Q. An object moves at a constant speed along a circular path in a horizontal $XY$ plane, with the centre at the origin. When the object is at $x = - 2m$, its velocity is $- (4m/s)\widehat{j}$. What is the object’s acceleration when it is $y = 2m$
A) $- (8m/s^{2})\widehat{j}$
B) $- (8m/s^{2})\widehat{i}$
C) $- (4m/s^{2})\widehat{j}$
D) $(4m/s^{2})\widehat{i}$
Explanation
(a) [image:4_image119.emf]
The radius of circular path is $2m$ and the speed of the object is $4m/s$
The magnitude of acceleration is
$a = \frac{v^{2}\ }{R} = \frac{16}{2} = 8m/s^{2}$
The acceleration is directed towards the centre
Therefore, when an object is at $y = 2m$, its acceleration is $- 8\widehat{j}\ m/s^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.