Kinematics

Definitions
• Vector is a physical quantity that has both magnitude and direction.
• Scalar is a physical quantity that has magnitude only.
• Resultant vector is the single vector which has the same effect as the original vectors
acting together.
• Distance is the length of path travelled.
• Displacement is a change in position.
• Velocity is the rate of change of position or the rate of displacement or the rate of
change of displacement.
• Acceleration is the rate of change of velocity.

Concepts
• Position of an object is relative to a reference point.
• Position is a vector quantity that points from the reference point as the origin.
• Displacement is a vector quantity that points from the initial to the final position.
• Average velocity is the change in the object's position (displacement) over a longer
period of elapsed time. It can be calculated by taking the total displacement divided
by the total time.

𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
𝑣𝑎𝑣𝑒 =
𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒

• Instantaneous velocity is the rate of change of an object's position in a specific

direction at a particular instant in time.

𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 Δ𝑥
𝑣𝑎𝑣𝑒 = =
𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑇𝑖𝑚𝑒 Δ𝑡

• Projectiles fall freely with gravitational acceleration 'g', where g = 9,8 m.s−2 near the
surface of the Earth.
• Free fall is the movement of an object under only the force of gravity.
 Formulae
Formula Description
𝑣𝑓 = 𝑣𝑖 + 𝑎Δ𝑡 𝑂𝑅 𝑣 = 𝑢 + 𝑎𝑡 Equation of motion without
displacement.
𝑣𝑓 (𝑣) − Final velocity.
𝑣𝑖 (𝑢) − Initial velocity.
𝑎 − Acceleration.
Δ𝑡 (𝑡) − Time

𝑣𝑓 + 𝑣𝑖 𝑣+𝑢 Equation of motion without

Δ𝑥 = ( ) Δ𝑡 𝑂𝑅 𝑠=( )𝑡 acceleration.
2 2
Δ𝑥 (𝑠) − Displacement.
𝑣𝑓 (𝑣) − Final velocity.
𝑣𝑖 (𝑢) − Initial velocity.
Δ𝑡 (𝑡) − Time

𝑣𝑓2 = 𝑣𝑖2 + 2𝑎Δ𝑥 𝑂𝑅 𝑣 2 = 𝑢2 + 2𝑎𝑠 Equation of motion without time.

Δ𝑥 (𝑠) − Displacement.
𝑣𝑓 (𝑣) − Final velocity.
𝑣𝑖 (𝑢) − Initial velocity.
𝑎 − Acceleration.

1 1 Equation of motion without final

Δ𝑥 = 𝑣𝑖 Δ𝑡 + 𝑎Δ𝑡 2 𝑂𝑅 𝑠 = 𝑢𝑡 + 𝑎𝑡 2 velocity.
2 2
Δ𝑥 (𝑠) − Displacement.
Δ𝑡 (𝑡) − Time
𝑣𝑖 (𝑢) − Initial velocity.
𝑎 − Acceleration.
 Vectors
Common vector and scalar quantities
Vectors Scalars
Displacement Distance
Velocity Speed
Acceleration Time
Force Mass
Weight Energy
Momentum Work
Impulse Power
Electric Field Electric Current
Gravitational Field Potential Difference (Voltage)

Resolving forces into components

Force at an Angle

When a force acts at an angle to the

horizontal, the force is resolved into its
horizontal and vertical components.

Angle given is with the horizontal

𝐹𝑥 = 𝐹 cos 𝜃
𝐹𝑦 = 𝐹 sin 𝜃
If 𝐹 sin 𝜃 > 𝐹𝑔 , the object will accelerate
𝐹
𝐹 sin 𝜃 up.
If 𝐹 sin 𝜃 = 𝐹𝑔 , the object will move up at a
 constant velocity.
𝐹 cos 𝜃 If 𝐹 sin 𝜃 < 𝐹𝑔 , the object will not move.

