Question

A point moves in a straight line so that its displacement $x$ metre at a time $t$ second is such that $t=(x^2-1{)}^{1/2}$. Its acceleration in $m/s^2$ at time $t$ second is:

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

Answer 1

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

Given,

$t=(x^2-1{)}^{1/2}$

Squaring on both sides

$t^2=x^2-1$

$\Rightarrow x^2=t^2+1$

Differentiating with respect to $t$

$2x\frac{dx}{dt}=2t$ as $\frac{dx}{dt}=v$

$v=\frac{t}{x}$

Differentiating both sides (w.r.t to $t$), we will get,

$a=\frac{1}{x}-\frac{tv}{x^2}$

$a=\frac{x^2-t^2}{x^3}=\frac{1}{x^3}$