For a particle moving on a straight line it is observed that the distance, ' $s$ ' at a time ' $t$ ' is…

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
  1. 3
  2. 6
  3. 9
  4. 12

Solution

$ \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} $ $V$ is maximum at $t=0$ $ V=6-\frac{3}{2} \times 0=6 $

Asked in: AP EAMCET 2022 (06 Jul Shift 2)

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