A monkey climbs up a slippery pole for $3$ and subsequently slips for $3$. Its velocity at time $t$ is given by $v (t) = 2t \,(3s -t)$ ;  $0 < t < 3$ and $v(t) =\,-\, (t -3)\,(6 -t)$ ; $3 < t < 6$ $s$ in $m/s$. It repeats this cycle till it reaches the height of $20\, m$.

$(a)$ At what time is its velocity maximum ?

$(b)$ At what time is its average velocity maximum ?

$(c)$ At what time is its acceleration maximum in magnitude ?

$(d)$ How many cycles (counting fractions) are required to reach the top ?

For maximum velocity, $\frac{d v}{d t}=0$

Given velocity, $v=2 t(3-t)=6 t-2 t^{2}$

$(a)$ For maximum velocity,

$\frac{d v}{d t}=0$

$\therefore \frac{d}{d t}\left(6 t-2 t^{2}\right)=0$

$\therefore 6-4 t=0$

$\therefore t=\frac{6}{4}=\frac{3}{2} \mathrm{~s}=1.5 \mathrm{~s}$

$(b)$ From equ. $(i)$ $v=6 t-2 t^{2}$

$\therefore \frac{d x}{d t}=6 t-2 t^{2}$

$\therefore d x=\left(6 t-2 t^{2}\right) d t$

where $x$ is displacement.

$\therefore$ Distance travelled in time interval 0 to $3 \mathrm{~s}$.

$x_{1} =\int_{0}^{3}\left(6 t-2 t^{2}\right) d t$

$=\left[\frac{6 t^{2}}{2}-\frac{2 t^{3}}{3}\right]_{0}^{3}=\left[3 t^{2}-\frac{2}{3} t^{3}\right]_{0}^{3}$

$=3 \times 9-\frac{2}{3} \times 3 \times 3 \times 3$

$=27-18=9 \mathrm{~m}$

$\text { Average velocity }=\frac{\text { Displacement }}{\text { Time }}$

$=\frac{9}{3}=3 \mathrm{~m} / \mathrm{s} ~\\ \text { Given, } x =6 t-2 t^{2}$

$3 =6 t-2 t^{2}$