Question

A car of mass m starts from rest and accelerates so that the instantaneous power delivered to the car has a constant magnitude ${{P}_{0}}$ . The instantaneous velocity of this car is proportional to –
A. ${{t}^{\dfrac{1}{2}}}$
B. ${{t}^{-\dfrac{1}{2}}}$
C. $t/\sqrt{m}$
D. ${{t}^{2}}{{P}_{0}}$

Answer 1

Hint: To solve this question we need to express the instantaneous velocity in terms of time t. Obtain power in terms of force and velocity. Then express it in terms of mass and velocity and integrate the equation. After integration, we will get the instantaneous velocity in terms of the time and we can define how velocity is proportional to time.

Complete step by step solution:
Power can be defined as the product of force and velocity.
${{P}_{0}}=\vec{F}.\vec{v}$
If the force and velocity are in the same direction the magnitude of power can be given as,
${{P}_{0}}=Fv$
Expressing the force as the product of mass and acceleration we get,
${{P}_{0}}=mav$
Expressing the acceleration as the rate of change of velocity per unit time,
$\begin{align}
  & {{P}_{0}}=m\left( \dfrac{dv}{dt} \right)v \\
 & {{P}_{0}}dt=mvdv \\
\end{align}$
Integrating on both side of the above equation, we get that
$\begin{align}
  & \int{{{P}_{0}}dt=\int{mvdv}} \\
 & {{P}_{0}}t=m\dfrac{{{v}^{2}}}{2} \\
 & {{v}^{2}}=\dfrac{2{{P}_{0}}t}{m} \\
 & v=\sqrt{\dfrac{2{{P}_{0}}t}{m}} \\
\end{align}$
So, we can write,
$v\propto {{t}^{\dfrac{1}{2}}}$
We can say that the instantaneous velocity of the car is directly proportional to ${{t}^{\dfrac{1}{2}}}$


The correct option is (A).




Note: The instantaneous power can be defined as the power at a given instant of time. Instantaneous power can be defined as the limiting value of average power. Average power is defined as the ratio of the total work done to the total time.
Average power is given by the mathematical expression,
${{P}_{av}}=\dfrac{\Delta W}{\Delta t}$
If we limit the time approximately to be zero, we get the instantaneous power i.e. the power of the system at an instant.
If the work done is uniform, then the average power and instantaneous power becomes equal.
Then we can write,
$P=\dfrac{\Delta W}{\Delta t}$