A) \[\frac{1}{x}-\frac{1}{{{x}^{2}}}\]
B) \[-\frac{t}{{{x}^{2}}}\]
C) \[-\frac{{{t}^{2}}}{{{x}^{3}}}\]
D) \[\frac{1}{{{x}^{3}}}\]
Correct Answer: D
Solution :
\[{{x}^{2}}={{t}^{2}}+1\] ...(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 \[\therefore \] \[xa=1-{{v}^{2}}\] \[\Rightarrow \] \[a=\frac{1-{{v}^{2}}}{x}\] \[=\frac{1-\frac{{{t}^{2}}}{{{x}^{2}}}}{x}=\frac{{{x}^{2}}-{{t}^{2}}}{{{x}^{3}}}\] \[=\frac{1}{{{x}^{3}}}\] \[(\because {{x}^{2}}={{t}^{2}}+1)\]
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