Scalar quantity
A quantity that only has magnitude and no direction like mass
Vector quantity
A quantity that has both magnitude and direction
Examples of vector quantities
Displacement
Velocity
Force
Weight
Acceleration
Examples of scalar quantities
Mass
Energy
Time
Distance
Speed
Displacement is defined as
A vector quantity measuring a change in position. How much an object has changed and what direction the motion takes place to be able to locate the new position of the object
Displacement is measured in
M
Mm to m÷1000
Cm to m÷100
An angle can move
Up down, left right, angle initial direction in angle, compass points or bearing.
North of East
You are heading to a north side from an east point.
Direction help
“North of East” and “East of North” may seem similar, but they have distinct meanings:
“North of East” (NE):
- Start facing East
- Turn slightly North (less than 45°)
“East of North” (EN):
- Start facing North
- Turn slightly East (less than 45°)
Key difference:
- “North of East” leans more towards East
- “East of North” leans more towards North
Visualize:
- Imagine a compass rose
- Draw a line from North to East
- “North of East” falls slightly below this line
- “East of North” falls slightly above this line
Mnemonic:
- “North of East” = Need Extra Energy
- “East of North” = Enter New Nation
Practice:
- Use online compass simulators
- Practice navigation with maps
- Familiarize yourself with directional notation
Now, you can confidently distinguish between “North of East” and “East of North”!
Pythagoras theorem and displacement
Pythagoras’ theorem helps find displacement when you know the horizontal and vertical components of the motion.
Pythagoras’ Theorem:
a² + b² = c²
where:
- a = horizontal displacement (or distance)
- b = vertical displacement (or distance)
- c = resultant displacement (hypotenuse)
Finding Displacement:
- Break down the motion into horizontal and vertical components.
- Calculate the horizontal (a) and vertical (b) displacements.
- Apply Pythagoras’ theorem to find the resultant displacement (c).
Example:
An object moves:
- 3 meters east (horizontal)
- 4 meters north (vertical)
Find the resultant displacement.
Solution:
a = 3 m (east)
b = 4 m (north)
c = ? (resultant displacement)
Using Pythagoras’ theorem:
c² = a² + b²
c² = 3² + 4²
c² = 9 + 16
c² = 25
c = √25
c = 5 m
Resultant Displacement: 5 meters
Direction: Use trigonometry (tan, cos, or sin) to find the direction.
tan(θ) = b/a
tan(θ) = 4/3
θ ≈ 53.13° north of east
Resultant Vector: 5 m, 53.13° north of east
Pythagoras’ theorem is essential in physics, engineering, and navigation for calculating displacements, velocities, and forces in two-dimensional space!
What about direction and trig ratios
Trigonometric ratios (trig ratios) help find the direction of the resultant vector.
Trigonometric Ratios:
- Sine (sin): opposite side / hypotenuse
- Cosine (cos): adjacent side / hypotenuse
- Tangent (tan): opposite side / adjacent side
Finding Direction:
- Identify the horizontal (adjacent) and vertical (opposite) components.
- Choose the appropriate trig ratio based on the given information.
Examples:
1. Finding angle (θ) using tangent:
Horizontal displacement (a) = 3 m
Vertical displacement (b) = 4 m
tan(θ) = b / a
tan(θ) = 4 / 3
θ = arctan(4/3)
θ ≈ 53.13° north of east
2. Finding angle (θ) using sine and cosine:
Resultant displacement (c) = 5 m
Vertical displacement (b) = 4 m
sin(θ) = b / c
sin(θ) = 4 / 5
θ = arcsin(4/5)
θ ≈ 53.13° north of east
cos(θ) = a / c
cos(θ) = 3 / 5
θ = arccos(3/5)
θ ≈ 53.13° north of east
Important:
- Use the correct trig ratio based on the given information.
- Consider the quadrant in which the angle lies.
Quadrants:
- I (0-90°): north-east
- II (90-180°): north-west
- III (180-270°): south-west
- IV (270-360°): south-east
Trigonometric ratios are essential in physics, engineering, and navigation for calculating angles and directions in two-dimensional space!
In a graph of motion a straight line tells us that
The position of an object changes by a constant amount in each second
The slope of a position time graph
Slope= change in position (final position - initial position =0)
So delta x over delta t equals velocity
A straight line graph
Has constant speed bc the velocity is also constant. If you cycle faster then your velocity increases.
Negative graph of motion
Slope goes down and negative gradient.
Position is in
Meters
Velocity is in
M/s - 1
Acceleration
Rate of change of velocity with time. A= delta x over delta t
Vf
Final velocity of an object. Meters per second to the power of negative 1.
Vi
Initial velocity. Meters epr second to the power of negative 1
A
Acceleration. Meters per second to the power of negative 2
Delta t
Time taken in seconds
First equation
A = vi- vf over delta t.
Uniformly accelerated motion
Equation 2 : displacement
Delta x= 1/2 ( initial velocity + final velocity) times delta x
