Question

A particle of mass $2\text{kg}$ starts moving in a straight line with an initial velocity of $2\text{m/s}$, at a constant acceleration of $2{\text{m/s}}^2$. Then, rate of change of kinetic energy

• A. is four times the velocity at any moment.
• B. is two times the displacement at any moment.
• C. is four times the rate of change of velocity at any moment.
• D. is constant throughout.

Answer 1

The correct option is A is four times the velocity at any moment.

We know that, Kinetic energy $K=\frac{1}{2}m{v}^2$
Differentiating on both sides with respect to time, we get
$\frac{dK}{dt}=mv.\frac{dv}{dt}$
$\frac{dK}{dt}=\left(m\frac{dv}{dt}\right)v=\left(ma\right)v$
From the data given in the question,
$m=2\text{kg}$ and $a=2{\text{m/s}}^2$
$\therefore \frac{dK}{dt}=4v.....\left(1\right)$
Since the given acceleration is constant, we can use kinematic formulae
Using, $v=u+at$ in $\left(1\right)$, we get
$\frac{dK}{dt}=4u+4at=8(1+t)$
So, option (d) is wrong.
From $\left(1\right)$, we can conclude that option (c) is also wrong.
Using $v^2=u^2+2as$ in $\left(1\right)$ and substituting the values, we get
$\frac{dK}{dt}=8\sqrt{1+2s}$
So, option (b) is also wrong.
Thus, option (a) is the correct answer.