TS EAMCET · Physics · Motion In One Dimension
The relation between time \(t\) and distance \(x\) of a particle is \(t=a x^2+b x\), where \(a\) and \(\mathrm{b}\) are constants.
If \(\mathrm{v}\) is the velocity of the particle, then its acceleration is
- A \(-2abv^2\)
- B \(2bv^3\)
- C \(-2av^3\)
- D \(2av^2\)
Answer & Solution
Correct Answer
(C) \(-2av^3\)
Step-by-step Solution
Detailed explanation
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