Question

In the given figure, $a =15\, m / s ^{2}$ represents the total acceleration of a particle moving in the clockwise direction in a circle of radius $R =2.5\, m$ 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)

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