JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Kinematics

The position of a particle as a function of time

t

, is given by

x (t) = a t + b t^{2} - c t^{3}

where

a

b

and

c

are constants. When the particle attains zero acceleration, then its velocity will be:

(A) $a + \frac{b ^{2}}{4 c}$
(B) $a + \frac{b ^{2}}{c}$
(C) $a + \frac{b ^{2}}{2 c}$
(D) $a + \frac{b ^{2}}{3 c}$ ✓ Correct

Answer: (D)

Solution

Velocity is

v = \frac{d x}{d t} = a + 2 b t - 3 c t^{2}

and acceleration is

\frac{d v}{d t} = 2 b - 6 c t

. The acceleration vanishes when

t = \frac{b}{3 c}

. Putting this time back into the velocity,

v = a + 2 b \cdot \frac{b}{3 c} - 3 c \cdot \frac{b ^{2}}{9 c ^{2}} = a + \frac{2 b ^{2}}{3 c} - \frac{b ^{2}}{3 c} = a + \frac{b ^{2}}{3 c}

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Single correct option

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