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# Chapter 3 Kinematics in Two Dimensions

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Chapter 3 Kinematics in Two Dimensions
Vector Addition and Projectile Motion Herriman High Honors Physics

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 Herriman High Honors Physics

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 Herriman High Honors Physics

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 Herriman High Honors Physics

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 Herriman High Honors Physics

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 Herriman High Honors Physics

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 Herriman High Honors Physics

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 Herriman High Honors Physics

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 Herriman High Honors Physics

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 Herriman High Honors Physics

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 Herriman High Honors Physics

Mathematical Addition
Step One: Draw Original Vectors 45º 125 Km West 100 Km 45° North of West Herriman High Honors Physics

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 Herriman High Honors Physics

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 Herriman High Honors Physics

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 Herriman High Honors Physics

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 Herriman High Honors Physics

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 Herriman High Honors Physics

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 Herriman High Honors Physics

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) Herriman High Honors Physics

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 Herriman High Honors Physics

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). Herriman High Honors Physics

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? Herriman High Honors Physics

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 Herriman High Honors Physics

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. Herriman High Honors Physics

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 Herriman High Honors Physics

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 Herriman High Honors Physics

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