Question

A body of mass $m$ accelerates uniformly from rest to a velocity ${{v}_{1}}$ in time interval ${{T}_{1}}$. The instantaneous power delivered to body as a function of time $t$ is
$A)\text{ }\dfrac{m{{v}_{1}}^{2}}{{{T}_{1}}^{2}}t$
$B)\text{ }\dfrac{m{{v}_{1}}}{{{T}_{1}}^{2}}t$
$C)\text{ }{{\left( \dfrac{m{{v}_{1}}}{{{T}_{1}}} \right)}^{2}}t$
$D)\text{ }\dfrac{m{{v}_{1}}^{2}}{{{T}_{1}}}{{t}^{2}}$

Answer 1

This problem can be solved by finding out the acceleration and the instantaneous velocity of the body at a specific time. The instantaneous power can be found out as the product of the instantaneous force and the instantaneous velocity of a body in one dimensional problems.

Formula used:
$F=ma$
$P=Fv$
$v=u+at$

Complete step-by-step answer:
We will try to find out the acceleration and the instantaneous velocity of the body at a specific time and form that write the expression for the instantaneous power delivered to the body.
The instantaneous power $P$ delivered to a body is the product of the instantaneous force $F$ and the instantaneous velocity $v$ of the body in that direction.
$\therefore P=Fv$ --(1)
Also, the force $F$ on a body of mass $m$ which produces an acceleration $a$ can be written as
$F=ma$ --(2)
For bodies moving with a constant acceleration $a$, the velocity $v$ at some time $t$ can be written as
$v=u+at$ --(3)
Where $u$ is the initial velocity of the body at time $t=0$.
Now, let us analyze the question.
The mass of the body is $m$.
Let the uniform acceleration of the body be $a$.
Since the body starts from rest, its initial velocity was $u=0$.
It reaches a velocity ${{v}_{1}}$ in time ${{T}_{1}}$.
Let the instantaneous velocity at any time $t$ be $v$.
Let the instantaneous force on the body at this time $t$ be $F$.
Let the instantaneous power delivered to the body at time $t$ be $P$.
Therefore, using (3), we get
${{v}_{1}}=0+a{{T}_{1}}=a{{T}_{1}}$
$\therefore a=\dfrac{{{v}_{1}}}{{{T}_{1}}}$ --(4)
Also, using (3), we get
$v=0+at=at$ --(5)
Putting (4) in (5), we get
$v=\dfrac{{{v}_{1}}}{{{T}_{1}}}t$ --(6)
Now, using (2), we get $F$ as
$F=ma$
Putting (4) in the above equation, we get
$F=m\dfrac{{{v}_{1}}}{{{T}_{1}}}$ --(7)
Now, using (1), we get the instantaneous power $P$ as
$P=Fv$
$\therefore P=m\dfrac{{{v}_{1}}}{{{T}_{1}}}\times \dfrac{{{v}_{1}}}{{{T}_{1}}}t=\dfrac{m{{v}_{1}}^{2}}{{{T}_{1}}^{2}}t$
Hence, we have got the expression for the instantaneous power delivered to the body at a time $t$ as $\dfrac{m{{v}_{1}}^{2}}{{{T}_{1}}^{2}}t$.

So, the correct answer is “Option A”.

Note: Students must note that since the acceleration is given to be uniform in the question, we could use the equation of motion for constant acceleration, that is, formula (3). However, if in the question, it is given that the acceleration is also changing as a function of some other physical quantity, for example time, then students must find the instantaneous force using the expression for instantaneous acceleration and not constant acceleration. If they do not do so, they will arrive at a completely wrong expression.