The distance $x$ covered by a particle in one dimensional motion varies with time $t$ as $x^{2}=a t^{2}+b t+c$. If the acceleration of the particle depends on $x$ as $x^{-n}$, where $n$ is an integer, the value of $n$ is ____.

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Kinematics

The distance

x

covered by a particle in one dimensional motion varies with time

t

x^{2} = a t^{2} + b t + c

. If the acceleration of the particle depends on

x

x^{- n}

, where

n

is an integer, the value of

n

is ____.

Answer: 3

Solution

Differentiating

x^{2} = a t^{2} + b t + c

once gives

2 x \overset{x}{˙} = 2 a t + b

, i.e.

xv = a t + b

. Differentiating again,

v^{2} + xα = a

, where

α

is the acceleration, so

α = \frac{a - v ^{2}}{x}

. Putting

v = \frac{a t + b}{x}

and using the original relation,

a - v^{2} = \frac{a x ^{2} - ( a t + b ) ^{2}}{x ^{2}} = \frac{a c - b ^{2}}{x ^{2}}

, and hence

α = \frac{a c - b ^{2}}{x ^{3}}

. The acceleration therefore varies as

x^{- 3}

, giving

n = 3

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Integer answer

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