JEE Mains · Physics · STD 11 - 2. motion in straight line
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 _______.
- A \(a=-2 \alpha v^3\)
- B \(a=-5 \alpha v^5\)
- C \(a=-3 \alpha v^2\)
- D \(a=-4 \alpha v^4\)
Answer & Solution
Correct Answer
(A) \(a=-2 \alpha v^3\)
Step-by-step Solution
Detailed explanation
\(\mathrm{t}=\alpha \mathrm{x}^2+\beta \mathrm{x} \text { (differentiating wrt time) }\) \(\frac{\mathrm{dt}}{\mathrm{dx}}=2 \alpha \mathrm{x}+\beta\) \(\frac{1}{\mathrm{v}}=2 \alpha \mathrm{x}+\beta\) \(\text { (differentiating wrt time) }\)…
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