Question

A particle moves in a circle of radius $4cm$ clockwise at constant speed $2cm{s}^{-1}$. If $\hat{x}$ and $\hat{y}$ are unit acceleration vectors along $X$ and $Y$ respectively $\left(incm{s}^{-2}\right)$, the acceleration of the particle at the instant half way between $P$ and $Q$ 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

Answer: (C)

$\frac{-(\hat{x}+\hat{y})}{\sqrt{2}}$

Let M be the mid-point, so it makes an angle of ${45}^o$ with the horizontal.

Centripetal acceleration at point M, $\overrightarrow{a}=\frac{v^2}{r}$ towards O

Given: $v=2cm/sr=4cm$

$|\overrightarrow{a}|=\frac{2^2}{4}=1cm/s^2$

Thus $\overrightarrow{a}=1(-cos{45}^o\hat{x}-sin{45}^o\hat{y})$

$\overrightarrow{a}=1(-\frac{1}{\sqrt{2}}\hat{x}-\frac{1}{\sqrt{2}}\hat{y})$

$⟹\overrightarrow{a}=-\frac{1}{\sqrt{2}}(\hat{x}+\hat{y})$