Question

A power $P=2t^3$ is applied to the particle initially at rest. If the kinetic energy of the particle is $K$ and velocity is $v$ then

• A. $K\propto t^2$
• B. $v\propto t^4$
• C. $K\propto t^4$
• D. $v\propto t^2$

Answer 1

Answer: (C), (D)

$v\propto t^2$

Since, the particle is initially at rest and also we know that power is the rate of change of kinetic energy.
So, $P=\frac{dK}{dt}$
$\Rightarrow K={\int }_0^{t}Pdt$
$\Rightarrow K={\int }_0^{t}2t^{3}dt$
$\Rightarrow K=\frac{2t^4}{4}=\frac{t^4}{2}$
$\Rightarrow K\propto t^4$
Now,
$K=\frac{t^4}{2}=\frac{1}{2}m{v}^2$
$\Rightarrow v^2=\frac{t^4}{m}$
$\Rightarrow v=\frac{t^2}{\sqrt{m}}$
$\Rightarrow v\propto t^2$
Hence option $\left(c\right)$ and $\left(d\right)$ is correct.