Q 11-03-080JEE MainJEE Main 2023 (29 Jan, Shift 2)Easy
An object moves at a constant speed along a circular path in a horizontal plane with centre at the origin. When the object is at $x=+2\ \text{m}$, its velocity is $-4\hat{j}\ \text{m s}^{-1}$. The object's velocity ($v$) and acceleration ($a$) at $x=-2\ \text{m}$ will be
Answer: (B) $v=4\hat{j}\ \text{m s}^{-1},\ a=8\hat{i}\ \text{m s}^{-2}$
The radius is $2\ \text{m}$ and the speed is $4\ \text{m s}^{-1}$. At the diametrically opposite point $x=-2\ \text{m}$ the velocity is reversed: $v=+4\hat{j}\ \text{m s}^{-1}$.
The acceleration points to the centre, i.e. along $+\hat i$, with magnitude $\dfrac{v^2}{r}=\dfrac{16}{2}=8\ \text{m s}^{-2}$. So $a=8\hat i\ \text{m s}^{-2}$.
Solution by Sreeraj P, M.Sc Physics