Angle given is with the vertical

𝐹𝑥 = 𝐹 sin 𝜃
𝐹𝑦 = 𝐹 cos 𝜃
𝐹
𝐹 sin 𝜃


𝐹 cos 𝜃
 Object on an inclined plane
When an object is on an inclined plane, the Component parallel to the plane, which
weight (𝐹𝑔 ) is resolved into two causes the object to slide down if the
components. frictional force is overcome.
𝐹𝑔 (𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙) = 𝐹𝑔 sin 𝜃

Component perpendicular which contributes

to the normal force.
Fg sin  𝐹𝑔 (𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟) = 𝐹𝑔 cos 𝜃
Fg cos 
Fg


Finding the resultant of two vectors

• Be able to determine the resultant vector of any two vectors by using the following
methods:

Tail to head method Tail to Tail Method Component method

Draw a vector diagram such Draw a vector diagram such Resolve vectors into
that the tail of one vector that the 2 tails of the vectors horizontal and vertical
meets the head of the other. meet at the same point. components.
The resultant vector is then Complete the parallelogram V2y= V2sin 
drawn from the tail of the the resultant vector is the
first to the head of the last diagonal of the
vector. parallelogram. V2

R V2x = V2cos 
V2 V2 R
V1

V1 V1
Find R by finding the
Use Pythagoras and trig to Use Pythagoras and trig to resultant of the horizontal
find R find R and vertical forces. If forces
are in equilibrium R = 0.

• Be able to answer unseen problems related to the above skills.

Vertical Projectile Motion
Sign Convention: Upward is taken as positive and downward is taken as negative, unless the
question specifies an alternate sign convention. Ground is taken as the reference point.
Object thrown up into the air from the Object thrown up into the air from a
ground building / Released from a hot air
balloon from a certain height.
VB= 0 VB = 0
B Going Up B
Going down
a = -9.8 a = -9.8
Going Up Going down v is positive vi
A C VC v is negative
a = -9.8 a = -9.8
v is positive v is negative
vi Vf
AC D Vf

Due to symmetry
• vf = -vi Due to symmetry
• Time taken from A to B = Time • vC = -vi
taken from B to C • Time taken from A to B = Time
taken from B to C
vf > vi

a a
(m.s-2) (m.s-2)
t (s) t (s)

-9.8 -9.8

A A
v v
(m.s-1) B B
(m.s-1)
t (s) t (s)
C
C
B
B D

y y A C
(m) (m)
D
A C t (s) t (s)
Object thrown down towards ground Rocket running out of fuel

vi C

Going down
a = -9.8
v is negative vi B D Vf
Vf
AE
If the object is released vi = 0
Rocket accelerates from A to B and runs
out of fuel at B.
Due to symmetry
• vf = -vi
• Time taken from B to C = Time
taken from C to D

a a
(m.s-2) (m.s-2)
t (s) t (s)

v v
(m.s-1) (m.s-1)
t (s)
A C t (s)

D
E
C

B D
y y
(m) E
(m)
t (s) A t (s)
 Ball dropped and bouncing twice Object falling through an obstacle like a
glass.

A A
C
E

B Glass
C

D
B D

a a
(m.s-2) (m.s-2)
t (s) t (s)

B
D
v v
(m.s-1)
A C E (m.s-1) A
t (s) t (s)
D C
B D
B

A C A
E
y y B C
(m) (m) D
B D t (s) t (s)
 4. Graphs of Motion (Grade 10)

Skills

• Be able to draw position vs time, velocity vs time and acceleration vs time graphs for
one dimensional motion.
• Interpret graphs of motion:
- Determine the velocity of an object from the gradient of a position (or displacement)
vs time graph.
- Determine the acceleration of an object from the gradient of a velocity vs time
Graph.
- Determine the displacement of an object by finding the area under a velocity vs
time graph.
- Determine the velocity of an object by finding the area under a acceleration vs time
graph
- Be able to describe the motion of an object from the graphs of motion.
• Be able to answer unseen problems related to the above skills.
Description of Graphs

Constant positive gradient Constant negative gradient Zero gradient

Increasing positive gradient Increasing negative gradient

Decreasing positive gradient Decreasing negative gradient

Summary of the shapes for sketching of graphs

Positive Negative
Decreasing Increasing

Negative
Positive
Decreasing
Increasing