Question

A particle moves in a circle of radius 4 cm clockwise at constant speed of 2 cm $s^{-1}$. If $\hat{x}$ and $\hat{y}$ are unit acceleration vectors along x and y axes, respectively, the acceleration of the particle at the instant half way between PQ is given by

• A. $-4(\hat{x}-\hat{y})$
• B. $4(\hat{x}+\hat{y})$
• C. $\frac{-(\hat{x}+\hat{y})}{\sqrt{2}}$
• D. $\frac{\hat{x}-\hat{y}}{4}$

Answer 1

The correct option is C $\frac{-(\hat{x}+\hat{y})}{\sqrt{2}}$

Mid point acceleration,
$a=\frac{v^2}{r}=\frac{4}{4}=1m{s}^{-1}$ at ${45}^{\circ }$ with x-axis
${\overrightarrow{a}}_x=-a\cos {45}^{\circ }\hat{x}$
$=-1\times \frac{1}{\sqrt{2}}\hat{x}=-\frac{1}{\sqrt{2}}\hat{x}$
${\overrightarrow{a}}_y=-a\sin {45}^{\circ }\hat{y}$
$=-1\times \frac{1}{\sqrt{2}}\hat{y}=-\frac{1}{\sqrt{2}}\hat{y}$
Thus $\hat{a}=\overrightarrow{a_x}+\overrightarrow{a_y}=-\frac{1}{\sqrt{2}}(\hat{x}+\hat{y})$