A) \[\frac{1}{{{x}^{4}}}\]
B) \[\frac{t}{{{x}^{2}}}\]
C) \[\frac{t}{x}-\frac{{{t}^{2}}}{{{x}^{3}}}\]
D) \[\frac{1}{x}-\frac{t}{{{x}^{2}}}\]
Correct Answer: C
Solution :
Given, \[{{x}^{2}}={{t}^{2}}+1\] Differentiate, w.r.t. r, we get \[2x=\frac{dx}{dt}=2t\] \[\therefore \] \[v=\frac{dx}{dt}=\frac{t}{x}\] Also, \[a=\frac{dv}{dt}=\frac{x-t\left( \frac{dx}{dt} \right)}{{{x}^{2}}}=\frac{x-\left( \frac{{{t}^{2}}}{x} \right)}{{{x}^{2}}}\] \[\Rightarrow \] \[a=\frac{1}{x}-\frac{{{t}^{2}}}{{{x}^{3}}}\]
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