Verified Solution • 2025–26 Session moderate

QJEEM20Ph8c92: JEE Main 2020 Physics - Class 11 JEEM20Ph8c92 Solution: The distance x covered by a particle in one dimensional motion…

The distance x covered by a particle in one dimensional motion varies with time t as
x2 = at2 + 2bt + c. If the acceleration of the particle depends on x as x–n, where n is an integer, the value of n is __________

Step-by-Step Verified Solution

x2 = at2 + 2bt + c .....(1)

⇒\Rightarrow 2xv = 2at + 2b .....(2)

⇒\Rightarrow v = at+bx{{at + b} \over x}

differentiating (2) w.r.t. time

⇒\Rightarrow xa' + v2 = a

Here a' is acceleration.

⇒\Rightarrow a'x = a - v2 = a - [at+bx]2{\left[ {{{at + b} \over x}} \right]^2}

⇒\Rightarrow a'x = ax2−(at+b)2x2{{a{x^2} - {{\left( {at + b} \right)}^2}} \over {{x^2}}}

⇒\Rightarrow a' = a(at2+2bt+c)−(at+b)2x2{{a\left( {a{t^2} + 2bt + c} \right) - {{\left( {at + b} \right)}^2}} \over {{x^2}}}

⇒\Rightarrow a' = ac−b2x3{{ac - {b^2}} \over {{x^3}}}

∴\therefore a' ∝\propto 1x3{1 \over {{x^3}}} ∝\propto x-3

∴\therefore n = 3

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