1. Gold: Important concept. Very likely to appear on an assessment.
Notes:Color Guide
Gold: Important concept. Very likely to appear on an assessment.
Blue: Supplemental information. Will not directly appear on an assessment.
Red: Example. Copy if needed.
White: Will be discussed by Mr. Williams.

2. Chapter 3: Two Dimensional Motion
Section 1: Motion in 2 directions

3. Objects don’t always move in perfectly straight lines that are in one dimension.
At this point, quantities such as velocity & force will be split into two dimensions.
Vectors are used to represent the value and direction of some quantity.
The size of the vector represents it’s quantity.
The arrowhead show its direction.

4. A vector can be split into the components that make it up.
For example… Sarah is walking southeast at 1.7 m/s.
Her motion has a SOUTHWARD component, and an EASTWARD component.
SKETCH THIS:
Vectors are resolved by separating them into their components.
South Component
East Component
“Resultant Vector”
Actual Motion

5. The combination of vectors gives a new vector: the resultant.
Combining vectors is called “vector addition”.
It isn’t quite like normal addition.
The dog’s leash pulls in two directions: UP and RIGHT.
The UP component and the RIGHT component give the resultant.
UP and RIGHT are in two different dimensions!
The vectors make a RIGHT TRIANGLE with each other.
They’re added using the PYTHAGOREAN THEOREM!

6. Vectors can be added graphically as well by drawing the vectors.
Suppose Shannon walked 400 meters north to the store, and then 700 meters east to school.
To resolve vectors graphically, we use the head to tail method.
2) Then slide the next vector’s tail to the first vector’s head.
∆x = 700m
head
∆y = 400m
d = 806m
tail
3) Once all vectors are added, draw the resultant.
1) ID the head & tail of the first vector.

7. ∆x = 700m
∆y = 400m
∆xresultant = 806m

8. Consider this scenario:
A toy car is moving south at .80 m/s straight across a walkway that moves at 1.4 m/s towards the east. What will the car’s resultant speed be?
Draw this scenario. Use the scale 1cm = .1 m/s.
Measure the resultant to find the answer.
Try to find the answer using math…

9. The car has a resultant speed of:
1.61 m/s (toward the southeast)

10. It is important to note that vectors can be moved parallel to themselves in a diagram.
Moving vectors will be vital to drawing the correct resultant vector!
Vectors can act at the same point at the same time.

11. Coordinate System
Please note that displacement may now be represented by “d”. y N -x x W E -y S

12. Chapter 3: Two Dimensional Motion
Section 2: Vector Operations

13. Vectors can also be determined using geometry 
Although finding vectors graphically is “fun”, it is not very practical.
Vectors can also be determined using geometry 
The Pythagorean Theorem for Rt. Triangles:
c2 = a2 + b2
c represents the hypotenuse; a & b represent legs
The Pythagorean theorem will only give you the magnitude of the resultant vector! c a b

14. Pythagorean’s theorem can only be used when a right triangle is present!

15. Direction is represented with angles.
Theta (θ) is used to represent the angle when working with vectors.
Trigonometry is applied in order to find the angles of right triangles. We’ll begin with tangent.
Tangent Function for Rt. Triangles:
tan θ = opp/adj
(opp = opposite leg of angle; adj = leg adjacent to angle)
hyp
opp θ
adj

16. C Vectors will often need to be broken down into components.
A vector component is the horizontal or vertical aspect of a vector.
The vector C can be broken down into its X component and Y component. C

17. Sine Function for Rt. Triangles: sin θ = opp/hyp
(opp = opposite leg of angle; adj = leg adjacent to angle)
Cosine Function for Rt. Triangles:
cos θ = adj/hyp
(opp = opposite leg of angle; adj = leg adjacent to angle)
hyp
opp θ

18. Chapter 3: Two Dimensional Motionvy Vx
Section 3: Projectile Motion

19. A projectile is any object launched into the air that accelerates vertically due to gravity.
Examples: A baseball, arrow, bullet, etc…
Projectiles follow a parabolic path.

20. Free fall motion is motion in which gravity is the only vertical force acting on an object.
Free fall includes objects dropped AND objects thrown upward. v
Non
Free Fall
Free Fall
Free Fall

21. The path of the projectile will always be a parabola.
Projectiles have an initial horizontal velocity that gives them a parabolic path.
Projectile motion is just freefall motion with a horizontal component.
The path of the projectile will always be a parabola.

22. Even with an initial velocity, a projectile still falls at an acceleration of -9.8 m/s2
The motion of the projectile can be split into two components.

23. Vertical Motion of Projectiles with no initial vertical velocity.
The equations of motion can be used to analyze projectiles that are dropped or launched horizontally:
Please note that these are not “new” equations. They are kinematic equations with vertical components.
Vertical Motion of Projectiles with no initial vertical velocity.
Vy,f = ay∆t
[ay=gravity, g]
Vy,f2 = 2g∆y
∆ y = ½g∆t2

24. Note the horizontal velocity (vx) of the object.
The horizontal velocity of a projectile remains constant.
Ignoring air resistance, there is no horizontal force to change a projectile’s motion.
Horizontal Motion of a Projectile
∆x = vx∆t
Note the horizontal velocity (vx) of the object.

25. Suppose a ball is pushed off the table, as shown.
If the table is .75 m high, and the ball’s displacement is .58 m from the edge, how fast was it moving when it left the table?
1. Use ∆ y = ½g∆t2 and solve for time. (.39 s)
2. Use ∆x = vx∆t and solve for vx.
Vx = 1.5 m/s
Given:
ay=g = m/s2
∆y =- .75m
∆x = .58 m
vx=?

26. In order to assess the motion of a projectile launched at an angle, isolate the components of its motion.
Use trigonometry to break down the motion into horizontal motion and vertical motion.
The projectile will continue in a parabolic path…
But we can assess a portion of it’s path with trig. vy vx

27. This diagram represents the initial velocity vectors of a projectile launched at an angle, θ.vi
vy,i = visinθ θ
vx = vicosθ

28. This diagram represents the initial velocity vectors of a projectile launched horizontally.
Determine the fall-time of the projectile. Use that time to determine the range of the projectile.
vx = vi
vyi = 0 Δy Δx