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JEE Mains · Physics · STD 11 - Motion in straight line
  • JEE Mains 2024
  • Hard
  • MCQ

The relation between time ' \(t\) ' and distance ' \(x\) ' is \(t=\) \(\alpha x^2+\beta x\), where \(\alpha\) and \(\beta\) are constants. The relation between acceleration \((a)\) and velocity \((v)\) is _______.

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Answer & Solution

Correct Answer

(A) \(a=-2 \alpha v^3\)

Step-by-step Solution

Detailed explanation

\(\mathrm{t}=\alpha \mathrm{x}^2+\beta \mathrm{x} \text { (differentiating wrt time) }\)

\(\frac{\mathrm{dt}}{\mathrm{dx}}=2 \alpha \mathrm{x}+\beta\)

\(\frac{1}{\mathrm{v}}=2 \alpha \mathrm{x}+\beta\)

\(\text { (differentiating wrt time) }\)

\(-\frac{1}{\mathrm{v}^2} \frac{\mathrm{dv}}{\mathrm{dt}}=2 \alpha \frac{\mathrm{dx}}{\mathrm{dt}}\)

\(\frac{\mathrm{dv}}{\mathrm{dt}}=-2 \alpha \mathrm{v}^3\)

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