The distance x covered by a particle in one-d | vedclass

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(D) Given $x^{2} = at^{2} + 2bt + c$.Differentiating with respect to $t$:$2x \frac{dx}{dt} = 2at + 2b$$x v = at + b$,where $v = \frac{dx}{dt}$.Differe

Vedclass · Accepted Answer

$3$

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(D) Given $x^{2} = at^{2} + 2bt + c$.Differentiating with respect to $t$:$2x \frac{dx}{dt} = 2at + 2b$$x v = at + b$,where $v = \frac{dx}{dt}$.Differentiating again with respect to $t$:$v \frac{dx}{dt} + x \frac{dv}{dt} = a$$v^{2} + x a' = a$,where $a' = \frac{dv}{dt}$ is the acceleration.From $x v = at + b$,we have $v = \frac{at+b}{x}$.Substituting $v$ into the equation:$x a' = a - v^{2} = a - \left(\frac{at+b}{x}\right)^{2}$$x a' = \frac{ax^{2} - (at+b)^{2}}{x^{2}}$Substitute $x^{2} = at^{2} + 2bt + c$:$x a' = \frac{a(at^{2} + 2bt + c) - (a^{2}t^{2} + 2abt + b^{2})}{x^{2}}$$x a' = \frac{a^{2}t^{2} + 2abt + ac - a^{2}t^{2} - 2abt - b^{2}}{x^{2}}$$x a' = \frac{ac - b^{2}}{x^{2}}$$a' = \frac{ac - b^{2}}{x^{3}}$Thus,$a' \propto x^{-3}$. Comparing with $x^{-n}$,we get $n = 3$.

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

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