Distance and displacement will have the same magnitude when the object moves along a straight line without change in its direction.

If a is the uniform acceleration
then,s = u +\(\frac { 1 }{ 2 } a\)(2n-l) where u is the initial velocity.

A straight line inclined to the time axis.

Let x be the displacement . Then,  x = at²
\(\therefore\)      Velocity of the object , v = \(\frac{dx}{dt}\)
Accleration  of the object , a = \(\frac{dv}{dt}\)
It means that a is constant.

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.

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.

It is true because if the velocity of the object is negative, then slope of v-t graph is negative.

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.

No, because the size of the beaker is not negligible as compared to the height of the table.

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

Relative speed = (34 + 38) kmh^-1 = 72 krnh^-1
= 72 x - ms^-1 = 20 ms^-1
Total distance = (109 + 91) m = 200 m
Time = \(\frac { 200m }{ 20ms^{ -1 } } \) =10 s.

Given equation of velocity,
v = 4 + 2 (c₁ + c₂t)
=> v = (4 + 2c₁) + 2c₂t
Compare the above equation with equation of motion
v = u + at
Initial velocity, u = 4 + 2c₁
Acceleration of the particle = 2c₂.

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.

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.

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.

For uniformly accelerated motion, we can derive some simple equations that relate displacement(x), time taken(t), Initial velocity(u), final velocity(v) and acceleration(a).
(i) Velocity attained after time t: The velocity-time graph for positive constant acceleration of a particle is shown in the figure.

Let u be the initial velocity of the particle at t = 0 and v is the final velocity of the particle after time t. Consider two points A and B on the curve corresponding to t = 0 and t = t respectively.
Draw BD perpendicular to time axis. Also draw AC perpendicular to BD.
ஃ OA = CD = u;
BC = (v - u) and OD = t
Now slope of v-t graph = acceleration (a)
∴ a = slope of v - t graph = tan θ = \(\frac{BC}{AC}=\frac{BC}{OD} \ \ \ [∵ AC=OD]\)
∴ \(a=\frac{v-u}{t}\)
or v - u = at
v = u + at
(ii) Distance travelled in time t:
Let  x₀ position of the particle at t = 0 from the origin.
x = position of the particle at t = t from the origin.
∴ (x - x₀) = S = distance travelled by the particle in the time interval (t - 0) = t
We know, distance travelled by a particle in the given time interval =  area under velocity-time graph 
∴ (x - x₀) = Area OABD (see fig. above)
= Area of trapezium OABD
= \(\frac{1}{2}\) [Sum of parallel sides x perpendicular distance between parallel side]
=\(\frac{1}{2}(OA+BD)\times AC=\frac{1}{2}(u+v)t\)
Since v = u + at
∴  (x- x₀) = \(\frac{1}{2}(OA+BD)\times AC = \frac{1}{2} (u+v) \times t\)
Since x - x₀= S
∴   S = ut + \(\frac{1}{2}\)at²
(iii) Velocity attained after travelling a distance S:
We know, distance travelled by a particle in time t is equal to the area under velocitytime graph. Therefore, the distance (S) travelled by a particle during time interval tis given by
S = Area under v - t graph (see fig.) or
S = area of trapenium OABD
= \(\frac{1}{2}\)(sum of parallel sides) x perpendicular distance between these parallel
or S = \(\frac{1}{2}\)(OA + BD) x AC    --- (i)
Now, acceleration, a = slope of v - t graph
or a = \(\frac{BC}{AC}=\frac{BD-CD}{AC}=\frac{v-u}{AC}\) 
or \(AC=(\frac{v-u}{a})\)     --- (ii)
Also OA = u and BD = v  --- (iii)
Using equations (ii) and (iii) in equation (i), we get
\(S=\frac{1}{2}(v+u)\frac{(v-u)}{a}=\frac{v^{2}-u^{2}}{2a}\)
or, v²-u² = 2aS

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