JEE Physics 2023 Question with Solution

Given: The object moves with constant speed in a circular path centered at the origin. At $x = +2 \, \text{m}$, the velocity is $-4\hat{j} \, \text{m/s}$.

Find: The velocity and acceleration at $x = -2 \, \text{m}$.

For uniform circular motion, the velocity is tangential to the path and the acceleration is centripetal, always directed toward the center.

At $x = +2 \, \text{m}$, the velocity is downward, so at the diametrically opposite point $x = -2 \, \text{m}$, the velocity will be upward:

$$v = +4\hat{j} \, \text{m/s}$$

The acceleration at $x = -2 \, \text{m}$ points toward the center, that is, along the positive $x$-axis:

$$a = 8\hat{i} \, \text{m/s}^2$$

Therefore, the correct option is B. The solution marks C, but its own working gives $v = +4\hat{j} \, \text{m/s}$ and $a = 8\hat{i} \, \text{m/s}^2$, which matches option B.