Question

A particle moves in a circle of radius $4\text{m}$ at a speed given by $v=4t$, where $v$ is in $\text{m/s}$ and $t$ in seconds. The total acceleration of particle at $t=1\text{s}$ is :

• A. $4{\text{m/s}}^2$
• B. $4\sqrt{3}{\text{m/s}}^2$
• C. $8{\text{m/s}}^2$
• D.
  $4\sqrt{2}{\text{m/s}}^2$

Answer 1

The correct option is D

$4\sqrt{2}{\text{m/s}}^2$
Given that particle is moving in circular path of radius ($r=4\text{m}$) with speed, $v=4t$. It means that it is a case of non-uniform circular motion.

$\therefore$The acceleration of particle will contain tangential acceleration $a_t$ as well as centripetal acceleration, $a_c$.




We know that acceleration is the rate of change of velocity per unit of time.

$\therefore a_t=\frac{dv}{dt}{a}_t=\frac{d}{dt}\left(4t\right)=4{\text{m/s}}^2$

at $t=1\text{s},v=4\text{m/s}$, hence centripetal acceleration,
$a_c=\frac{v^2}{r}\therefore a_c=\frac{4^2}{4}=4{\text{m/s}}^2$
$\Rightarrow$Total acceleration $a_T$ of particle given by:
$a_T=\sqrt{{a_t}^2+{a_c}^2}\therefore a_T=\sqrt{4^2+4^2}=4\sqrt{2}{\text{m/s}}^2$

Hence option D is the correct answer.