JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Kinematics

A particle is moving in a straight line. The variation of position '

x

' as a function of time '

t

' is given as

x = (t^{3} - 6 t^{2} + 20 t + 15)

m. The velocity of the body when its acceleration becomes zero is:

(A) $4 m / s$
(B) $8 m / s$ ✓ Correct
(C) $10 m / s$
(D) $6 m / s$

Answer: (B)

Solution

Position:

x = t^{3} - 6 t^{2} + 20 t + 15

.

Velocity:

v = \frac{d x}{d t} = 3 t^{2} - 12 t + 20

Acceleration:

a = \frac{d v}{d t} = 6 t - 12

Setting

a = 0

6 t - 12 = 0 ⟹ t = 2

s.

At

t = 2

v = 3 (2)^{2} - 12 (2) + 20 = 12 - 24 + 20 = 8 m/s

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Single correct option

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