Question

A body is moving in a circular path of radius $0.125\text{m}$. If the body covers a distance of one quarter of the circumference of the path in $\sqrt{2}\text{s}$, what is the velocity of the body in that distance, in ${\text{m min}}^{-1}$?

• A. 7.5

Answer 1

The correct option is A 7.5
Given:
The body is moving in a circular path of radius, $\text{r}=0.125\text{m}=\frac{1}{8}\text{m}$
The body covers one quarter of the circumference of the path in time, $\text{t}=\sqrt{2}\text{s}$, i.e. from the point $\text{A}$ to $\text{B}$ through the curvature as shown:

The displacement, $\text{AB}$ can be calculated using the Pythagorean theorem, as:

$\text{AB}=\sqrt{{\text{OA}}^2+{\text{OB}}^2}=\sqrt{{\text{r}}^2+{\text{r}}^2}=\text{r}\sqrt{2}$

$\text{AB}=\frac{1}{8}\times \sqrt{2}=\frac{1}{4\sqrt{2}}\text{m}$

The velocity from $\text{A}$ to $\text{B}$, ${\text{v}}_{\text{AB}}=\frac{\text{AB}}{\text{t}}$
$⟹\frac{1}{4\sqrt{2}\times \sqrt{2}}=\frac{1}{8}{\text{m s}}^{-1}$

$1\text{min}=60\text{s}$
$⟹\frac{1}{8}{\text{m s}}^{-1}=(\frac{1}{8}\times 60){\text{m min}}^{-1}$

$\therefore {\text{v}}_{\text{AB}}=\frac{60}{8}=\frac{15}{2}=7.5{\text{m min}}^{-1}$