAIN - Physics (2021 - 20th July Evening Shift - No. 15)
A body at rest is moved along a horizontal straight line by a machine delivering a constant power. The distance moved by the body in time 't' is proportional to :
$${t^{{3 \over 2}}}$$
$${t^{{1 \over 2}}}$$
$${t^{{1 \over 4}}}$$
$${t^{{3 \over 4}}}$$
Explanation
P = constant
$${1 \over 2}$$mv2 = Pt
$$\Rightarrow$$ v $$\propto$$ $$\sqrt t $$
$${{dx} \over {dt}} = C\sqrt t $$ [C = constant]
by integration.
$$x = C{{{t^{{1 \over 2} + 1}}} \over {{1 \over 2} + 1}}$$
$$x \propto {t^{3/2}}$$
$${1 \over 2}$$mv2 = Pt
$$\Rightarrow$$ v $$\propto$$ $$\sqrt t $$
$${{dx} \over {dt}} = C\sqrt t $$ [C = constant]
by integration.
$$x = C{{{t^{{1 \over 2} + 1}}} \over {{1 \over 2} + 1}}$$
$$x \propto {t^{3/2}}$$
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