AP EAMCET · Maths · Application of Derivatives
If the displacement \(S\) of a particle travelling along a straight line in \(t\) seconds is given by \(\mathrm{S}=2 \mathrm{t}^3+2 \mathrm{t}^2-2 \mathrm{t}-3\), then the time taken (in seconds) by the particle to change its direction is
- A \(\frac{1}{3}\)
- B \(2\)
- C \(3\)
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{3}\)
Step-by-step Solution
Detailed explanation
\( V = \frac{dS}{dt} = 6t^2 + 4t - 2 \) \( 6t^2 + 4t - 2 = 0 \Rightarrow 3t^2 + 2t - 1 = 0 \) \( t = \frac{-(2) \pm \sqrt{(2)^2 - 4(3)(-1)}}{2(3)} = \frac{-2 \pm \sqrt{4 + 12}}{6} = \frac{-2 \pm 4}{6} \) \( t = \frac{2}{6} = \frac{1}{3} \text{ s} \text{ (since } t > 0) \)
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