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Question

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:

A
3
Solution
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Correct option is A. 3
$$x^2=at^2+2bt+c$$

Differentiating w.r.t. $$'t'$$

$$2x.\dfrac{dx}{dt}=2at+2b$$............(1)

$$2\left[\left(\dfrac{dx}{dt}\right)^2+x\dfrac{d^2x}{dt^2}\right]=2a$$

Put the value of $$\dfrac{dx}{dt}$$ form eq (1)

$$2\left[\left(\dfrac{2at+2b}{2x}\right)^2+x.\dfrac{d^2x}{dt^2}\right]=2a$$
$$2\left[\left(\dfrac{at+b}{x}\right)^2+x.\dfrac{d^2x}{dt^2}\right]=2a$$

$$x.\dfrac{d^2x}{dt^2}=a-\left(\dfrac{at+b}{x}\right)^2$$

$$x.\dfrac{d^2x}{dt^2}=a-(at+b)^2x^{-2}$$

$$\dfrac{d^2x}{dt^2}=\dfrac{a}{x}-(at+b)^2 x^{-3}$$

$$\dfrac{d^2x}{dt^2}=ax^{-1}-(at+b)^2 x^{-3}$$

$$=(ax^2+(at+b)^2) x^{-3}$$

$$accl^x=\dfrac{d^2x}{dt^2}[(at^2+2bt+c)+(at+b)^2]x^{-3}$$

$$acc\,\propto x^{-3}\therefore \boxed{n=3}$$

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