Question

A particle moves along a line $suc\Uparrow t\Uparrow at$ its displacement $x$ changes $wit\Uparrow$ time $t$ as $x=\sqrt{a{t}^2+2bt+c}$ where $a,b$ and $c$ are constants, $t\Uparrow ent\Uparrow e$ acceleration varies as:

• A. $\frac{1}{x}$
• B. $\frac{1}{x^2}$
• C. $\frac{1}{x^3}$
• D. $\frac{1}{x^4}$

Answer 1

The correct option is C $\frac{1}{x^3}$

$\begin{array}{c}x=\sqrt{a{t}^2+2bt+c}\\ x^2=a{t}^2+2bt+c\end{array}$
differentiate w.r.t. time t
$\begin{array}{c}2x\frac{dx}{dt}=2at+2b\\ or,xv=at+b\end{array}$
again differentiate
$\begin{array}{c}x\frac{dv}{dt}+v\frac{dx}{dt}=a\\ x\left(accelartion\right)+v^2=a\\ accelartion=\frac{a-v^2}{x}\\ v\propto \frac{1}{x}\\ so,v^2\propto \frac{1}{x^2}\\ accelartion\propto \frac{v^2}{x}\propto \frac{\frac{1}{x^2}}{x}\\ accelaration\propto \frac{1}{x^3}\end{array}$