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

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Question

$\mathrm{x}=\sqrt{\mathrm{at}^{2}+2 \mathrm{bt}+\mathrm{c}}$

Differentiating w.r.t. time

$\frac{d x}{d t}=v=\frac{1}{2 \sqrt{a t^{2}+2 b t+c}} \

Vedclass · Accepted Answer

$3$

Vedclass · Answer

$\mathrm{x}=\sqrt{\mathrm{at}^{2}+2 \mathrm{bt}+\mathrm{c}}$

Differentiating w.r.t. time

$\frac{d x}{d t}=v=\frac{1}{2 \sqrt{a t^{2}+2 b t+c}} 	imes(2 a t+2 b)$

$\Rightarrow \mathrm{v}=\frac{\mathrm{at}+\mathrm{b}}{\mathrm{x}}$

$\Rightarrow \mathrm{vx}=\mathrm{at}+\mathrm{b}$

Differentiating w.r.t. $\mathrm{x}$

$\Rightarrow \frac{d v}{d x} 	imes x+v=a 	imes \frac{d t}{d x}$

Multiply both side by v

$\Rightarrow\left(v \frac{d v}{d x}ight) x+v^{2}=a$

$\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\mathrm{a}-\mathrm{v}^{2}$ [Here $a'$ is acceleration]

$\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\mathrm{a}-\left(\frac{\mathrm{at}+\mathrm{b}}{\mathrm{x}}ight)^{2}$

$\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\frac{\mathrm{ax}^{2}-(\mathrm{at}+\mathrm{b})^{2}}{\mathrm{x}^{2}}$

$\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\frac{\mathrm{a}\left(\mathrm{at}^{2}+2 \mathrm{bt}+\mathrm{c}ight)-(\mathrm{at}+\mathrm{b})^{2}}{\mathrm{x}^{2}}$

$\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\frac{\mathrm{ac}-\mathrm{b}^{2}}{\mathrm{x}^{2}}$

$\Rightarrow \mathrm{a}^{\prime}=\frac{\mathrm{ac}-\mathrm{b}^{2}}{\mathrm{x}^{3}}$

$	herefore a^{\prime} \propto \frac{1}{x^{3}} \quad 	herefore n=3$

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