Question

A point moves in a straight line so that its displacement $x$ at time $t$ is given by equation $x^2=t^2+1$. Its acceleration is:

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

Answer 1

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

Differentiating w.r.t ${}^{\prime }{t}^{\prime }$ on both sides,
$2x\frac{dx}{dt}=2t+0$
$xv=t\Rightarrow v=\frac{t}{x}$
Differentiating again w.r.t $t$,
$x\frac{dv}{dt}+v\frac{dx}{dt}=1$
$xa+v^2=1$
$xa=1-v^2$
$xa=1-\frac{t^2}{x^2}(\because xv=t)$
$xa=1-\left[\frac{x^2-1}{x^2}\right](\because x^2=t^2+1)$
$xa=1-1+\frac{1}{x^2}$
$\Rightarrow a=\frac{1}{x^3}$