Question

A body of mass m accelerates uniformly from rest to ${v_1}$ in time ${t_1}$ As a function of t, the instantaneous power delivered to the body is:
A). $\dfrac{{m{v_1}t}}{{{t_2}}}$
B). $\dfrac{{m{v^2}_1t}}{{{t_2}}}$
C). $\dfrac{{m{v_1}{t^2}}}{{{t_2}}}$
D). $\dfrac{{m{v_1}^2t}}{{{t_2}^2}}$

Answer 1

Hint- This problem is related to the kinematics in which it’s given that acceleration is uniform therefore, the three equations of motion will be valid and used here. We will proceed by finding the final velocity and acceleration at the given time and then will find the power at that time.

Complete step-by-step solution -
And one more formula of power will be used i.e.
$\text{Power} = \text{force} \times \text{velocity}$.
Where, instantaneous power is defined as the instantaneous work done per unit time.
Let m be the mass of the object
a be the acceleration of the object
And v be the velocity of the object at time ${t_1}$
According to the equation of motions
$v = u + at$
It is given in the question that the initial velocity of the object is zero.
$u = 0$
Substituting this value in the above equation
$v = at............\left( 1 \right)$
Also we know that the power is given as
$P = FV...............\left( 2 \right)$
Also $F = ma$
$v = at$
Using equation (1) and (2)
Therefore
$P = ma \times at \\
  P = m{a^2}t \\ $
Also, At time $t_1$ velocity is ${v_1}$
$
  {v_1} = a{t_1} \\
  a = \dfrac{{{v_1}}}{{{t_1}}} \\
$
Now substituting this value of acceleration in above equation
$
  P = m{\left( {\dfrac{{{v_1}}}{{{t_1}}}} \right)^2}t \\
  P = \dfrac{{m{v_1}^2t}}{{{t_1}^2}} \\
$
Hence, the correct option is “D”.

Note- In order to solve these types of questions, you must remember the laws of motion and the equation of motions. Also keep in mind that if the force or acceleration is not uniform then we can’t use the equations of motion directly, in that case we have to use the integral format of equations of motion. Equations of motion can only be applied when we have constant acceleration.