$${\text{Average}}\,\,{\text{speed}} = \frac{{{\text{ Total distance travelled }}}}{{{\text{ Time taken }}}}$$
Let $${t_1}$$ and $${t_2}$$ be times taken by the car to go from $$X$$ to $$Y$$ and then from $$Y$$ to $$X$$ respectively.
$${\text{Then,}}\,{t_1} + {t_2} = \frac{{XY}}{{{v_u}}} + \frac{{XY}}{{{v_d}}} = XY\left( {\frac{{{v_u} + {v_d}}}{{{v_u}{v_d}}}} \right)$$
Total distance travelled $$ = XY + XY = 2XY$$
Therefore, average speed of the car for this round trip is
$$\eqalign{ & {v_{av}} = \frac{{2XY}}{{XY\left( {\frac{{{v_u} + {v_d}}}{{{v_u}{v_d}}}} \right)}}\, \cr & {\text{or}}\,{v_{av}} = \frac{{2{v_u}{v_d}}}{{{v_u} + {v_d}}} \cr} $$

When particle comes to rest,
$$\eqalign{ & V = 0 = \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {40 + 12t - {t^3}} \right) \cr & \Rightarrow 12 - 3{t^2} = 0 \cr & \Rightarrow {t^2} = \frac{{12}}{3} = 4 \cr & \therefore t = 2\,\sec \cr} $$
Therefore distance travelled by particle before coming to rest,
$$x = 40 + 12t - {t^3} = 40 + 12 \times 2 - {\left( 2 \right)^3} = 56\,m$$

$${\text{Given}}:r = 30\,cm = 0.3\,cm = 0.3\,m\,{\text{and}}\,V = 2t$$
Radial acceleration at $$t = 3\,\sec$$
$${a_r} = \frac{{{v^2}}}{r} = \frac{{4{t^2}}}{{0.3}} = \frac{{4 \times {{\left( 3 \right)}^2}}}{{0.3}} = 120\,m/{s^2}$$
and tangential acceleration
$${a_t} = \frac{{dv}}{{dt}} = 2\,m/{s^2}$$

On differentiating, acceleration $$= 0.2\,t$$
$$ \Rightarrow a = f\left( t \right)$$

$$\eqalign{ & {\text{Linear speed}} = {\text{radius}} \times {\text{angular speed}} \cr & {\text{or}}\,v = r\omega \cr & {\text{Here,}}\,r = 20\,cm = 0.20\,m \cr & \omega = 10\,rad/s \cr & \therefore v = 0.20 \times 10 \cr & \Rightarrow v = 2\,m/s \cr} $$

Let the body falls through the height of tower in $$t$$ seconds.
From $${s_n} = u + \frac{a}{2}\left( {2n - 1} \right),$$
we have
Total distance travelled in last $$2\,s$$ of fall is
$$\eqalign{ & s = {s_t} + {s_{\left( {t - 1} \right)}} \cr & = \left[ {0 + \frac{g}{2}\left( {2t - 1} \right)} \right] + \left[ {0 + \frac{g}{2}\left( {2\left( {t - 1} \right) - 1} \right)} \right] \cr & = \frac{g}{2}\left( {2t - 1} \right) + \frac{g}{2}\left( {2t - 3} \right) \cr & = \frac{g}{2}\left( {4t - 4} \right) = \frac{{10}}{2} \times 4\left( {t - 1} \right) \cr & {\text{or}}\,\,40 = 20\left( {t - 1} \right){\text{ }} \cr & {\text{or}}\,\,t = 2 + 1 = 3\,s \cr} $$
Distance travelled in $$t$$ sec is
$$\eqalign{ & s = ut + \frac{1}{2}a{t^2} \cr & = 0 + \frac{1}{2} \times 10 \times {3^2} = 45\,m \cr} $$

On double differentiation of displacement equation gives acceleration of body i.e. $$a = \frac{{{d^2}x}}{{d{t^2}}}$$
The displacement of a particle along a straight line is
$$s = 3{t^3} + 7{t^2} + 14t + 5\,......\left( {\text{i}} \right)$$
Differentiating Eq. (i) w.r.t. time, which gives the velocity
$$\eqalign{ & v = \frac{{ds}}{{dt}} = \frac{d}{{dt}}\left( {3{t^3} + 7{t^2} + 14t + 5} \right) \cr & = \frac{d}{{dt}}\left( {3{t^3}} \right) + \frac{d}{{dt}}\left( {7{t^2}} \right) + \frac{d}{{dt}}\left( {14t} \right) + \frac{d}{{dt}}\left( 5 \right) \cr & v = 3\frac{d}{{dt}}\left( {{t^3}} \right) + 7\frac{d}{{dt}}\left( {{t^2}} \right) + 14\frac{d}{{dt}}\left( t \right) + 0\,......\left( {{\text{ii}}} \right) \cr} $$
(as differentiation of a constant is zero)
$$\eqalign{ & {\text{Now}}\,{\text{use}}\,\frac{d}{{dt}}\left( {{x^n}} \right) = n{x^{n - 1}} \cr & {\text{So}}\,\,v = 3\left( 3 \right){t^{3 - 1}} + 7\left( 2 \right)\left( {{t^{2 - 1}}} \right) + 14\left( {{t^{1 - 1}}} \right) \cr & \Rightarrow v = 9{t^2} + 14t + 14\,......\left( {{\text{iii}}} \right) \cr & \left( {\because {t^0} = 1} \right) \cr} $$
Again differentiating Eq. (iii) w.r.t. time, which gives the acceleration
$$\eqalign{ & a = \frac{{dv}}{{dt}} = \frac{d}{{dt}}\left( {9{t^2} + 14t + 14} \right) \cr & = 18t + 14 + 0 = 18t + 14 \cr & {\text{At}}\,t = 1\,s, \cr & a = 18\left( 1 \right) + 14 = 18 + 14 = 32\,m/{s^2} \cr} $$

The required vector is,
$$\eqalign{ & = 3\frac{{\left( {\vec A \times \vec B} \right)}}{{\left| {\vec A \times \vec B} \right|}} = 3\frac{{\left[ {\left( {2\hat i + 2\hat j + \hat k} \right) \times \left( {2\hat i - 2\hat j + 3\hat k} \right)} \right]}}{{\left| {\vec A \times \vec B} \right|}} \cr & = 3\frac{{\left( {8\hat i - 4\hat j - 8\hat k} \right)}}{{\sqrt {{8^2} + {4^2} + {8^2}} }} \cr & = 2\hat i - \hat j - 2\hat k. \cr} $$

Time taken for vertical direction motion
$$t = \sqrt {\frac{{2h}}{g}} = \sqrt {\frac{{2 \times 490}}{{9.8}}} = \sqrt {100} = 10\,s$$
The same time is for horizontal direction.
$$\therefore x = vt = \left( {60 \times \frac{5}{{18}}} \right) \times 10 = \frac{{500}}{3}m$$

Maximum velocity point means, the point at which $$\frac{{dx}}{{dt}}$$ i.e, the slope of the graph is maximum. At point $$C,$$  slope is maximum.