Question

Power supplied to a particle of mass $2kg$ varies with time as $P=\frac{3t^2}{2}$ watt, where $t$ is in second. If the velocity of particle at $t=0isv=0$, the velocity of particle at time $t=2s$ will be

• A. $2\sqrt{2}m/s$
• B. $4m/s$
• C. $1m/s$
• D. $2m/s$

Answer 1

The correct option is D

$2m/s$

Step1: Given data

The velocity of the particle at $t=0isv=0$

Time is from $0sto2s$

Hence, $t_1=0s$ and $t_2=2s$

Step2: Formula used

Work,

$w=p\times t[wherep=power,t=time]$

The work-energy theorem establishes a link between work, power, and energy. Work done by a net force on an object equals a change in kinetic energy, according to the work-energy theorem.

${\int }_{t_1}^{t_2}Pdt=K{E}_i-K{E}_f$

Step3: Calculating the velocity of the particle

Let the mass of the body be $m$.

It is being said that particles start from rest, it means that initial velocity, $u=0m/s$.

Substituting all the given values in the formula,

$$\begin{gathered} {\int }_{t_1}^{t_2}Pdt=K{E}_i-K{E}_f \\ {\int }_{t_1}^{t_2}\frac{3t^2}{2}dt=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2 \\ {\int }_{t_1}^{t_2}\frac{3t^2}{2}dt=\frac{1}{2}m{v}^2-0[Asu=0m/s] \\ \frac{3}{2}{\left[\frac{t^3}{3}\right]}_0^2=\frac{1}{2}\times 2\times v^2 \\ \frac{3}{2}\times \frac{8}{3}=v^2 \\ {v}^2=4 \\ v=2m/s \end{gathered}$$

Hence, the velocity of a particle at time $t=2s$ is given as $v=2m/s$.

Hence, option (C) is the correct answer.