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 $(5,5)$ in $m/s^2$ is

• 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 $-\hat{i}-\hat{j}$



The radius of the circle is $5\sqrt{2}m.(5,5)$ is on the circle.

Since the speed is constant, the particle only experiences centripetal acceleration.

At $(5,5),\overrightarrow{a_c}$ is in the third quadrant at ${45}^{\circ }$. By calculating rectangular components of $\overrightarrow{a_c}$, we get
$\overrightarrow{a_c}=\frac{v^2}{R}\left(\frac{-\hat{i}-\hat{j}}{\sqrt{2}}\right)$

Here, $\left(\frac{-\hat{i}-\hat{j}}{\sqrt{2}}\right)$ is the unit vector along ${45}^{\circ }$ in the 3rd quadrant.

$\overrightarrow{a_c}=\frac{10}{5\sqrt{2}}\left(\frac{-\hat{i}-\hat{j}}{\sqrt{2}}\right)$
$\overrightarrow{a_c}=(-\hat{i}-\hat{j})m{s}^{-1}$