A particle moves in a semicircular path of radius $R$ from $O$ to $A$. Then it moves parallel to $z$-axis covering a distance $R$ upto $B$. Finally it moves along $B C$ parallel to $y$-axis through a distance $2 R$. Find the ratio of $D / s$, where $D$ is the length of OABC and $s$ is the magnitude of $\vec{s}$ shown in the diagram.

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Aicharya · Accepted Answer

The distance $D$, that is, length of the actual path covered by the particle $\left(O A^{\prime} A B C\right)$ as shown in figure. $D=$ length of the semicircle $O A^{\prime} A+$ length $A B+$ length $B C$. This gives $D=\pi R+R+2 R=(\pi+3) R$ Since $O A=2 R, A B=R$, and $B C=2 R$, the coordinates of $C$ can be given as $C \equiv(2 R, R, 2 R)$. Then the position of $C$ is expressed as: $\vec{r}_{C}=2 R \hat{i}+2 \hat{R}+R \hat{k}$ As $\vec{s}(=\overrightarrow{O C})=\vec{r}_{C}-\vec{r}_{0}$, substituting $\vec{r}_{C}$ and $\vec{r}_{0}=0 \hat{i}+0 \hat{j}+0 \hat{k}$, we obtain $\vec{s}=(2 \hat{i}+2 \hat{j}+\hat{k}) R$. Its magnitude $|\vec{s}|=\left(\sqrt{2^{2}+2^{2}+1^{2}}\right) R=3 R$ Hence, $\frac{D}{s}=\frac{(\pi+3) R}{3 R}=\frac{\pi+3}{3}$ ,

A particle moves in a semicircular path of radius \(R\) from \(O\) to \(A\). Then it moves parallel to…

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