AP EAMCET · Maths · Application of Derivatives
For a particle moving on a straight line it is observed that the distance, ' \(s\) ' at a time ' \(t\) ' is given by \(S=6 t-\frac{t^3}{2}\). The maximum velocity during the motion is
- A 3
- B 6
- C 9
- D 12
Answer & Solution
Correct Answer
(B) 6
Step-by-step Solution
Detailed explanation
\[ \begin{aligned} \text { } V & =\frac{d S}{d t}=\frac{d}{d t}\left(6 t-\frac{t^3}{2}\right) \\ V & =6-\frac{3}{2} t^2 \Rightarrow \frac{d V}{d t}=-3 t \\ \frac{d V}{d t} & =0 \Rightarrow-3 t=0 \\ \Rightarrow t & =0 \Rightarrow \frac{d^2 V}{d t^2}=-3 < 0 \end{aligned} \] \(V\)…
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