1. Chapter 3 Kinematics in Two Dimensions
Vector Addition and Projectile Motion
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2. Herriman High Honors Physics
Vectors and Scalars
Vector vs. Scalars
Vectors have both magnitude and direction
Displacement, Velocity, Acceleration, Force, and Momentum
Scalars have only magnitude
Mass, Time, and Temperature
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3. Herriman High Honors Physics
Vector Addition
Two Methods
Graphical Method
Requires a ruler and a protractor
Process
Convert each vector into a line that fits a scale of your choosing
Draw Vectors head to tail, measuring the angle exactly
Draw a resultant
Measure the resultant and the resulting angle
Convert the measurement back into a vector answer
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4. Herriman High Honors Physics
Graphical Addition
A man drives 125 km west and then turns 45º North of West and travels an additional 100 km. What is his displacement?
Step one: Set Scale
1 cm = 25 Km
100 km = 4 cm
45°
125 km = 5 cm
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5. Herriman High Honors Physics
Graphical Addition
Step Two: Move Vectors Head to Tail
Step Three: Draw Resultant from Tail of First Vector to the head of the last.
100 km = 4 cm
45°
125 km = 5 cm
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6. Herriman High Honors Physics
Graphical Addition
Step Four: Measure the Measure the Resultant and the resulting angle.
Step Five: Using the Scale, convert the measurement into an answer.
100 km = 4 cm
8.3 cm
8.3 cm * 25 Km/cm = 208 Km
45°
125 km = 5 cm
19.9°
Final Answer: His displacement is ° North of West
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7. Mathematical Addition
Mathematical Addition of Vectors Requires a basic knowledge of Geometry – You must know:
Pythagorean Theorem a2 + b2 = c2
Sin θ = Opposite/Hypotenuse
Cos θ = Adjacent/Hypotenuse
Tan θ = Sin θ/ Cos θ =Opposite/Adjacent
ArcSin, ArcCos, ArcTan
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8. Mathematical Additions Vectors at Right Angles (Easiest)
A plane flies due north at 200 m/s while the wind is blowing due west at 25 m/s. What is the resulting velocity of the plane with respect to the ground?
To Solve:
Draw a rough sketch of the original vectors
Draw a parallelogram
Draw the Resultant
Use pythagorean theorem to find the magnitude of the velocity
Use Arctan to find the direction 1
200 m/s
25 m/s
2 & 3
200 m/s θ
25 m/s
4 & 5
Tanθ=(200/25)
θ=82.9° N of W
Try: P. 89
Practice A
Problems 2 & 4
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9. Vectors at Right Angles Sample Problems
An archaeologist climbs the Great Pyramid in Giza, Egypt. The pyramid’s height is 136 meters and its with is 230 meters. What is the magnitude and direction of the displacement of the archaeologist after she has climbed from the bottom of the pyramid to the top?
Answer: Δx = ° with respect to the ground.
Try: P. 89
Practice A
Problems 1 & 3
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10. Resolving Vectors into Components
Cos θ = W/100 km
W = 100 Cos 45° =70.7 Km
The opposite of adding to vectors at right angles is resolving a single vector into two components at right angles.
This process uses the trig functions Sin and Cosine
For example, you have a displacement vector of ° N of W and you want to know how far north and how far west you are of your starting point.
Solution
Draw original Vector
Draw Parallelogram
Label the sides North and West
Use the definition of Sine and Cosine to solve W N
45°
Sin θ = N/100 km
N = 100 Sin 45° =70.7 Km
Try: Practice B, P. 92
Problems 2 & 4
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11. Herriman High Honors Physics
Putting it all Together Mathematical Addition when Vectors are Not at Right Angles
Draw a rough sketch of the original vectors
Resolve each vector into its Components using Sine and Cosine.
