Q. A particle moves in a straight line so that its displacement $x$ metre at time $t$ second is given by $t = \sqrt{x^{2} - 1}$ Its acceleration in $ms^{- 2}\ $at time $t$ second is
A) $\frac{1}{x^{3}}$
B) $\frac{t^{2}}{x^{3}}$
C) $\frac{1}{x} - \frac{t^{2}}{x^{3}}$
D) $\frac{t^{2}}{x^{3}} - \frac{1}{x^{2}}$
Explanation
(a) $t^{2} = x^{2} - 1$ or ${\ x}^{2} = t^{2} + 1$
Or $2x\frac{dx}{dt} = 2t$ or $xv = t$
Or $2x\frac{dx}{dt} + v\frac{dx}{dt} = 1$
Or $x\frac{dv}{dt} = 1 - v^{2}$
Or $\frac{dv}{dt} = \frac{1 - v^{2}}{x} = \frac{1 - \frac{t^{2}}{x^{2}}}{x} = \frac{x^{2} - t^{2}}{x^{3}}$
But $x^{2} - t^{2} = 1$
$\therefore\ \ \frac{dv}{dt} = \frac{1}{x^{3}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.