a simple note to understand kinematics.

1. Unit 3 Kinematics
Kinematics in 1-D: is part of mechanics
that studies relationship among
displacement, velocity, acceleration and
time of a moving body. Kinematics
describes motion without its cause, i.e.
force.

2. Motion in a straight line : -it can be
described using one coordinate axis
Motion:-is a change of position relative
to frame of reference, it depends on the
choice of frame of reference

3. Frame of reference
a rigid framework or coordinate
system that can be used to measure
the motion of an object
Displacement:-is the change in position of
the moving object. Therefore, displacement
can be written:
∆x = xf –xi is the change in an
object’s position.

4. Average velocity: - is the displacement
divided by the time taken in which the change in
position occurs.
⇒ vav =
𝚫𝒔
𝚫𝒕
=
𝑺𝟐−𝑺𝟏
𝒕𝟐−𝒕𝟏
Average speed:- is the total distance
travelled divided by the total time.
⇒ Vav = =
𝑺𝟏+𝑺𝟐
𝒕𝟏+𝒕𝟐

5. Instantaneous velocity
 is the velocity of a body at a particular instant of time,
or at a specific point. Is the limit of the ratio
⧋ s
⧋t
as ⧋t
approach zero. This limit is called the derivative of
displacement with respect to time.
 The slope of the line tangent to the curve at that
particular point on the graph of displacement vs. time.
Vinst. =
𝚫𝑺
𝚫𝒕
, as t0 ⇒ v ins =
𝒍𝒊𝒎
⧋𝐭 →𝟎
[
⧋𝒔
⧋𝐭
] =
𝑠 𝑡+⧋ 𝑡 −𝑠(𝑡)
⧋t
=
𝒅𝑺
𝒅𝒕

6. INSTANTANEOUS ACCELERATION :
IS THE ACCELERATION OF A BODY AT A PARTICULAR INSTANT OF TIME
→AINST = ∆V/ ∆T AS ∆T APPROACH TO ZERO
THE SLOPE OF THE LINE TANGENT TO THE CURVE AT THAT PARTICULAR POINT ON THE
GRAPH OF V VS. T
IS THE LIMIT OF THE RATIO AS ⧋T APPROACH ZERO. THIS LIMIT IS CALLED THE
DERIVATIVE OF VELOCITY WITH RESPECT TO TIME
IS THE RATE OF CHANGE OF VELOCITY AT A SPECIFIC POINT OR AT A PARTICULAR
TIME.
⟹ 𝒂INST. =
𝚫𝑽
𝚫𝒕
, AS T0 OR = 𝐥𝐢𝐦
𝒕𝟎
𝒗 𝒕+ᴧ𝒕 −𝒗(𝒕)
𝒕
=
𝒅𝑽
𝒅𝒕
Average acceleration is the change in velocity
divided by the time interval in which the change
occurs. ⇒ aav =
𝚫𝑽
𝚫𝒕
=
𝑽𝟐−𝑽𝟏
𝒕𝟐−𝒕𝟏

7. Deceleration This word usually
means slowing down or a reduction
in speed
When the object’s velocity and acceleration are in
the same direction, the speed of the object
increases with time.
When the object’s velocity and acceleration are in
opposite directions, the speed of the object
decreases with time. .

8. For accelerated motion, For deceleration motion
V = u + at v = u - at
S = ut + 1/2 at2 S = ut - 1/2 at2
V2 =u2+ 2as V2= u2- 2as
S (𝑡𝑡ℎ) = u + a/2(2t- 1) S (𝑡𝑡ℎ) = u - a/2( 2t- 1)
Where: s = displacement v = final velocity u =
initial velocity a = acceleration

9. Freely falling bodies
Free fall: when an object falls under the
influence of gravity alone under the assumption of
neglecting the effect of air resistance
All bodies’ falls with the same down
ward acceleration regardless of their
weight & size.

