AP EAMCET · PHYSICS · Motion In One Dimension
The motion of a particle along a straight line is described by the function, \(x=(2 t-3)^2\) where \(x\) is in metres and \(t\) is in seconds. The acceleration of the particle at \(t=2 \mathrm{~s}\) is
- A \(1 \mathrm{~ms}^{-2}\)
- B \(4 \mathrm{~ms}^{-2}\)
- C \(8 \mathrm{~ms}^{-2}\)
- D \(7 \mathrm{~ms}^{-2}\)
Answer & Solution
Correct Answer
(C) \(8 \mathrm{~ms}^{-2}\)
Step-by-step Solution
Detailed explanation
Given, \(x=(2 t-3)^2\) where, \[ \begin{aligned} & x=\text { displacement and } t=\text { time. } \\ & x=4 t^2+9-12 t \end{aligned} \] Differentiate w.r.t. time, we get Velocity, \(v=\frac{d x}{d t}=8 t-12\) Acceleration, \(a=\frac{d v}{d t}=8 \mathrm{~m} / \mathrm{s}^2\)
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