A body of mass m starts from rest and is acce | vedclass

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(B) Given that the body starts from rest,initial velocity $u = 0$.Using the equation of motion $v = u + at$,we have $v_1 = 0 + a t_1$,which gives acce

Vedclass · Accepted Answer

$\frac{m v_1^2 t}{t_1^2}$

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(B) Given that the body starts from rest,initial velocity $u = 0$.Using the equation of motion $v = u + at$,we have $v_1 = 0 + a t_1$,which gives acceleration $a = \frac{v_1}{t_1}$.At any time $t$,the velocity of the body is $v(t) = at = \left( \frac{v_1}{t_1} ight) t$.The force acting on the body is $F = ma = m \left( \frac{v_1}{t_1} ight)$.Instantaneous power $P$ is given by $P = F \cdot v$.Substituting the values,$P = \left( m \frac{v_1}{t_1} ight) 	imes \left( \frac{v_1}{t_1} t ight) = \frac{m v_1^2 t}{t_1^2}$.

$A$ body of mass $m$ starts from rest and is accelerated uniformly to a velocity $v_1$ in time $t_1$. The instantaneous power delivered to the body as a function of time $t$ is:

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