AP EAMCET · Maths · Application of Derivatives
The displacement S of a particle measured from a fixed point O on a line is given by \(\mathrm{S}=\mathrm{t}^3-16 \mathrm{t}^2+64 \mathrm{t}-16\). Then the time at which displacement of the particle is maximum is
- A 8
- B 4
- C \(\frac{8}{3}\)
- D \(\frac{4}{3}\)
Answer & Solution
Correct Answer
(C) \(\frac{8}{3}\)
Step-by-step Solution
Detailed explanation
\( \frac{dS}{dt} = 3t^2 - 32t + 64 \) \( 3t^2 - 32t + 64 = 0 \) \( t = \frac{32 \pm \sqrt{(-32)^2 - 4(3)(64)}}{2(3)} = \frac{32 \pm \sqrt{1024 - 768}}{6} = \frac{32 \pm 16}{6} \) \( t = \frac{48}{6} = 8 \) or \( t = \frac{16}{6} = \frac{8}{3} \) \( \frac{d^2S}{dt^2} = 6t - 32 \)…
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