Question

A car drives along a straight frictionless road by an engine delivering constant power. Then the velocity is directly proportional to
A. $t$
B. $\dfrac{1}{{\sqrt t }}$
C. $\sqrt t$
D. None of the above

Answer 1

The power in terms of velocity is given as the dot product of the applied force and the velocity. Mathematically this is expressed as $P = F.v$. We shall use this expression and consider the power to be a constant while calculating. We will further substitute $F = ma$ where m is the mass of the body and a is the acceleration of the body and $a = \dfrac{{dv}}{{dt}}$ where v is the velocity of the object and t tis the time post which we shall integrate the expression to get a relation between the velocity and time.

Complete step by step answer:
We know that $P = F.v$
Also, $F = ma$ and $a = \dfrac{{dv}}{{dt}}$
Making all substitutions we get,
$P = ma.v$
$\Rightarrow P = m\dfrac{{dv}}{{dt}}.v$
Now rearranging the terms, we get,
$Pdt = mdv.v$
Now integrating it, we get,
$\int {Pdt} = \int {mdv.v}$
Here $P$ and $m$ are constants, so
$P\int {dt} = m\int {dv.v}$

Further solving the expression,
$Pt = \dfrac{{m{v^2}}}{2}$
Arranging all constants on one side,
$\dfrac{{2P}}{m}t = {v^2}$
Now since we need the relation between the velocity and time, we need to remove the square on $v$.Taking square root both sides,
$\sqrt {\dfrac{{2P}}{m}t} = \sqrt {{v^2}}$
$\Rightarrow \sqrt {\dfrac{{2P}}{m}} \sqrt t = v$
Since $P$ and m are constants, $\sqrt {\dfrac{{2P}}{m}}$ is also a constant.
We can rewrite the equation as
$\therefore v \propto \sqrt t$

Hence, C is the correct option.

Note: We need to keep in mind that we cannot consider a differential equation as a physical relation between two physical quantities. We must integrate the terms to remove the differential terms. The expression thus obtained would be the physical relation between the quantities and any proportionalities are then decided on the basis of the obtained relation only.