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 __________.

Answer 1

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