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 $\mathrm{x}=+2 \mathrm{~m}$, its velocity is $-4 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s}$. The object's velocity (v) and acceleration (a) at $\mathrm{x}=-2 \mathrm{~m}$ will be

(B) $\mathrm{v}=4 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s}, \mathrm{a}=8 \hat{\mathrm{i}} \mathrm{m} / \mathrm{s}^{2}$

(A) $\mathrm{v}=4 \hat{\mathrm{i}} \mathrm{m} / \mathrm{s}, \mathrm{a}=8 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s}^{2}$

(C) $\mathrm{v}=-4 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s}, \mathrm{a}=8 \hat{\mathrm{i}} \mathrm{m} / \mathrm{s}^{2}$

(D) $\mathrm{v}=-4 \hat{\mathrm{i}} \mathrm{m} / \mathrm{s}, \mathrm{a}=-8 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s}^{2}$

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Laws of Motion

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 m

, its velocity is

- 4 \hat{j} m / s

. The object's velocity (v) and acceleration (a) at

x = - 2 m

will be

Answer: (B)

Since the path is a circle centred at the origin and the object is at

x = + 2

m, the radius is

r = 2

m. The velocity there is

- 4 \hat{j}

m/s, tangent to the circle and pointing in the

- y

direction, which fixes the sense of rotation. The point

x = - 2

m is diametrically opposite, so the tangent there runs the opposite way along

y

, giving a velocity of magnitude

4

m/s directed as

4 \hat{j}

m/s. The acceleration is purely centripetal, always directed from the particle towards the centre; at

x = - 2

m this points along

+ x

, with magnitude

a = \frac{v ^{2}}{r} = \frac{4 ^{2}}{2} = 8

m/s

^{2}

. At

x = - 2

m the velocity is

4 \hat{j}

m/s and the acceleration is

8 \hat{i}

m/s

^{2}

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Single correct option