AP EAMCET · PHYSICS · Motion In One Dimension
If the displacement (s in metre) of a moving particle in terms of time (t in second) is \(s=t^3-6 t^2+18 t+9\), then the minimum velocity attained by the particle is
- A \(29 \mathrm{~m} \mathrm{~s}^{-1}\)
- B \(5 \mathrm{~ms}^{-1}\)
- C \(6 \mathrm{~ms}^{-1}\)
- D \(12 \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(C) \(6 \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
\(v = \frac{ds}{dt} = 3t^2 - 12t + 18\) \(\frac{dv}{dt} = 6t - 12\) \(6t - 12 = 0 \Rightarrow t = 2 \mathrm{~s}\) \(v_{min} = 3(2)^2 - 12(2) + 18 = 12 - 24 + 18 = 6 \mathrm{~m} \mathrm{~s}^{-1}\)
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