If a body travels equal displacements in equal intervals of time, then the velocity of body is uniform velocity. On the other hand, if the body covers unequal displacements in equal intervals of time, then its velocity is variable velocity.

When two bodies have equal veloci.ties (i.e., \(\overrightarrow { v_{ a } } =\overrightarrow { v_{ b } } =\overrightarrow { v } \)), then their relative velocity is zero i.e., \(\overrightarrow { v_{ ab } } =\overrightarrow { v_{ a } } -\overrightarrow { v_{ b } } =\overrightarrow { v } -\overrightarrow { v } =0.\)

Yes, it is possible. In motion under gravity at the highest point of its motion velocity is zero but acceleration a = g in downward direction.

An object is said to be point object if its dimensions are negligible as compared to the distance travelled by it. For example, an aeroplane which flies from Delhi to London.

(i) u = 0, a = 0.1 m / s² and t = 2 min 120 m / s
v = u + at 
= 0 + 0.1 \(\times\)120  
= 12 m / s
(ii)  u = 0, a = 0.1 m / s² and t = 2 min 120 m / s
v = ut +\( \frac{1}{2}\) at²
= 0 +\( \frac{1}{2}\) \(\times\) 0.1\(\times\)( 120 )²
= \( \frac{1}{2}\) \(\times\) 0.1\(\times\)120\(\times\)120
= 720 m

The velocity-time graph for uniformly accelerated motion has been shown in Fig. with initial velocity at t = 0 as u and final velocity at = 5  time t as v. Then area under the v-t graph gives the value of total displacement in the given time. Hence, displacement of moving particle in time t
=area of trapezium ∆ ABC
s = \(\frac { 1 }{ 2 } \)(OA + CB) x OC,
= \(\frac { 1 }{ 2 } \)(U + V) x t
However, from definition of acceleration, we know that \(a=\frac { v-u }{ t } \)or \(t=\frac { v-u }{ a } \)
Substituting this value of time t in equation (i), we get
Displacement s =\(\frac { 1 }{ 2 } (u+v)\times \frac { v-u }{ a } \) or \(\frac { v^{ 2 }-u^{ 2 } }{ 2a } \) 
\(\Rightarrow \)2as = v²-u² of v² = u²+ 2as

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.

(a) We are given that u = 0 and velocity after 8 s is 42 m/s, so we can use v = u + at to find acceleration
a = \(\frac { v-u }{ t } =\frac { 42.0-u }{ 8.0 } \)= 5.25ms^-2
(b) distance travelled in 8.0 s, we can use , s = ut +\(\frac { 1 }{ 2 } \)at²
= 0 + \(\frac { 1 }{ 2 } \)\(\times \)5.25\(\times \)8² = 168m
(c) distance travelled in 8th second, we have, S_n =u+(2n-1)\(\frac { a }{ 2 } \)
= (2 x 8 - 1) \(\times \frac { 5.25 }{ 2 } \) = 39.375 m.

(i) True, when a body is thrown vertically upwards in the space, then at the highest point, the body has zero speed but has downward acceleration equal to the acceleration due to gravity.
(ii) False, because velocity is the speed of body in a given direction. When speed is zero, the magnitude of velocity of body is zero, hence velocity is zero.
(iii) True, when a particle is moving along a straight line with a constant speed, its velocity remains constant with time. Therefore, acceleration (i,e. change in velocity/time) is zero.
(iv) False, if the initial velocity of a body is negative, then even in the case of positive acceleration, the body speeds down. A body speeds up when the acceleration acts in the direction of motion.

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.

(iii) What will be the position-time curve for dropped packet?
In position-time curve, when a packet is dropped. Let us consider origin at the point where packet was dropped.

Juggler throws the ball into the air with ball having initial velocity = u
At the highest point of its path velocity will be zero as at maximum height velocity =0
So final velocity=0
According to question juggler throws a ball each second so for "n" balls time taken to reach highest position (t)= 1/n
v=u+gt
0=u-g/n
-u=-g/n
u=g/n
S= ut-gt²/2
S=u/n-g/2n²
But u=g/n
S= g/n²-g/2n²
S=g/2n²

(i) Draw position-time graph for
(a) Accelerated motion  
(b) Retarded motion



(ii) Juggler throws the ball into the air with ball having initial velocity = u
At the highest point of its path velocity will be zero as at maximum height velocity =0
So final velocity=0
According to question juggler throws a ball each second so for "n" balls time taken to reach highest position (t)= 1/n
v=u+gt
0=u-g/n
-u=-g/n
u=g/n
S= ut-gt²/2
S=u/n-g/2n²
But u=g/n
S= g/n²-g/2n²
S=g/2n²