Question

In the given figure, $a=15{\text{m/s}}^2$ represents the total acceleration of a particle moving in the clockwise direction in a circle of radius $R=2.5\text{m}$ at a given instant of time. The speed of particle at that instant will be

• A. $4.5\text{m/s}$
• B. $5\text{m/s}$
• C. $5.7\text{m/s}$
• D. $6.2\text{m/s}$

Answer 1

The correct option is C $5.7\text{m/s}$

The centripetal acceleration of a particle moving on a circular path is given by $a_c=\frac{v^2}{R}$.


$v$ represents the speed of the particle.

From figure,
$a_c=a\cos {30}^{\circ }$

or $a_c=15\times \frac{\sqrt{3}}{2}{\text{m/s}}^2$
Hence,

$\frac{v^2}{R}=\frac{15\sqrt{3}}{2}$

$v^2=\frac{15\sqrt{3}}{2}\times 2.5$

$v^2=32.48$

$\therefore v=\sqrt{32.48}\approx 5.7\text{m/s}$

$$\begin{array}{c}\text{Why this question?}\\ \text{Tip: The component of net acceleration}\\ \text{along the line joining centre and point on}\\ \text{circumference will give the centripetal}\\ \text{acceleration.}\end{array}$$