Question

Figure shows a particle starting from point $A$, travelling upto $B$ with a speed $s$, then upto point $C$ with a speed $2s$ and finally upto $A$ with a speed of $3s$. If the average speed is $ns$, the value of $n$ is

Answer 1

Angle $\angle AOB={90}^{\circ }=\pi /2$, $\angle BOC={120}^{\circ }=2\pi /3$
$\therefore \angle AOC=2\pi -(\frac{\pi }{2}+\frac{2\pi }{3})=\frac{5\pi }{6}$
$t_1=\frac{AB}{s}=\frac{\frac{\pi }{2}R}{s},t_2=\frac{BC}{2s}=\frac{\frac{2\pi }{3}R}{2s},t_3=\frac{CA}{3s}=\frac{\frac{5\pi R}{6}}{3s}$
$\text{Average Speed =}\frac{2\pi R}{t_1+t_2+t_3}=\frac{2\pi R}{\frac{\pi R}{2s}+\frac{\pi R}{3s}+\frac{5\pi R}{18s}}=\frac{2\times 18s}{20}=1.8s$
$\Rightarrow n=1.8$