1. Mr. Rockensies – Regents Physics
AIM – How do we add vectors?
DO NOW – Where have you heard the word vector aside from Physics class?
HW - Textbook p. 26 #67(a-d), 68(a-d), 72(a-d)
Vector Addition
Mr. Rockensies – Regents Physics

2. Vectors Vectors Scalars (no direction)
Quantities with magnitude (amount, size) and direction.
Example: 20 m North or 20 m West
Vectors
Displacement
Velocity
Acceleration
Force
Momentum
Impulse
Electric Field
Scalars (no direction)
Distance
Speed
Time
Mass
Energy/Work
Charge
Power

3. NEVER FORGET TO DRAW THE ARROWS!!
Drawing Vectors
Vectors can be drawn to graphically represent magnitude as well as direction.
Length indicates magnitude, and therefore must be drawn to scale using a ruler and protractor.
The angle indicates direction, represented by θ (theta).
length θ
Horizontal Axis = +X direction
NEVER FORGET TO DRAW THE ARROWS!!

4. Resultant – Adding Vectors
Resultant – the result of 2 or more displacements (vectors)
20 m West
R = resultant displacement
θ = direction
20 m North R θ
R = 28 m, 45° determined by measuring with a ruler and protractor

5. Mathematical Techniques
When vectors are at right angles, we can use the Pythagorean Theorem and SOHCAHTOA:
a2 + b2 = c2
20 m West
R2 = (20m)2 + (20m)2
R = √800 m2
R = 28.2 m
20 m North R θ
tan θ = opp/adj = 20/20 = 1
θ = tan-1 (1) = 45°

6. Vector Addition (cont.)
Same Direction: simply add
= 11 m
4 m
7 m
Opposite Direction: subtract
5 m
= 4 m
9 m

7. Practice
A plane flies 1500 miles East and 200 miles South. What is the magnitude and direction of the plane’s final displacement?
A hiker walks 80 m North, 20 m East, 50 m South, and 30 m West. What is the magnitude and direction of the hiker’s displacement?

8. Practice Problem #1
A plane flies 1500 miles East and 200 miles South. What is the magnitude and direction of the plane’s final displacement?
**not drawn to scale**
1500 miles θ
200 miles
Resultant miles
a2 + b2 = c2
(1500 m)2 + (200 m)2 = R2
R = √ (1500 m)2 + (200 m)2
R = m
tan θ = opp/adj
θ = tan-1 (200/1500)
θ = °

9. Practice Problem #2
A hiker walks 80 m North, 20 m East, 50 m South, and 30 m West. What is the magnitude and direction of the hiker’s displacement?
By subtracting the opposing directions from each other, we find the hiker’s displacement along the y-axis to be 30 m North, and the displacement on the x-axis to be 10 m West.
a2 + b2 = c2
= R2
R = √
R = m
tan θ = opp/adj
θ = tan-1 (10/30)
θ = °

10. Mr. Rockensies – Regents Physics
AIM – What are the components of the resultant?
DO NOW – A car drives 4 miles North, 3 miles East, and 2 miles South, what is its total displacement?
HW - Textbook p. 53 #50, 51, 53
Vector Addition
Mr. Rockensies – Regents Physics

11. Velocity Vectors Occur at the same time – concurrent
Displacement vectors occurred sequentially – one after the other
Boat velocity
Boat
How do we find the resultant velocity?
Stream velocity
River

12. Resultant velocity found by drawing the vectors head to tail – just as with displacement
8 m/s
Boat
Boat
6 m/s
8 m/s
VR2 = = 100
VR = 10 m/s
tan θ = 6/8
θ = tan-1 (6/8) = 37° θ
6 m/s VR
Velocity Resultant

13. Vector Components
If R = A + B, then we can say that A and B are components of R B R
Two or more components add to make a resultant A
Rectangular Components – components which lie on the x and y axes
A resultant can also be resolved back into components!!

14. Japanese Vector Video
Japanese Vector Video - Launching a Ball from a moving truck

15. Practice Problems
A ball is kicked 30° above the horizontal at a velocity of 65 m/s. (a) What is the horizontal component of the velocity? (b) What is the vertical component of the velocity?
A frisbee is tossed 12° above the horizontal at a velocity of 23 m/s. (a) What is the horizontal component of the velocity? (b) What is the vertical component of the velocity?

16. Practice Problem #1 Horizontal Component:
Vertical Component
Horizontal Component:
Since we have the hypotenuse and the angle, we need to find this piece by using the cosine trig function
Ax=Acos(θ)
Ax=(65m/s)cos(30°)
Ax=56.29 m/s
(b) Vertical Component:
Since we have the hypotenuse and the angle, we need to find this piece by using the sine trig function
Ay=Acos(θ)
Ay=(65m/s)sin(30°)
Ay=32.5 m/s
65 m/s
30°
Horizontal Component

17. Practice Problem #2 Horizontal Component:
Since we have the hypotenuse and the angle, we need to find this piece by using the cosine trig function
Ax=Acos(θ)
Ax=(23m/s)cos(12°)
Ax=22.50 m/s
(b) Vertical Component:
Since we have the hypotenuse and the angle, we need to find this piece by using the sine trig function
Ay=Acos(θ)
Ay=(23m/s)sin(12°)
Ay=4.78 m/s
23 m/s
12°