Question

A particle is moving with constant speed over circle $x^2+y^2=50$, with speed $\sqrt{10}m/s$. Acceleration of the particle, when it is at point $(5,5)$ is (is m/s${}^2$)

• A. $\hat{i}-\hat{j}$
• B. $-\hat{i}-\hat{j}$
• C. $\frac{1}{\sqrt{2}}\hat{i}-\frac{1}{\sqrt{2}}\hat{j}$
• D. $\frac{-1}{\sqrt{2}}\hat{i}-\frac{1}{\sqrt{2}}\hat{j}$

Answer 1

The correct option is B $-\hat{i}-\hat{j}$

$\left|ac\right|=\frac{V^2}{R}=\frac{10}{\sqrt{50}}$

$=\frac{10}{5\sqrt{2}}=\sqrt{2}$

$\therefore \overrightarrow{ac}=\sqrt{2}\cos 45\left(-\hat{i}\right)+\sqrt{2}\sin 45\left(-\hat{j}\right)$

$=-\hat{i}-\hat{j}$

Hence,

option $\left(B\right)$ is correct answer.