Question

A particle moves in a semicircular path of radius R from O to A as shown in the Fig. Then it moves parallel to z-axis covering a distance R upto B. Finally it moves along BC parallel to y-axis through a distance 2R. Find the ratio of D/s.

Answer 1

The distance D, that is the length of the actual path covered by the particle(0A'ABC) as shown in Fig.
D = length of the semicircle OA'A + length AB + length BC. This gives D = $\pi$R + R +
2R = ($\pi$ + 3)R.
Since OA = 2R,AB = R, and BC = 2R, the coordinates of C can be given as C = (2R, R,2R). Then the position of C is expressed as:
${\overrightarrow{r}}_c$ = 2R$\hat{i}$ + R$\hat{j}$ + 2R$\hat{k}$
As $\overrightarrow{s}$(=$\overrightarrow{OC}$) = ${\overrightarrow{r}}_c$ - ${\overrightarrow{r}}_0$, substituting ${\overrightarrow{r}}_c$ and ${\overrightarrow{r}}_0$ = 0$\hat{i}$ +0$\hat{j}$ +0$\hat{k}$,
we obtain $\overrightarrow{s}$ = (2$\hat{i}$ + $\hat{j}$ + 2$\hat{k}$)R.
Its magnitude |$\overrightarrow{s}$| = ($\sqrt{2^1+1^2+2^2}$R = 3R
Hence, $\frac{D}{s}$ = $\frac{(\pi +3)R}{3R}$ = $\frac{\pi +3}{3}$