Reduce the component vectors into a system of two vectors by adding horizontal and vertical vectors
Draw a parallelogram
Draw the Resultant
Use Pythagorean Theorem to find the magnitude, and tangent to find the direction of the resultant
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12. Mathematical Addition
Step One: Draw Original Vectors
45º
125 Km West
100 Km 45° North of West
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13. Mathematical Addition
Step Two: Resolve Vectors into
components
Cos θ = W/100 km
W = 100 Cos 45° =70.7 Km
Sin θ = N/100 km
N = 100 Sin 45° =70.7 Km
125 Km West
100 Km 45° North of West
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14. Mathematical Addition
Step Three: Reduce to a system of two
vectors
195.7 Km West
70.7 Km North
70.7 Km North
70.7 Km West = +
125 Km West
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15. Mathematical Addition
Step 4: Draw Parallelogram
Step 5: Draw Resultant
Step 6:Use Pythagorean Theorem to find magnitude and tangent to find the direction of the resultant. B
Pythagorean Theorem: C2 = A2 + B2
C = SQRT(A2 + B2) = SQRT{(195.7)2+(70.7)2} =
208 km
Tan θ = 70.7/197.5 =
Θ= ArcTan = 19.9° North or West A
TRY: P. 94
Practice C
Problems 1 & 3
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16. Graphical vs. Mathematical Methods
Graphical Method
Less Complicated Math
Good Approximation
Not as exact as you would like
Mathematical Method
No need for ruler or protractor
Less time dedicated to sketch
More Accurate result
Mathematics is more difficult
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17. Herriman High Honors Physics
Projectile Motion
A projectile is any object thrown or launched
Uses the same kinematic equations you have already learned
Requires that you use two sets of equations; one in the horizontal and one in the vertical
These two sets are related only by time
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18. Projectile Motion (Launched or Thrown Horizontally)
A projectile which is dropped or thrown horizontally has an initial velocity in the Y direction (V0y) = 0
Acceleration in the horizontal direction (ax) = 0; hence Velocity in the horizontal direction (Vx) is constant
This means that only one kinematic equation applies in the horizontal direction:
Xx = Vxt
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19. Projectile Motion (Launched or Thrown Horizontally)
A ball is thrown horizontally off a building which is 200 m high with a velocity of 10 m/s
How long does it take to reach the ground?
How far from the building will it land?
What is its velocity just before it hits the ground (remember magnitude and direction)
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20. Solution X Direction Vx= 10 m/s The only Equation is: Xx=VxT
b) Solve for displacement in the X direction
Xx=(10 m/s)(6.4 s) = 64 m
Y Direction
Viy= 0 m/s
Ay = 9.8 m/s2
Xy= 200 m
A) Find Time
Xy= ViyT + ½ayT2
200 m = 0(T)+½(9.8 m/s2)T2
C) 2. Combine X & Y Components to get final velocity magnitude and direction s
C) 1. Find final velocity in the Y direction
Vfy = aT = (9.8 m/s2)(6.4 s) = 62.7 m/s
Tan θ=62.7/10 =6.27
θ= 81° above horizontal
Try: p. 99, Practice D
Problems 1& 3
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21. Projectile Motion (Launched at an Angle)
Similar to Launching Horizontally however:
Initial Velocity in the Y direction is Not zero.
You must break the initial velocity into components using Sine and Cosine
Final velocity, if the projectile falls to the same level, will have the same magnitude and direction as at the launch (assuming no air resistance).
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22. Projectile Motion (Launched at an Angle)
An arrow is fired into the air with a velocity of 25 m/s at 30º above the horizontal.
How high will it rise?
How long does it take to reach the ground?
How far from the building will it land?
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23. Herriman High Honors Physics
Solution
Y Direction
Viy= (25 m/s)(sin 30°)
=12.5 m/s
Ay = -9.8 m/s2
Vfy = 0 m/s
Vfy2 = Viy2 + 2AyXy
X= -(12.5 m/s)2/(2)(-9.8 m/s2)
= 9.97 m
Vfy= Viy + AT
T = -(12.5 m/s)/-9.8 m/s2
= 1.28 seconds
X Direction
Xx= VxT
Xx= (21.7 m/s)(1.28 sec) = m
Vix= (25 m/s)(cos 30°) = 21.7 m/s
Try: P. 101
Practice E
Problems 2 & 4
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24. Frame of Reference Relative Velocity
Velocity Measurements differ in different frames of reference.
For example if a dummy is dropped from a low flying airplane a person in the plane would have a different frame of reference than an observer watching from the ground.
The observer in the plane, because they are moving at the same horizontal speed as the dummy, would see the dummy fall straight down.
The observer on the ground, because they are standing still, would see the dummy following parabolic trajectory, moving forward while falling.
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25. Frame of Reference (Moving along the same path)
Two cars are traveling in the same direction one at 60 km/h and the other at 90 km/h. How fast is the faster car moving with respect to the slower one?
Vse= 60 km/h (se = slower w/respect to earth)
Vfe = 90 km/h (fe = faster w/respect to earth)
Vfs = (fs = faster w/respect to slower)
Solve
Vfs = Vfe- Vse = 90 km/h – 60 km/hr= 30 km/h
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26. Frame of Reference (Moving along different paths)
A boat heading north crosses a wide river with a velocity of 10 km/h relative to the water. The river has a uniform velocity of 5 km/h due east. Determine the boat’s resultant velocity relative to an observer on the shore.
Solve using vectors (demo)
Pythagorean Theorem (magnitude)
Tangent (direction)
Try: p. 105
Practice F
Problems 1 & 3
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