JEE Mains · Physics · STD 11 - 2. motion in straight line
The relation between time \(t\) and distance \(x\) for a moving body is given as \(t=m x^{2}+n x\), where \({m}\) and \({n}\) are constants. The retardation of the motion is - (Where \(v\) stands for velocity)
- A \(2 n^{2} v^{3}\)
- B \(2 {mv}^{3}\)
- C \(2 n v^{3}\)
- D \(2 {mnv}^{3}\)
Answer & Solution
Correct Answer
(B) \(2 {mv}^{3}\)
Step-by-step Solution
Detailed explanation
\(t =m x^{2}+n x\) \(\frac{1}{v}=\frac{d t}{d x}=2 m x+n\) \(v=\frac{1}{2 m x+n}\) \(\frac{d v}{d t}=-\frac{2 m}{(2 m x+n)^{2}}\left(\frac{d x}{d t}\right)\) \(a=-(2 m) v^{3}\)
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