Question

A point moves such that its displacement as function of time is given by $ {{x}^{2}}={{t}^{2}}+1. $ Its acceleration at time $ t $ is

• A. $ \frac{1}{{{x}^{4}}} $
• B. $ -\frac{t}{{{x}^{2}}} $
• C. $ \frac{1}{x}-\frac{{{t}^{2}}}{{{x}^{3}}} $
• D. $ \frac{1}{x}-\frac{t}{{{x}^{2}}} $

Answer 1

Answer: (C)

$ {{x}^{2}}={{t}^{2}}+t $ Differentiate wrt x $ 2\times \frac{dx}{dt}=2t $ $ v=\frac{dx}{dt}=\frac{t}{x} $ $ a=\frac{dv}{dt} $ $ =\frac{x-t\left( \frac{dx}{dt} \right)}{{{x}^{2}}} $ $ =\frac{x-\left( \frac{{{t}^{2}}}{x} \right)}{{{x}^{2}}} $ $ a=\frac{1}{x}-\frac{{{t}^{2}}}{{{x}^{3}}} $