AP EAMCET · PHYSICS · Motion In One Dimension
The velocity of a particle is given by \(v=2 t^2-8 t+15 \mathrm{~ms}^{-1}\). Find its instantaneous acceleration at \(t=5 \mathrm{~s}\).
- A \(18 \mathrm{~ms}^{-2}\)
- B \(20 \mathrm{~ms}^{-2}\)
- C \(5 \mathrm{~ms}^{-2}\)
- D \(12 \mathrm{~ms}^{-2}\)
Answer & Solution
Correct Answer
(D) \(12 \mathrm{~ms}^{-2}\)
Step-by-step Solution
Detailed explanation
Given, velocity, \(v=2 t^2-8 t+15\) \[ \text { time, } t=5 \mathrm{~s} \] Let acceleration be \(a\). \[ \begin{aligned} \because \quad & a=\frac{d v}{d t} \\ \therefore \quad & a=4 t-\left.8\right|_{t=5} \\ & =4 \times 5-8 \\ & =20-8 \\ & =12 \mathrm{~ms}^{-2} \end{aligned} \]
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