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) $