Question

A body of mass $m$ accelerates uniformly from rest to velocity ${v_1}$ in time interval ${T_1}$ . The instantaneous power delivered to the body as a function of time $t$ is:
(A) $m\dfrac{{v_1^2}}{{T_1^2}}t$
(B) $m\dfrac{{{v_1}}}{{T_1^2}}t$
(C) ${\left( {m\dfrac{{{v_1}}}{{{T_1}}}} \right)^2}t$
(D) $m\dfrac{{v_1^2}}{{{T_1}}}{t^2}$

Answer 1

The force acting on the body delivering the power can be quantified using the mass and acceleration as in Newton’s law. The velocity to be used in the power formula is the instantaneous velocity after a particular time.

Formula used: In this solution we will be using the following formulae;
$P = Fv$ where $P$ is the power delivered to a body, $F$ is the force exerted on the body and $v$ is the instantaneous velocity of the body.
$F = ma$ where $m$ is the mass of a body and $a$ is the acceleration of the body.
$a = \dfrac{{v - u}}{t}$ where $v$ is the final velocity of an accelerating body, $u$ is the initial velocity, and $t$ is the time taken to accelerate from the initial velocity to the final velocity.

Complete Step-by-Step solution
To find the instantaneous power, we shall note that the instantaneous power delivered to a body can be defined by
$P = Fv$ where $F$ is the force exerted on the body and $v$ is the instantaneous velocity of the body.
The force acting on the body can be calculated from
$F = ma$ where $m$ is the mass of a body and $a$ is the acceleration of the body, and acceleration is given by
$a = \dfrac{{v - u}}{t}$ (since $u = 0$ ) where $v$ is the final velocity and $u$ is the initial velocity, and $t$ is the time taken to accelerate from the initial velocity to the final velocity.
Hence,
$a = \dfrac{{{v_1}}}{{{T_1}}}$ (since $u = 0$ )
Now, the velocity after any particular time $t$ can thus be given as
$v = at$
$\Rightarrow v = \dfrac{{{v_1}}}{{{T_1}}}t$
By inserting all known expressions into the power formula
$P = Fv = mav = m\dfrac{{{v_1}}}{{{T_1}}}\left( {\dfrac{{{v_1}}}{{{T_1}}}t} \right)$
$\Rightarrow P = m\dfrac{{v_1^2}}{{T_1^2}}t$
Hence, the correct option is A.

Note
Alternatively, we could use a simple unit check, to check which of the options are dimensionally equal to power as all options have different dimensions.
For option A, we have
$m\dfrac{{v_1^2}}{{T_1^2}}t$ , hence by unit,
$kg\dfrac{{{{\left( {m{s^{ - 1}}} \right)}^2}}}{{{s^2}}}s = \dfrac{{kg{m^2}{s^{ - 2}}}}{s}$
$Nm{s^{ - 1}}$
Which is the same as the unit of force times velocity, which is the formula for velocity.
Hence, option A is the answer.