It this problem to calculate maximum velocity we will use $\frac{d v}{d t}=0$, then the time corresponding to maximum velocity will be obtained.

Given velocity

$$ v(t)=2 t(3-t)=6 t-2 t^2 \,\,\,\,....(i) $$

(a) For maximum velocity $\frac{d v(t)}{d t}=0$

$$ \begin{aligned} &\Rightarrow \frac{d}{d t}\left(6 t-2 t^2\right) =0 \\\\ & \Rightarrow 6-4 t =0 \\\\ & \Rightarrow t =\frac{6}{4}=\frac{3}{2} s=1.5 \mathrm{~s} \end{aligned} $$

(b) From Eq. (i) $v=6 t-2 t^2$

$$ \begin{array}{ll} \Rightarrow & \frac{d s}{d t}=6 t-2 t^2 \\ \Rightarrow & d s=\left(6 t-2 t^2\right) d t \end{array} $$

where, s is displacement

$\therefore$ Distance travelled in time interval 0 to 3 s .

$\begin{aligned} s & =\int_0^3\left(6 t-2 t^2\right) d t \\\\ & =\left[\frac{6 t^2}{2}-\frac{2 t^3}{3}\right]_0^3=\left[3 t^2-\frac{2}{3} t^3\right]_0^3 \\\\ & =3 \times 9-\frac{2}{3} \times 3 \times 3 \times 3 \\\\ & =27-18=9 \mathrm{~m} . \\\\ \text { Average velocity } & =\frac{\text { Distance travelled }}{\text { Time }} \\ & =\frac{9}{3}=3 \mathrm{~m} / \mathrm{s}\end{aligned}$

Given,

$$ \begin{aligned} &x =6 t-2 t^2 \\\\ &\Rightarrow 3 =6 t-2 t^2 \\\\& \Rightarrow 2 t^2-6 t-3 =0 \end{aligned} $$

$$ \begin{aligned} \Rightarrow \quad t & =\frac{6 \pm \sqrt{6^2-4 \times 2 \times 3}}{2 \times 2}=\frac{6 \pm \sqrt{36-24}}{4} \\\\ & =\frac{6 \pm \sqrt{12}}{4}=\frac{3 \pm 2 \sqrt{3}}{2} \end{aligned} $$

Considering positive sign only

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

(c) In a periodic motion when velocity is zero acceleration will be maximum putting $v=0$ in Eq. (i)

$$ \begin{aligned} 0 & =6 t-2 t^2 \\\\ 0 & =t(6-2 t) \\\\ & =t \times 2(3-t)=0 \\\\ t & =0 \text { or } 3 \mathrm{~s} \end{aligned} $$

$$ \Rightarrow \quad 0=t(6-2 t) $$

$$ \Rightarrow \quad t=\text { 0 or 3s } $$

(d) Distance covered in 0 to $3 \mathrm{~s}=9 \mathrm{~m}$

Distance covered in 3 to $6 \mathrm{~s}=\int_3^6\left(18-9 t+t^2\right) d t$

$$ \begin{aligned} & =\left(18 t-\frac{9 t^2}{2}+\frac{t^3}{3}\right)_3^6 \\ & =18 \times 6-\frac{9}{2} \times 6^2+\frac{6^3}{3}-\left(18 \times 3-\frac{9 \times 3^2}{2}+\frac{3^3}{3}\right) \\ & =108-9 \times 18+\frac{6^3}{3}-18 \times 3+\frac{9}{2} \times 9-\frac{27}{3} \\ & =108-18 \times 9+\frac{216}{3}-54+4.5 \times 9-9=-4.5 \mathrm{~m} \end{aligned} $$

$\therefore$ Total distance travelled in one cycle $=s_1+s_2=9-4.5=4.5 \mathrm{~m}$

Number of cycles covered in total distance to be covered $=\frac{20}{4.5} \approx 4.44 \approx 5$.

Let the speeds of the two balls (1 and 2) be $v_1$ and $v_2$ where

if $v_1 = 2v, v_2 = v$

If $y_1$ and $y_2$ and the distance covered by the balls 1 and 2, respectively, before coming to rest, then

$y_1 = \frac{v_1^2}{2g} = \frac{4v^2}{2g}$ and $y_2 = \frac{v_2^2}{2g} = \frac{v^2}{2g}$

$y_1 - y_2 = 15m$

$\frac{4v^2}{2g} - \frac{v^2}{2g} = 15m$ or $ v^2 = \frac{15m \times (2 \times 10)}{3} $

or $v = 10 \; m/s$

Clearly, $v_1 = 20 m/s$ and $v_2 = 10 m/s$

as $y_1 = \frac{v_1^2}{2g} = \frac{(20m)^2}{2 \times 10m/s^2} = 20m$

$y_2 = y_1 - 15m = 5m$

If $t_2$ is the time taken by the ball 2 to ner a distance of 5 m, then from $y_2 = v_2t - \frac{1}{2}gt^2$

$5 = 10t_2 - \frac{5t_2^2}{2}$ or $t_2^2 - 2t_2 + 1 = 0,$

where $t_2 = 1s$

Since $t_1 $ (time taken by ball 1 to cover distance of 20 m ) is 2s, time interval between the two thvnws

$ = t_1 - t_2 = 2s - 1s = 1s $

Note We should be very careful, when we are applying the equation of rectilinear motion.