JEE MAIN - Physics (2005 - No. 65)
A body of mass $$m$$ is accelerated uniformly from rest to a speed $$v$$ in a time $$T.$$ The instantaneous power delivered to the body as a function of time is given by
$${{m{v^2}} \over {{T^2}}}.{t^2}$$
$${{m{v^2}} \over {{T^2}}}.t$$
$${1 \over 2}{{m{v^2}} \over {{T^2}}}.{t^2}$$
$${1 \over 2}{{m{v^2}} \over {{T^2}}}.t$$
Explanation
$$u = 0;v = u + aT;v = aT$$
Instantaneous power $$ = F \times v = m.\,a.\,at = m.{a^2}.t$$
$$\therefore$$ Instantaneous power $$ = {{m{v^2}t} \over {{T^2}}}$$
Instantaneous power $$ = F \times v = m.\,a.\,at = m.{a^2}.t$$
$$\therefore$$ Instantaneous power $$ = {{m{v^2}t} \over {{T^2}}}$$
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