JEE Mains · Physics · STD 11 - 2. motion in straight line
A particle moves in a straight line so that its displacement \(x\) at any time \(t\) is given by \(x^2=1+t^2\). Its acceleration at any time \(\mathrm{t}\) is \(\mathrm{x}^{-\mathrm{n}}\) where \(\mathrm{n}=\) _______.
- A \(5\)
- B \(2\)
- C \(3\)
- D \(1\)
Answer & Solution
Correct Answer
(C) \(3\)
Step-by-step Solution
Detailed explanation
\(x^2=1+t^2\) \(2 x \frac{d x}{d t}=2 t\) \(x v=t\) \(x \frac{d v}{d t}+v \frac{d x}{d t}=1\) \(x \cdot a+v^2=1\) \(a=\frac{1-v^2}{x}=\frac{1-t^2 / x^2}{x}\) \(a=\frac{1}{x^3}=x^{-3}\)
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