JEE Main · 2006 · 30 Apr 2006 · Offline · Kinematics

A particle located at

x = 0

at time

t = 0

, starts moving along the positive x-direction with a velocity '

v

' that varies as

v = α x

. The displacement of the particle varies with time as

(A) $t^{3}$
(B) $t^{2}$ ✓ Correct
(C) $t$
(D) $t^{1/2}$

Answer: (B)

Solution

Since the speed depends on position as

v = α x

, I write

v = \frac{d x}{d t} = α x

and separate the variables as

\frac{d x}{x} = α d t

. Integrating from the start, where

x = 0

t = 0

, gives

2 x = α t

, so

x = \frac{α t}{2}

and

x = \frac{α ^{2} t ^{2}}{4}

. The displacement therefore grows in proportion to

t^{2}

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option

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