JEE MAIN - Physics (2021 - 18th March Morning Shift - No. 26)
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 :-
t2/3
t3/2
t
t1/2
Explanation
$$P = F.v = mav$$
$$P = {{mvdv} \over {dt}}$$
$$\int\limits_0^t {Pdt} = m\int\limits_0^v {vdv} $$
$$Pt = {{m{v^2}} \over 2}$$
$$v = \sqrt {{{2Pt} \over m}} $$
$${{dx} \over {dt}} = \sqrt {{{2Pt} \over m}} $$
$$\int {dx} = \int {\sqrt {{{2Pt} \over m}} } dt$$
$$x \propto {t^{3/2}}$$
$$P = {{mvdv} \over {dt}}$$
$$\int\limits_0^t {Pdt} = m\int\limits_0^v {vdv} $$
$$Pt = {{m{v^2}} \over 2}$$
$$v = \sqrt {{{2Pt} \over m}} $$
$${{dx} \over {dt}} = \sqrt {{{2Pt} \over m}} $$
$$\int {dx} = \int {\sqrt {{{2Pt} \over m}} } dt$$
$$x \propto {t^{3/2}}$$
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