WBJEE · Physics · Motion In One Dimension
A particle moves along \(x\) -axis and its displacement at any time is given by \(x(t)=2 t^{3}-3 t^{2}+4 t\) in SI units. The velocity of the particle when its acceleration is zero, is
- A \(2.5 \mathrm{ms}^{-1}\)
- B \(3.5 \mathrm{ms}^{-1}\)
- C \(4.54 \mathrm{ms}^{-1}\)
- D \(8.5 \mathrm{ms}^{-1}\)
Answer & Solution
Correct Answer
(A) \(2.5 \mathrm{ms}^{-1}\)
Step-by-step Solution
Detailed explanation
Given, \(x(t)=2 t^{3}-3 t^{2}+4 t\) So, velocity, \(v=\frac{d x}{d t}=\left(6 t^{2}-6 t+4\right)\) and acceleration \(a=\frac{d v}{d t}=(12 t-6)\) when acceleration is zero, \(i . e . .(12 t-6)=0\) or \[ t=\frac{6}{12}=\frac{1}{2} \mathrm{s} \] \(\therefore\) Velocity of the…
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