VECTORS Level Physics Objectives and Essential Questions  Objectives  Distinguish between basic trigonometric functions (SOH CAH TOA)  Distinguish between vector and scalar quantities  Add vectors using graphical and analytical methods  Essential Questions  What is a vector quantity? What is a scalar quantity? Give examples of each SCALAR A SCALAR quantity is any quantity in physics that has MAGNITUDE ONLY Number value with units Scalar Example Magnitude Speed 35 m/s Distance 25 meters Age 16 years VECTOR A VECTOR quantity is any quantity in physics that has BOTH MAGNITUDE and DIRECTION ρr r r x, v , a, F Vector Example Magnitude and Direction Velocity 35 m/s, North Acceleration 10 m/s2, South Force 20 N, East An arrow above the symbol illustrates a vector quantity It indicates MAGNITUDE and DIRECTION VECTOR APPLICATION ADDITION: When two (2) vectors point in the SAME direction, simply add them together EXAMPLE: A man walks 46.5 m east, then another 20 m east Calculate his displacement relative to where he started 46.5 m, E + 66.5 m, E 20 m, E MAGNITUDE relates to the size of the arrow and DIRECTION relates to the way the arrow is drawn VECTOR APPLICATION SUBTRACTION: When two (2) vectors point in the OPPOSITE direction, simply subtract them EXAMPLE: A man walks 46.5 m east, then another 20 m west Calculate his displacement relative to where he started 46.5 m, E 20 m, W 26.5 m, E NON-COLLINEAR VECTORS When two (2) vectors are PERPENDICULAR to each other, you must use the PYTHAGOREAN THEOREM Example: A man travels 120 km east then 160 km north Calculate his resultant displacement FINIS H the hypotenuse is called the RESULTANT 160 km, N c = a2 + b2 → c = a2 + b2 VERTICAL COMPONENT c = resultan t = [(120) + (160) ] c = 200 km 120 km, E HORIZONTAL COMPONENT WHAT ABOUT DIRECTION? In the example, DISPLACEMENT asked for and since it is a VECTOR quanti we need to report its direction N W of N N of E E of N N of E N of W W NOTE: When drawing a right triangle that conveys some type of motion, you MUST draw your components HEAD TO TOE E S of W S of E W of S E of S S NEED A VALUE – ANGLE! Just putting N of E is not good enough (how far north of east ?) We need to find a numeric value for the direction 200 km θ N of E 120 km, E 160 km, N To find the value of the angle we use a Trig function called TANGENT opposite side 160 Tanθ = = = 1.333 adjacent side 120 θ = Tan −1 (1.333) = 53.1o So the COMPLETE final answer is : 200 km, 53.1 degrees North of East What are your missing components? Suppose a person walked 65 m, 25 degrees East of North What were his horizontal and vertical components? H.C = ? The goal: ALWAYS MAKE A RIGHT TRIANGLE! 25 To solve for components, we often use the trig functions since andadjacent cosine side opposite side V.C = ? 65 m cosineθ = hypotenuse adj = hyp cos θ sineθ = hypotenuse opp = hyp sin θ adj = V C = 65 cos 25 = 58.91m, N opp = H C = 65 sin 25 = 27.47m, E Example A bear, searching for food wanders 35 meters east then 20 meters north Frustrated, he wanders another 12 meters west then meters south Calculate the bear's displacement - 12 m, W - = m, S 14 m, N 20 m, N 35 m, E R = 14 + 232 = 26.93m 14 m, N R θ 23 m, E = 23 m, E 14 = 6087 23 θ = Tan −1 (0.6087) = 31.3 Tanθ = The Final Answer: 26.93 m, 31.3 degrees NORTH or EAST Example A boat moves with a velocity of 15 m/s, N in a river which flows with a velocity of 8.0 m/s, west Calculate the boat's resultant velocity with respect to due north Rv = 82 + 152 = 17 m / s 8.0 m/s, W 15 m/s, N Rv θ Tanθ = = 0.5333 15 θ = Tan −1 (0.5333) = 28.1 The Final Answer : 17 m/s, @ 28.1 degrees West of No Example A plane moves with a velocity of 63.5 m/s at 32 degrees South of East Calculate the plane's horizontal and vertical velocity components H.C =? 32 63.5 m/s adjacent side cosineθ = hypotenuse adj = hyp cos θ V.C = ? opposite side sineθ = hypotenuse opp = hyp sin θ adj = H C = 63.5 cos 32 = 53.85 m / s, E opp = V C = 63.5 sin 32 = 33.64 m / s, S Example A storm system moves 5000 km due east, then shifts course at 40 degrees North of East for 1500 km Calculate the storm's resultant displacement 1500 km 40 5000 km, E H.C adjacent side hypotenuse V.C adj = hyp cos θ cosineθ = opposite side hypotenuse opp = hyp sin θ sineθ = adj = H C = 1500 cos 40 = 1149.1 km, E opp = V C = 1500 sin 40 = 964.2 km, N 2 R = 6149.1 + 964.2 = 6224.2 km 5000 km + 1149.1 km = 6149.1 km 964.2 Tanθ = = 0.157 6149.1 R 964.2 km θ = Tan −1 (0.157) = 8.92 o θ 6149.1 km The Final Answer: 6224.2 km @ 8.92 degrees, North of East [...]...Example A bear, searching for food wanders 35 meters east then 20 meters north Frustrated, he wanders another 12 meters west then 6 meters south Calculate the bear's displacement - 12 m, W - = 6 m, S 14 m, N 20 m, N 35 m, E R = 14 2 + 232 = 26.93m 14 m, N R θ 23 m,... 0.5333 15 θ = Tan −1 (0.5333) = 28.1 The Final Answer : 17 m/s, @ 28.1 degrees West of No Example A plane moves with a velocity of 63.5 m/s at 32 degrees South of East Calculate the plane's horizontal and vertical velocity components H.C =? 32 63.5 m/s adjacent side cosineθ = hypotenuse adj = hyp cos θ V.C = ? opposite side sineθ = hypotenuse opp = hyp sin θ adj = H C = 63.5 cos 32 = 53.85 m / s, E