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

Given, $x=\sqrt{a{t}^2+2bt+c}$
Differentiating with respect to time
$\frac{dx}{dt}=v=\frac{1}{2\sqrt{a{t}^2+2bt+c}}\times (2at+2b)$

$\Rightarrow v=\frac{at+b}{x}$
$\Rightarrow vx=at+b$

Differentiating with respect to $x$
$\Rightarrow \frac{dv}{dx}\times x+v=a\times \frac{dt}{dx}$

Multiply both side by $v$

$\Rightarrow \left(v\frac{dv}{dx}\right)x+v^2=a$

$\Rightarrow a^{\prime }x=a=v^2$

$a^{\prime }x=a-{\left(\frac{at+b}{x}\right)}^2$

$a^{\prime }x=\frac{a{x}^2-(at+{)}^2}{x^2}$

$\Rightarrow a^{\prime }x=\frac{a(a{t}^2+2bt+c)-(at+b{)}^2}{x^2}$

$\Rightarrow a^{\prime }x=\frac{ac-b^2}{x^2}$

$\Rightarrow a^{\prime }=\frac{ac-b^2}{x^3}$

$\therefore a^{\prime }\propto \frac{1}{x^3}$

$\therefore n=3$