Question

A particle moves along a straight line in such a way that the product $a{v}^2=constant$, where a is acceleration and v is velocity of the particle. Assuming that the particle is started from rest, then the distance moved by the particle in time $t$ varies as:

• A. $t$
• B. $t^2$
• C. $t^{3/2}$
• D. $t^{4/3}$

Answer 1

The correct option is D $t^{4/3}$

It is given that $a{v}^2=K.....\left(1\right)$

Where a is acceleration and v is velocity of particle.

$a=\frac{dv}{dt}$

So, equation (1) becomes

$\int v^{2}dt=\int Kdt$

$\frac{v^3}{3}=Kt$

$v^3=3Kt$

$v={\left(3Kt\right)}^{\frac{1}{3}}...........\left(2\right)$

$\because v=\frac{dx}{dt}wheredxisdis\tan ce$

So, equation (2) becomes

$\frac{dx}{dt}={\left(3Kt\right)}^{\frac{1}{3}}$

$\int dx=\int {\left(3Kt\right)}^{\frac{1}{3}}dt$

$x={\left(3K\right)}^{\frac{1}{3}}\frac{4}{3}{t}^{\frac{4}{3}}$

$x\propto t^{\frac{4}{3}}$