10.  The constant acceleration of free fall is acceleration due to
gravity, g = 9.8m/s2 down, it is –ve.
 All vectors directed vertically upwards are +ve.
 in the absence of air resistance, all objects dropped near
the Earth’s surface fall toward the Earth with the same
constant acceleration under the influence of the Earth’s
gravity.
Equations for free fall motion in three cases
I / thrown up ii/ thrown down Iii/ dropped/released/
v = u – gt v = u + gt v = – gt
y = ut -
1
2
gt2 y = ut +
1
2
gt2 y = -
1
2
gt2
V2 = u2 - 2gy V2 = u2 + 2gy V2 = 2gy

11. Terminal velocity as a particle falls, the force from
air resistance increases during its fall. This force
increases as the velocity of the particle increases. If
the particle is falling for long enough, the force from
air resistance will be the same as the force from
gravity. As there is no net force on the particle,
there is no net acceleration on it and its velocity will
not increase any more. This velocity is known as the
terminal velocity.

12. 3.2 MOTION IN A PLANE
Projectile motion
Uniform Circular motion,
Non-Uniform Circular motion
3.2.1 Projectile motion
Projectile is any object moving through the air
without an engine or other motive force.
Example:- a bullet that is fired from a gun, a
kicked soccer ball in to the air
The horizontal and vertical components of a
projectile’s motion are completely independent of
each other and can be treated separately

13. A projectile launched horizontally
. is a Projectile motion where the projectile initially
travelling horizontally
Example:-a package released from the underside of an aircraft
flying horizontally.
 Horizontal motion - constant velocity i.e. there are no
horizontal forces acting, so there is no acceleration
along the horizontal direction, ax=0 ⇒ ux = vx
 Vertical motion - velocity changes linearly, because the
accleration due to gravity is constant

14. It is interesting to note that the time it takes to hit the floor is the same in
both cases for an object thrown horizontally and for an object released from
the same height simultaneously
 vertical displacement
⟹ y=1/2 gt2
 The time of flight in horizontal projection is given by
⟹ t =
2y
g
 horizontal displacement
⟹ X =Vx t = Vx
2y
g
 vertical velocity ⟹ Vy = gt

15. Object launched at an angle
Projectile at angle Ѳ0 is a motion of a projectile thrown with an initial
velocity u at an angle θ0 above the horizontal as shown below. In this
motion both vertical and horizontal initial velocity are non-zero. Therefore
the velocity must resolve into horizontal and vertical component.
 The initial velocity “u” along in “x” and “y” components:
ux = u cos and uy = u sin

16. The vertical motion of the projectile is an
example of uniformly accelerated
motion.
Vertical motion – the same as equation of
free fall

17. 1. PROVIDED AIR RESISTANCE IS NEGLIGIBLE, THE HORIZONTAL COMPONENT OF THE
VELOCITY VX REMAINS CONSTANT BECAUSE THERE IS NO HORIZONTAL COMPONENT OF
ACCELERATION. UX = U COS Ѳ UX =VX IS CONSTANT
DX =UXT =(U COS Ѳ)T
2. THE VERTICAL COMPONENT OF THE ACCELERATION IS EQUAL TO THE FREE-FALL
ACCELERATION .
3. THE VERTICAL COMPONENT OF THE VELOCITY VY AND THE DISPLACEMENT IN THE Y-
DIRECTION ARE IDENTICAL TO THOSE OF A FREELY FALLING BODY.
Projectile motion can be described as two independent motions in
the x- and y-directions.

