AP EAMCET · Maths · Application of Derivatives
A particle moves along a straight line according to the law \(s=\frac{1}{3} t^3-3 t^2+9 t+17\), where \(s\) is in metre and \(t\) is in second. Its velocity decreases in
- A \(0 < t < 5\)
- B \(0 < t < 3\)
- C \(t>5\)
- D \(t>3\)
Answer & Solution
Correct Answer
(B) \(0 < t < 3\)
Step-by-step Solution
Detailed explanation
We have, \[ \begin{aligned} & S=\frac{1}{3} t^3-3 t^2+9 t+17 \\ & V=\frac{d s}{d t}=t^2-6 t+9 \\ & V=t^2-6 t+9 \\ & \frac{d V}{d t}=2 t-6 \end{aligned} \] Velocity is decreasing, if \(\frac{d V}{d t} 0] \end{array} \]
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