(i) The shaded portion of the car represents the distance travelled by the car in the first four seconds

The car travels with a non-uniform speed which is accelerated in nature.
(ii) The straight line portion of the graph represents the uniform motion of the car i.e. from point A to B.

Relative velocity of B W.Lt. A, v_AB = v_A - V_B

In the above x-t graph, body is at rest at position x₀

(a) Let the car accelerate at a rate a for time t₁ and attain a maximum velocity v. Then the car decelerates at a constant rate \(\beta\) for remaining time (t - t₁) and again comes to rest. Then from adjoining figure, it is clear that
 \(v=\alpha t_1=\beta(t-t_1)\)
\(\therefore t_1=\frac{v}{\alpha}\) and \((t-t_1)=\frac{v}{\beta}\)
Adding these two, we have
\(t=\frac{v}{\alpha}+\frac{v}{\beta}=v(\frac{\alpha+\beta}{\alpha\beta})\)
\(\Rightarrow v=\frac{\alpha\beta t}{(\alpha+\beta)}\)
(b) Total distance travelled by the car s = area OAB = \(\frac{1}{2}(t)\times (v)\)
\(\therefore s=\frac{1}{2}t\times \frac{\alpha\beta t}{(\alpha+\beta)}=\frac{1}{2}\frac{\alpha\beta}{2(\alpha+\beta)}t^2\)

Let x bet the distance travelled in time t. Then , 
 x \(\propto\) t²
x = kt² [ here, k = constant of proportionality ]
We know that velocity  is gigven 
v = \( \frac{dx}{dt}\)
= 2kt
and acceleration is given by
a = \(\frac{dv}{dt}\) = 2 k  [ constant ]
thus, the body has uniform accelerated motion.

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   \(\omega t=\frac { \pi }{ 2 } \Rightarrow t=\left( \frac { \pi }{ 2\omega } \right) \)

The effective distance travelled by drunked in 8 steps = 5-3 = 2 m
Therefore, he takes 32 steps to move 8 m
Now he will have to cover 5 m more to reach the pit, for which he has to take only 5 forward steps.
Therefore, he will have to take = 32 + 5 = 37 steps to move 13 m. Thus, he will fall into the pit after taking 37 steps. i.e., after 37 s from the start.

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.

Acceleration is 9.8 m/s² (downwards) and velocity is zero at the highest point.

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

Acceleration means that velocity is non - uniform. So, x - t graph will be curved.

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.

\(\frac { { v }_{ y } }{ v_{ x } } =\frac { tan60^{ o }\ }{ tan30^{ o } } =\frac { \sqrt { 3 } }{ 1/\sqrt { 3 } } =3:1\)

Any object can be considered as a point object if the distance travelled by it is very large in comparison to its dimensions.
Man along with monkey is cycling smoothly which indicates that the distance travelled by the man is very large, therefore monkey can be taken as a point object.

The distance and the magnitude of displacement of an object have the same values, when the body is moving along a straight line path in a fixed direction.

No, the displacement of the objects does not depend on the choice of the position of the origin.

An object can be considered as a point object if the distance travelled by it is very large than its size.

The problem here may be stated as
Find x when v = 70.0 m/s
It contains the single unknown x, as well as aand v, which are known with u = 0, v_x² = 2axx
Solving for x, we obtain
x = \( \frac{v^2}{2a} \)
= \(\frac{(70.00 m/s)^2}{2(4.00 m/s^2)}\)
= 613 m

\(u=50 \mathrm{~km} / \mathrm{h}=50 \times \frac{5}{18}=\frac{125}{9} \mathrm{~m} / \mathrm{s}, v=0, x=6 \mathrm{~m}\)
Using the relation, v ² -u² = 2ax
\(0-\left(\frac{125}{9}\right)^{2}=2 a \times 6\)
\(\Rightarrow \quad a=\frac{-125 \times 125}{81 \times 2 \times 6}=-16.27 \mathrm{~m} / \mathrm{s}^{2}=16 \mathrm{~m} / \mathrm{s}\)
In second case
\(u=100 \mathrm{~km} / \mathrm{h}=100 \times \frac{5}{18}=\frac{250}{9} \mathrm{~m} / \mathrm{s}\)
v = 0, a = -16m/s ² and x = ?
Using the relation, v ² -u² = 2ax
\(0-\left(\frac{250}{9}\right)^{2}=2 \times(-16) x\)
\(\Rightarrow \quad x=\frac{250 \times 250}{81 \times 2 \times 16}=24.1 \mathrm{~m}\)

(i) Magnitude of dusplacement of a particle in motion for a given time is the shortest distance between the initial and final position of the particle in that time, wheras the total legth of the path covered by particle is the length of actual path traversed by the particle in the given time.
(ii) Let us consider an example,
 
A car starts from 0 and moves along positive x-axis, The car stops at B and starts moving towards negative x-axis, Finally, the car reaches at A.
For the motion from O to B.
So, |Displacement| = Distance covered
|Average velocity| = Average speed                 
For the motion from \(O\rightarrow B\rightarrow A\)
So,|Displacement|    < Distance covered
|Average velocity| < Average speed

In notation of differential calculus, the velocity is
\(v=\frac { dx }{ dt }= \frac { d }{ dt } (a+{ bt }^{ 2 })=2bt=5.0t\ m/s^{-1}\)
At t = 0 s, v = 0 m s^–1 and at t = 2.0 s, v = 10 m s^-1
Average velocity \(=\frac { x(4.0)-x(2.0) }{ 4.0-2.0 } =\frac { a+16b-a-4b }{ 2.0 } = 6.0 \times b\)
= 6.0 × 2.5 = 15 m s^–1

Strategy: While solving such problems a very useful method is to assume that the event (catching the bus) happened at time t. Solve for t and find the condition that t is not a real value.
Let the passenger catches the bus after time t. Distance travelled by the bus in time t.
\(s=0+\frac{1}{2} a t^2=\frac{1}{2} a t^2\)
Distance travelled by the passenger, \(s_2=v t+0=v t\)
The passenger will catch the bus if, \(d+s_1=s_2\)
or, \(d+\frac{1}{2} a t^2=v t\) or \(\frac{1}{2} a t^2-v t+d=0\)
or, \(t=\frac{v \pm \sqrt{v^2-2 a d}}{a}\)
The passenger will catch the bus it t is real, i.e., \(v^2 \leq 2 a d\) or \(v \geq \sqrt{2 a d}\). Thus, minimum speed or passenger for catching the bus is \(\sqrt { 2ad } \)

(i) Since, the ball is moving under the effect of gravity, the direction of acceleration due to gravity is always vertically downwards.
(ii) At the highest point, the velocity of the ball becomes zero and acceleration is equal to the acceleration due to gravity=9.8ms^-2 in vertically downward direction.
(iii) When the highest point is chosen as the location for x=0 and t=0 and vertically downward direction to be the positive direction of x-axis and upward direction as negative direction of x-axis.
During upward motion, sign of position is negative, sign of velocity is negative and sign of acceleration is positive.During downward motion, sign of position is positive, sign of velocity is positive and sign of acceleration is also positive.
(iv) 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
\(\therefore 0=-29.4+9.8\times t\ or\ t=\frac { 29.4 }{ 9.8 } =3s\)
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.

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