In this problem equations related to one dimensional motion will be applied for acceleration positive sign will be used and for retardation negative sign will be used.

Given, speed of car as well as truck = 72 km/h

$= 72 \times \frac{5}{18} \text{ m/s} = 20 \text{ m/s}$

Retarded motion for truck

$v = u + a_t t$

$0 = 20 + a_t \times 5$

or $a_t = -4 \text{ m/s}^2$

Retarded motion for the car

$$ \begin{aligned} v & =u+a_c t \\\\ 0 & =20+a_c \times 3 \\\\ \text{or}\,\, a_c & =-\frac{20}{3} \mathrm{~m} / \mathrm{s}^2 \end{aligned} $$

Let car be at a distance $x$ from truck, when truck gives the signal and $t$ be the time taken to cover this distance.

As human response time is 0.5 s , therefore, time of retarded motion of car is $(t-0.5) \mathrm{s}$. Velocity of car after time $t$,

$$ \begin{aligned} v_c & =u-a t \\ & =20-\left(\frac{20}{3}\right)(t-0.5) \end{aligned} $$

Velocity of truck after time $t$,

$$ v_t=20-4 t $$

To avoid the car bump onto the truck,

$$ \begin{aligned} 20-\frac{20}{3} & (t-0.5)=20-4 t \\ 4 t & =\frac{20}{3}(t-0.5) \\ t & =\frac{5}{3}(t-0.5) \\ 3 t & =5 t-2.5 \\ \Rightarrow \quad t & =\frac{2.5}{2}=\frac{5}{4} \mathrm{~s} \end{aligned} $$

Distance travelled by the truck in time $t$,

$$ \begin{aligned} s_t & =u_t t+\frac{1}{2} a_t t^2 \\ & =20 \times \frac{5}{4}+\frac{1}{2} \times(-4) \times\left(\frac{5}{4}\right)^2 \\ & =21.875 \mathrm{~m} \end{aligned} $$

$$ \begin{aligned} & \text { Distance travelled by car in time } t=\text { Distance travelled by car in } 0.5 \mathrm{~s} \text { (without retardation) } \\ & + \text { Distance travelled by car in }(t-0.5) \mathrm{s} \text { (with retardation) } \\\\ & s_c=(20 \times 0.5)+20\left(\frac{5}{4}-0.5\right)-\frac{1}{2}\left(\frac{20}{3}\right)\left(\frac{5}{4}-0.5\right)^2 \\\\ & =23.125 \mathrm{~m} \\\\ & s_c-s_t=23.125-21.875=1.250 \mathrm{~m} \\\\ & \end{aligned} $$

Therefore, to avoid the bump onto the truck, the car must maintain a distance from the truck more than 1.250 m .

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.