Question

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

Answer 1

Answer: 3.00

$x=\sqrt{a{t}^2+2bt+e}$
Differentiating $w.r.t.$ time
$\frac{dx}{dt}=v=\frac{1}{2\sqrt{a{t}^2+2bt+c}}\times \left(2at+2b\right)$
$\Rightarrow v=\frac{at+b}{x}$
$\Rightarrow vx=at+b$
Differentiating $w.r.t.x$
$\Rightarrow \frac{dx}{dx}\times x+v=a\times \frac{dt}{dx}$
Multiply both side by $v$
$\Rightarrow \left(v\frac{dv}{dx}\right)x+v^2=a$
$\Rightarrow a^{\prime }x=a-v^2$ [Here a' is acceleration]
$\Rightarrow a^{\prime }x=\frac{a{x}^2-{\left(at+b\right)}^2}{x^2}$
$\Rightarrow a^{\prime }x=\frac{a\left(a{t}^2+2bt+c\right)-{\left(at+b\right)}^2}{x^2}$
$\Rightarrow a^{\prime }x=\frac{ac-b^2}{x^2}$
$\Rightarrow a^{\prime }=\frac{ac-b^2}{x^3}$
$\therefore a^{\prime }\propto \frac{1}{x^3}\therefore n=3$