Question

The figure shows a particle starting from point A, travelling up to B with a speed s, then up to point C with a speed 2 s, and finally to A with a speed of 3 s. Determine its average speed.

Answer 1

(Refer to Image)

${90}^o$	${120}^o$	${150}^o$
$D_1=\frac{90}{360}\times 2\pi r$ $=\frac{2\pi r}{4}$ $\Rightarrow t_1=\frac{2\pi r}{4s}$	$D_2=\frac{120}{360}\times 2\pi r$ $=\frac{2\pi r}{3}$ $\Rightarrow t_2=\frac{2\pi r}{6s}$	$D_3=\frac{150}{360}\times 2\pi r$ $=\frac{5}{12}\times 2\pi r$ $\Rightarrow t_3=\frac{10\pi r}{36s}$

Average speed= $\frac{\text{Total Distance}}{\text{Total Time}}$

$=\frac{2\pi r(\frac{1}{4}+\frac{1}{3}+\frac{5}{12})}{2\pi r(\frac{1}{4s}+\frac{1}{6s}+\frac{5}{36s})}$

$=\frac{(3+4+5)36}{12(9+6+5)}s$

$=\frac{12\times 3}{20}=\frac{18}{10}$

$=1.8s$