TS EAMCET · Maths · Application of Derivatives
If a particle is moving in a straight line so that after \(t\) seconds its distance S (in cms ) from a fixed point on the line is given by \(\mathrm{S}=f(\mathrm{t})=\mathrm{t}^3-5 \mathrm{t}^2+8 \mathrm{t}\) then the acceleration of the particle at \(t=5 \mathrm{sec}\) is \(\left(\mathrm{in} \mathrm{cm} / \mathrm{sec}^2\right)\)
- A \(10\)
- B \(30\)
- C \(20\)
- D \(40\)
Answer & Solution
Correct Answer
(C) \(20\)
Step-by-step Solution
Detailed explanation
\(v(t) = \frac{dS}{dt} = \frac{d}{dt}(t^3-5t^2+8t) = 3t^2-10t+8\) \(a(t) = \frac{dv}{dt} = \frac{d}{dt}(3t^2-10t+8) = 6t-10\) \(a(5) = 6(5)-10 = 30-10 = 20 \mathrm{cm/sec^2}\)
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