AP EAMCET · PHYSICS · Motion In One Dimension
If the velocity of a particle moving along a straight line with uniform acceleration, is given by \(\mathrm{V}=(\sqrt{196-16 \mathrm{X}}) \mathrm{ms}^{-1}\), then its acceleration is ( \(\mathrm{x}\) is displacement of the particle)
- A \(8 \mathrm{~ms}^{-2}\)
- B \(14 \mathrm{~ms}^{-2}\)
- C \(-8 \mathrm{~ms}^{-2}\)
- D \(-16 \mathrm{~ms}^{-2}\)
Answer & Solution
Correct Answer
(C) \(-8 \mathrm{~ms}^{-2}\)
Step-by-step Solution
Detailed explanation
Velocity of a particle is given by \(\begin{aligned} & v=\sqrt{196-16 x} \Rightarrow v^2=196-16 x \\ & 2 v \frac{d v}{d t}=-16 \frac{d x}{d t} \\ & v \frac{d v}{d t}=-8 v \end{aligned}\) Acceleration, \(a=-8 \mathrm{~m} / \mathrm{s}^2\)
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