Question

Engine of an automobile of mass $m$ supplies a constant power $P$ starting from rest. At an instant of time $t$
A. The velocity $v$ is proportional to $\sqrt P$
B. The velocity $v$ is proportional to $\sqrt t$
C. The displacement is proportional to $\dfrac{1}{{\sqrt m }}$
D. The displacement is proportional to ${t^{\dfrac{3}{2}}}$

Answer 1

To solve this question, we need to understand the relation between power and time. For that we will start with the definition of power in relation with force and velocity. Then the force van can be converted into mass and acceleration and as we know that acceleration is a function of time. Therefore, with the help of integration, we will find the velocity and displacement to determine the correct answer.

Formulas used:
$P = Fv$,
where, $P$ is the power, $F$ is the force and $v$ is the velocity
$F = ma$,
where,$F$ is the force, $m$ is the mass and $a$is the acceleration
$a = \dfrac{{dv}}{{dt}}$,
where, $a$is the acceleration, $v$ is the velocity and $t$ is the time
$v = \dfrac{{dx}}{{dt}}$,
where, $v$ is the velocity, $x$ is the displacement, $t$ is the time

Complete step by step answer:
We know that the power is given by:
$P = Fv$
We will put $F = ma$
$\Rightarrow P = mav$
Now, $a = \dfrac{{dv}}{{dt}}$
\Rightarrow P = m\dfrac{{dv}}{{dt}}v \\\\\ \Rightarrow \dfrac{{dv}}{{dt}}v = \dfrac{P}{m} \\\ \Rightarrow vdv = \dfrac{P}{m}dt \\\
We will integrate both the sides

$$\Rightarrow \int\limits_0^v {vdv} = \int\limits_0^t {\dfrac{P}{m}dt} \\\ \Rightarrow \dfrac{{{v^2}}}{2} = \dfrac{{Pt}}{m} \\\ \Rightarrow v = \sqrt {\dfrac{{2Pt}}{m}} \\\$$

From this equation we can say that the velocity $v$ is proportional to $\sqrt P$and the velocity $v$ is proportional to $\sqrt t$
Thus, options A and B are correct.
We have determined that
$v = \sqrt {\dfrac{{2Pt}}{m}}$
We know that $v = \dfrac{{dx}}{{dt}}$

$$\Rightarrow \dfrac{{dx}}{{dt}} = \sqrt {\dfrac{{2Pt}}{m}} \\\ \Rightarrow dx = \sqrt {\dfrac{{2P}}{m}} \sqrt t dt \\\$$

Integrating both the sides,

$$\Rightarrow \int\limits_0^x {dx} = \sqrt {\dfrac{{2P}}{m}} \int\limits_0^t {\sqrt t dt} \\\ \Rightarrow x = \sqrt {\dfrac{{2P}}{m}} \left( {\dfrac{2}{3}{t^{\dfrac{3}{2}}}} \right) \\\ \Rightarrow x = \sqrt {\dfrac{{8P}}{{9m}}} {t^{\dfrac{3}{2}}} \\\$$

So from this equation, we can say that the displacement is proportional to $\dfrac{1}{{\sqrt m }}$ and the displacement is proportional to ${t^{\dfrac{3}{2}}}$. Thus, options C and D are also correct.

Hence, in this question, all the four options are the correct answer.

Note: In this question, we have determined the answer by using the concept of integration. Whenever we are asked to find the value of quantity at any instant of time, we can use this method. For example, we have determined the value of velocity and displacement by considering them as a function of time and then integrating accordingly.