(b) For maximum average velocity net displacement should be maximum.

Step 2: Calculate net displacement in 0 < t < 3 sec.

 $As \frac{\mathrm{ds}}{\mathrm{dt}} = \mathrm{v}\left(\mathrm{t}\right)$
$\Rightarrow \mathrm{ds} = \mathrm{v}\left(\mathrm{t}\right)\mathrm{dt}$
$\Rightarrow \mathrm{s} = {\int }_0^3\left(6\mathrm{t}-2{\mathrm{t}}^2\right)d\mathrm{t}$
$= {\left[\frac{6{\mathrm{t}}^2}{2} - \frac{2{\mathrm{t}}^3}{3}\right]}_0^3 = {\left[3{\mathrm{t}}^2 - {\mathrm{t}}^3\right]}_0^3$
$= 3\times 9 - \frac{2}{3}\times 3\times 3\times 3$
$= 27 - 18 = 9 \mathrm{m}$

where, s is displacement

$\begin{array}{l}Average velocity = \frac{Net displacement}{Total time taken}\\ = \frac{9}{3} = 3 \mathrm{m}/\mathrm{s}.\end{array}$

Step 3: Calculate the time at which average velocity is maximum

Given,

$\begin{array}{l} \mathrm{v}\left(\mathrm{t}\right) = 6\mathrm{t}-2{\mathrm{t}}^2\\ \Rightarrow 3 = 6\mathrm{t}-2{\mathrm{t}}^2\\ \Rightarrow 2{\mathrm{t}}^2 - 6\mathrm{t}-3=0\\ \Rightarrow \mathrm{t} = \frac{6\pm \sqrt{6^2-4\times 2\times 3}}{2\times 2} = \frac{6\pm \sqrt{36-24}}{4}\\ = \frac{6\pm \sqrt{12}}{4} = \frac{3\pm 2\sqrt{3}}{2}\end{array}$

Considering positive, sign only

$\mathrm{t}=\frac{3+2\sqrt{3}}{2}=\frac{3+2\times 1.732}{2}=\frac{9}{4} \mathrm{s}$

(c) In a periodic motion when velocity is zero acceleration will be maximum putting v = 0. $\begin{array}{l} 0=6\mathrm{t}-2{\mathrm{t}}^2\\ \Rightarrow =\mathrm{t}(6-2\mathrm{t})\\ =\mathrm{t}\times 2(3-\mathrm{t})=0\\ \Rightarrow =0\text{ or }3\mathrm{s}\end{array}$

(d) Number of cycles = $\frac{\mathrm{Height} \mathrm{of} \mathrm{pole}}{\mathrm{Net} \mathrm{upward} \mathrm{displacement} \mathrm{per} \mathrm{cycle}}.$

Step 4: Find the distance covered in 0 to 3s 

As calculated above distance covered in 0 to 3s = 9 m

Step 5: Find the distance covered in 3 to 6 sec.

$\begin{array}{l}\text{ Distance covered in }3\text{ to }6\mathrm{s}={\int }_3^6(18-9\mathrm{t}+{\mathrm{t}}^2)\mathrm{dt}\\ = {\left[18\mathrm{t}-\frac{9{\mathrm{t}}^2}{2}+\frac{{\mathrm{t}}^3}{3}\right]}_3^6\\ = 18\times 6 - \frac{9}{2}\times 6^2 + \frac{6^3}{3} - \left[18\times 3 - \frac{9\times 3^2}{2} + \frac{3^3}{3}\right]\\ =108 - 9\times 18 + \frac{6^3}{3} - 18\times 3 + \frac{9}{2}\times 9 - \frac{27}{3}\\ =108 - 18\times 9 + \frac{216}{3} - 54 + 4.5\times 9 - 9=- 4.5\mathrm{m}\end{array}$

Step 6: Find net upward displacement in one cycle

$\therefore$ Net upward displacement traveled in one cycle  $= s_1\mathit{ }+ s_2\mathit{ }= 9 - 4.5 = 4.5 \mathrm{m}$

Step 7: Calculate the number of cycles required to reach the top 

The number of cycles $= \frac{20}{4.5} = 4.44 = 5.$