Question

A car of mass m starts from rest and accelerates so that the instantaneous power delivered to the car has a constant magnitude P0. The instantaneous velocity of this car is proportional to-
A. \[{t^{\dfrac{1}{2}}}\]
B. \[{t^{ - \dfrac{1}{2}}}\]
C. \[\dfrac{t}{{\sqrt m }}\]
D. \[{t^2}{P_0}\]

Answer 1

Hint:In this question mass and instantaneous power of a car is given to us so we will use the instantaneous power formula then we will write this formula in terms of velocity and by further integrating the equation with respect to time and the velocity we will find the instantaneous velocity of this car.

Formula Used:
Instantaneous power is given by the formula \[{P_{instantaneous}} = F \cdot v\].

Complete step by step answer:
Given, the mass of the car is \[ = m\]. Instantaneous power delivered to the car \[ = {P_0}\].Now as we know instantaneous power is the product of the force and the velocity, hence we can write \[{P_0} = Fv - - (i)\]
Where force F is directly proportional to the product of the acceleration a produced on a body of mass m given as\[F = ma\], hence by substituting this in equation (i) we can write
\[{P_0} = mav - - (ii)\]
Now since acceleration a is the rate of change of the velocity of a body so we can write
\[a = \dfrac{{dv}}{{dt}}\], hence by substituting this in equation (ii), we can further write
\[{P_0} = m\dfrac{{dv}}{{dt}}v - - (iii)\]
This can also be written as
\[mvdv = {P_0}dt - - (iv)\]
Now by integrating the equation (iv), we get
\[\int {mvdv} = \int {{P_0}dt} \\
\Rightarrow m\int {vdv} = {P_0}\int {dt} \\
\Rightarrow m\dfrac{{{v^2}}}{2} = {P_0}t \\ \]
Hence we can further write this obtained equation as
\[\therefore v = \sqrt {\dfrac{{2{P_0}t}}{m}} \]
Therefore we get \[v = \sqrt t \]

Hence we can say the instantaneous velocity of this car is proportional \[{t^{\dfrac{1}{2}}}\] and option A is correct.

Note: Instantaneous power is the multiple of the product of force and the velocity by the cosine of the angle between the force and the velocity given by the formula \[{P_{ins\tan \tan eous}} = F \cdot v \cdot \cos \theta \]
When the velocity of a body is in the same direction of the force applied i.e. the angle between the force applied and the direction of velocity becomes same so the instantaneous power will be\[{P_{ins\tan \tan eous}} = F \cdot v\].