Question

A particle moves in a straight line so that its displacement $x$ (metre) in time t (second) is given by $x^2=t^2+1$ . Its acceleration in $m/s^{-2}$ , is

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

Answer 1

Answer: (D)

$x^2=t^2+1$ ...(i)
Differentiating Eq. (i) w.r.t. t, we have
$2x=\frac{dx}{dt}=2t$ or $xv=t$ ...(ii)
where, $v=\frac{dx}{dt}=$
velocity Differentiating Eq. (ii), w.r.t. t, we have
$x\frac{dv}{dt}+v\frac{dx}{dt}=1$
$xa+v.v=1$ where, $a=\frac{dv}{dt}=$ acceleration
$\therefore$ $xa=1-v^2$
$\Rightarrow$ $a=\frac{1-v^2}{x}$
$=\frac{1-\frac{t^2}{x^2}}{x}=\frac{x^2-t^2}{x^3}$
$=\frac{1}{x^3}$ $(\because x^2=t^2+1)$