An object moves at a constant speed along a circular path in a horizontal plane with center at the origin. When the object is at x = +2 m, its velocity is -4j m/s. The object's velocity (v) and acceleration (a) at x = -2 m will be (a) v = 4i m/s, a = 8j m/s² (b) v = 4j m/s, a = 8i m/s² (c) v = -4i m/s, a = -8j m/s² (d) v = -4j m/s, a = 8i m/s²

Answer: (b). O-4jx=+2v = +4ja = +8ix=-2 Radius r = 2 m, speed v = 4 m/s. At (2, 0) velocity is -j, so the motion is clockwise. At (-2, 0), half a turn later, the velocity is reversed: v = +4j m/s. a = v²/r = 16/2 = 8 m/s², pointing to the centre, i.e. along +i. So the answer is (b) v = 4j, a = 8i.

An object moves at a constant speed along a circular path in a… | JEE Main 2023 (29 Jan, Shift 2)

Q21JEE Main 2023 · 29 JAN · SHIFT 2Medium

Centripetal Acceleration & Force

An object moves at a constant speed along a circular path in a horizontal plane with center at the origin. When the object is at $x = +2\text{ m}$ , its velocity is $-4\hat{j}$ m/s. The object's velocity (v) and acceleration (a) at $x = -2\text{ m}$ will be

Show answer and solution

Answer: (b)

Radius $r = 2\text{ m}$ , speed $v = 4\text{ m/s}$ . At $(2, 0)$ velocity is $-\hat{j}$ , so the motion is clockwise.
At $(-2, 0)$ , half a turn later, the velocity is reversed: $\vec v = +4\hat{j}\text{ m/s}$ .
$a = \frac{v^2}{r} = \frac{16}{2} = 8\text{ m/s}^2$ , pointing to the centre, i.e. along $+\hat{i}$ .
So the answer is (b) $v = 4\hat{j}$ , $a = 8\hat{i}$ .

Report a mistake

Centripetal Acceleration & Force question from JEE Main 2023 (29 Jan, Shift 2)

More questions on Centripetal Acceleration & Force