The distance x covered by a particle in one dimensional motion varies

Question:

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

Solution:

(3) Distance $\mathrm{X}$ varies with time $\mathrm{t}$ as $x^{2}=a t^{2}+2 b t+c$

$\Rightarrow 2 x \frac{d x}{d t}=2 a t+2 b$

$\Rightarrow x \frac{d x}{d t}=a t+b \Rightarrow \frac{d x}{d t}=\frac{(a t+b)}{x}$

$\Rightarrow x \frac{d^{2} x}{d t^{2}}+\left(\frac{d x}{d t}\right)^{2}=a$

$\Rightarrow \frac{d^{2} x}{d t^{2}}=\frac{a-\left(\frac{d x}{d t}\right)^{2}}{x}=\frac{a-\left(\frac{a t+b}{x}\right)^{2}}{x}$

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

$\Rightarrow a \propto x^{-3}$

Hence, $n=3$

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