Question

Instantaneous velocity of a particle moving in a straight line is given as $v=(4+4\sqrt{t})$$\text{m/s}$, for the first $5\text{sec}$ of its motion, there after velocity becomes constant. Find the acceleration and displacement of the particle at time $t=3.0\text{sec}$.

• A. $\frac{2}{\sqrt{3}}{\text{m/s}}^2,35.50\text{m}$
• B. $\frac{2}{\sqrt{3}}{\text{m/s}}^2,25.85\text{m}$
• C. $5{\text{m/s}}^2,25.85\text{m}$
• D. $10{\text{m/s}}^2,15.85\text{m}$

Answer 1

The correct option is B $\frac{2}{\sqrt{3}}{\text{m/s}}^2,25.85\text{m}$

Given,
velocity of particle for the first $5\text{sec}$, $v=(4+4\sqrt{t})\text{m/s}$

As we know that acceleration of a particle is time derivative of its instantaneous velocity.$$a=\frac{dv}{dt}$$ $\Rightarrow a=\frac{d(4+4\sqrt{t})}{dt}$

$\Rightarrow a=\frac{2}{\sqrt{t}}$

And at time $t=3.0\text{sec}$, acceleration is

$a=\frac{2}{\sqrt{3}}{\text{m/s}}^2$

To find displacement: We know that velocity is the time derivative of displacement, we have to integrate as follows

$v=\frac{dx}{dt}=4+4\sqrt{t}$

$\Rightarrow dx=(4+4\sqrt{t})dt$

Integrating the above expression from time $t=0$ to $t=3.0\text{sec}$, we get the displacement, which varies from $x_i=0$ to $x_f=x$, for these time instants.

${\int }_{x_i=0}^{x_f=x}dx={\int }_{t=0}^{t=3}(4+4\sqrt{t})dt$

$\Rightarrow [x-0]={[4t+\frac{8}{3}{t}^{3/2}]}_0^3$

$\Rightarrow x=[12+13.85]$

$\therefore x\approx 25.85\text{m}$

Hence, option (b) is the correct answer.