The given equation is v = 4 + 2( C₁ + C₂t )
v  = ( 4 + 2C₁ ) + 2C₂t
Comparing the above equations with equation of motion 
v = u + at
Intial velocity,  u = 4 + 2C₁
Acceleration of the particle = 2C₂

The value of position x will be maximum, when the value of sin (wt) is maximum for this
\(sin(\omega t)=1=sin\quad \pi /2\)
or   \(\omega t=\frac { \pi }{ 2 } \Rightarrow t=\left( \frac { \pi }{ 2\omega } \right) \)

An object could be considered as a point object if it covers a distance much larger than its own size.
e.g., If a bus of 5 m in size move 100 km, then the bus can be considered as a point object.

Yes,the particle has a uniform acceleration because it does not depend on time t

Given, acceleration = as \( \int { dv }  =  \int { \beta dv }\)
On integrating , we get  \(v =  \frac{\beta t^2}{2} +C=\frac{\beta t^2}{2}\)    [∵ C = 0]

No, because velocity is a speed with direction. Therefore, abody having constant velocity cannot have a varying speed.

Yes, a body could have constant speed but varying velocity is equal to the speed for a particular motion changes.

False,because velocity is the speed of body in a given condition. When speed is zero, the magnitude of velocity of body is zero. Thus, velocity is zero.

Yes,when the velocity is negative, the x-t graph has a negative slope.

The velocity of the paritcle is zero because the slope of (x-t )graph is zero.

(i) The given equation x = at² - bt³
If t = 0, x_o=0
if t = 2s,x₂ = 4a - 8b
\(\triangle x={ x }_{ 2 }-{ x }_{ 0 }=4a-8b-0=4a-8b\)
Average speed in the given interval of time.
\({ v }_{ av }=\frac { \triangle x }{ \triangle t } =\frac { 4a-8b }{ 2 } =2a-4b\)
(ii) Instantaneous speed
\(v=\frac { dx }{ dt } =\frac { d }{ dt } ({ at }^{ 2 }-{ bt }^{ 3 })=2at-3b{ t }^{ 2 }\)
At t = 2s, v = 4a - 12 b m/s

Let t be the time taken by the ball to reach the highest point where height from ground be s.
Taking vertical upward motion of the ball, we have u =- 29.4ms^-, a = 9.8ms^-, v = 0, s = S,t = ?
As, v²-u² = 2 as
0 -(-29.4)² = 2 x 9.8 x S
or \(S=\frac { -{ (29.4) }^{ 2 } }{ 2\times 9.8 } =-44.1m\)
Here, negative sign shows that the distance is covered in upward direction.
As, v = u + at
\(\therefore \quad 0=-29.4+9.8\times t\quad or\quad t=\frac { 29.4 }{ 9.8 } =3s\)
It means time of ascent = 3s
When an object moves under the effect of gravity alone, the time of ascent is always equal to the time of descent.
Therefore, total time after which the ball returns to the player's hand = 3 + 3 = 6s.

For accelerated motion,
u = 0,a = f,s = s
As v²-u² = 2 as,
\(\therefore \quad { v }_{ 1 }^{ 2 }-{ o }^{ 2 }=2fs\Rightarrow \sqrt { 2fs } ,\)
Distance travelled,
\({ S }_{ 2 }={ v }_{ 1 }t=t\sqrt { 2fs } \)
For decelerated motion,
\(u=\sqrt { 2fs } ,\quad a=-f/2,v=0\)
As v²-u² = 2 as,
\(\therefore \quad { o }^{ 2 }-{ (\sqrt { 2fs } ) }^{ 2 }=2\times (-f/2){ s }_{ 3 }\)
Distance travelled, s₃ = 2s
Given, s + s₂ + s₃ = 5s
\(\Rightarrow s+\sqrt { 2fs } +2s=5s\Rightarrow t\sqrt { 2fs } =2s\ \)
\( \Rightarrow s=\frac { 1 }{ 2 } f{ t }^{ 2 }\) 

Taking South to North direction as the positive direction 
i.e., x-axis, we have
Let velocity  of monkey with respect to ground = v _m
\(\therefore \) Relative velocity of monkey with respect to train A= v_m-v_A=-18kmh^- = -5ms^-1
v_m= v-5 = 15 - 5 = 10ms^-1