18.  The velocity “v” at any time “t” along “x” and “y”
components:
vx = ux =u cos and vy = u sin -gt
 The magnitude and direction of the velocity at any time
“t” is:
v = 𝑉𝑥2 + 𝑉𝑦2 and tan = vy / vx
 The “x” and “y” component of the displacement at any
time “t” is:
x = ux t and y = uy t –
1
2
gt2
 The displacement “S” magnitude and direction at any time
is:
S = 𝑥2 + 𝑦2 and tan =
𝑦
𝑥

19.  Maximum height
At the maximum height the vertical velocity of the
projectile will be zero.
The maximum height of projection at an angle is given by
hmax =
𝒖𝒚 𝟐
𝟐𝒈
or hmax =
( u2 sin2 θ) 𝟐
𝟐𝒈
 Flight time is the total time in which the projectile
stayed in air. t =
𝟐𝒖𝒔𝒊𝒏θ
𝒈
 Projectile range is the distance that the projectile
travels in the horizontal direction, it is given by
⇒ R= vx tT =
(2u2sinθcosθ)
𝒈
=
(u2 sin2θ)
𝒈
The range is maximum when the angle of projection is 45°.
i.e ⇒ Rmax =
u2
𝒈

20. 3.2.2Circular motion
1. Uniform Circular motion
2. Non-Uniform Circular motion
1. Uniform circular motion is the
motion of a body around a circular
path with constant speed.

21.  Centripetal acceleration:- is an acceleration of a body that always
point to the center of the circle and always perpendicular to the
path. This acceleration is always ⊥ to the path and points towards
the center of the circle.
 Centripetal acceleration =tangential speed2/ radius of curvature
⇒ a = ν2 / r = rω2
where v = speed and r = radius
 Centripetal Force ⇒ F = mν2 / r

22. Non-Uniform Circular motion
In this case there is a tangential acceleration
arises from a change in speed.
Ft = mg sin Ѳ, at = g sin Ѳ, Tangential
acceleration
⇒ at = ∆v/∆ t = r𝜶
Since a C ⊥ a t ⇒ a R = 𝒂𝒄
𝟐 + 𝒂𝒕
𝟐

23. 3.2.3Gravitation
Newton’s law of universal gravitation states that every
particle in the Universe attracts every other particle with a
force that is directly proportional to the product of their
masses and inversely proportional to the square of the
distance r between them F =
𝑮𝑴𝒎
𝒓 𝟐
Let us consider the force on a particle of mass m at the surface of the earth which is
given by
F =
𝑮𝑴𝒎
𝒓 𝟐
⇒
mg =
𝑮𝑴𝒎
𝒓 𝟐
The gravitational acceleration at the surface of the earth
⇒
ge =
𝑮𝑴𝒎
𝒓
The gravitational acceleration at any distance r from the center of Earth is
⇒
gr =
𝐆𝐌
(𝐫𝐞 + 𝐡) 𝟐

24. Satellites
Any objects in space that orbit or revolving around the
planet (sun)
The FG on satellite of mass m is equal to the FC needed to keep it
moving in a circle FG = FC
⇒ 𝐆𝐌𝐦
𝐫𝟐 =
𝒎𝒗𝟐
𝒓
= mrω2,
Orbital velocity The satellites experiences a FC of magnitude given by:
FC =
𝒎𝒗𝟐
(𝑹𝒆+𝒉) 𝟐 -------(1)
This force holds the satellites in orbit, it is provided by the FG of the earth
FG =
𝑮𝑴𝒎
𝑹𝒆+𝒉 𝟐 -------(2)
Equating 1 by 2 vorb = √
𝑮𝑴
𝑹𝒆+𝒉
The orbital velocity can also be expressed as
v =
𝟐𝝅𝒓
𝑻
=
𝟐𝝅(𝑹+𝒓)
𝑻

25. Geosynchronous satellites to facilitate
intercontinental communication network, navigators,
and weather forecasting
They have the same period of rotation of earth about
its axis T = 24hrs=8.64 X 104seconds
V =
𝑮𝑴
𝑹𝒆+𝒉
=
𝟐𝝅(𝑹+𝒓}
𝑻
Escape velocity vesc =
2𝐺𝑀𝑒
𝑅𝑒
= √2ge Re

26. 3.2.4 Relative velocity in one and two dimensions
Motion of a moving object ( velocity) described
using
1/ stationary reference frame or
2/ moving reference frame
Observers in different frame of reference measures
different displacements and velocities