Question

In the given figure, $a=15m/s^2$ represents the total acceleration of a particle moving in the clockwise direction in a circle of radius $R=2.5m$ at a given instant of time. The speed of the particle is -

• A. 4.5 m/s
• B. 5.0 m/s
• C. 5.7 m/s
• D. 6.2 m/s

Answer 1

Answer: (C)

$\mathrm{Tan}\theta =\frac{dV/dt}{V^2/R}$
$\mathrm{Tan}30^{\circ }=\frac{1}{\sqrt{3}}=\frac{dV/dt}{V^2/R}$
$dV=\frac{1}{\sqrt{3}}\times \frac{V^2}{R}$
$a=15=\sqrt{{\left(\frac{dV}{dt}\right)}^2+{\left(\frac{V^2}{R}\right)}^2}$
$15=\sqrt{\frac{4}{3}{\left(\frac{V^2}{R}\right)}^2}$
$15=\frac{2}{\sqrt{3}}\left(\frac{V^2}{R}\right)=\frac{2}{\sqrt{3}}\times \frac{V^2}{2.5}$
$\Rightarrow V^2=\frac{15\times \sqrt{3}\times 2.5}{2}=32.476=5.7m/s$