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 _______.
Pick an option to check your answer.
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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