Scalars & Vectors
Vectors are quantities that have both a direction and a magnitude (size). Ex. km, 30 ο north of east Examples of Vectors used in Physics Displacement Velocity Acceleration Force Scalars are quantities that have only a magnitude(size) are called. Scalar Example Speed Distance Age Heat Magnitude 0 m/s 10 m 15 years 1000 calories
Vectors can be represented by words Take your team clicks (km) north US Air 45, new course 30 o at 500 mph. Vectors can be represented by symbols In the text, boldface indicates vectors. Examples: F a Vav Δx t Vectors can be represented graphically using arrows The direction of the arrow is the direction of the vector. The length of the arrow tells the magnitude Vectors can be moved parallel to themselves and still be the same vector Vectors only tell amount and direction, so a vector doesn t care where it starts.
The sum of two vectors is called the resultant. To add vectors graphically, draw each vector to scale. Place the tail of the second vector at the tip of the first vector. Vectors can be added in any order. To subtract a vector, add its opposite.
VECTOR ADDITION If similar vectors point in the SAME direction, add them. Example: A man walks 54.5 meters east, then another 30 meters east. Calculate his displacement relative to where he started? 54.5 m, E + 30 m, E 84.5 m, E Notice that the SIZE of the arrow conveys MAGNITUDE and the way it was drawn conveys DIRECTION.
VECTOR SUBTRACTION - If vectors are going in opposite directions, you SUBTRACT. Example: A man walks 54.5 meters east, then 30 meters west. Calculate his displacement relative to where he started? 54.5 m, E 30 m, W 4.5 m, E
When vectors are perpendicular, you must use the Pythagorean Theorem. Example: A man walks 95 km, East then 55 km, north. Calculate his RESULTANT DISPLACEMENT The hypotenuse in Physics is called the RESULTANT. Finish c c c 55 km, N a b Resultant 1050 c 95 109.8 km a b 55 Horizontal Component Vertical Component 95 km,e Start The LEGS of the triangle are called the COMPONENTS
In the previous example, DISPLACEMENT was asked for and since it is a VECTOR we should include a DIRECTION on our final answer. N W of N E of N N of E W N of W S of W N of E S of E E NOTE: When drawing a right triangle that conveys some type of motion, you MUST draw your components HEAD TO TOE. W of S S E of S
Just putting North of East on the answer is NOT specific enough for the direction. We MUST find the VALUE of the angle. 109.8 km 55 km, N To find the value of the angle we use a Trig function called TANGENT. N of E 95 km, E Tan oppositeside adjacent side 550.5789 95 Tan1(0.5789) 30o So the COMPLETE final answer = 109.8 km, 30 degrees North of East
Resolve each vector into x and y components, using sin and cos. Add the x components together to get the total x component. Add the y component together to get the total y component. Find the magnitude of the resultant using Pythagorean theorem. Find the direction of the resultant using the inverse tan function.
Any vector can be resolved, that is, broken up, into two vectors, one that lies on the x- axis and one on the y-axis.
An arrow is shot from a bow at an angle of 5 ο above the horizontal, with an initial speed of 45 m/s. Find the horizontal and vertical components of the arrow s initial velocity. v 45 m/s 5 o v x vx cos v v cos v (45 m/s) cos(5 v x 40.78 x ) 41m/s v x v y v v x y vy sin v v sin v 5?? o (45 m/s) sin(5 v y ) 19.01 19 m/s y v y
Suppose a person walked 65 m, 5 degrees East of North. What were his horizontal and vertical components? V.C =? H.C. =? 5 65 m cos adj The goal: ALWAYS MAKE A RIGHT TRIANGLE! To solve for components, we often use the trig functions since and cosine. adjacent side hypotenuse hyp cos sin opp opposite side hypotenuse hyp sin adj V. C. 65cos 5 58.91m, N opp H. C. 65sin 5 7.47m, E
A bear, searching for food wanders 35 meters east then 0 meters north. Frustrated, he wanders another 1 meters west then 6 meters south. Calculate the bear's displacement. 1 m, W - = 3 m, E 6 m, S 0 m, N - = 14 m, N 35 m, E R 14 3 6.93m R 3 m, E 14 m, N 14 Tan.6087 3 1 Tan (0.6087) 31.3 The Final Answer: 6.93 m, 31.3 degrees NORTH or EAST
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. 8.0 m/s, W 15 m/s, N R v R v 8 15 17 m 8 Tan 0.5333 15 1 Tan (0.5333) / s 8.1 The Final Answer : 17 m/s, @ 8.1 degrees West of North
A plane moves with a velocity of 63.5 m/s at 3 degrees South of East. Calculate the plane's horizontal and vertical velocity components. H.C. =? 3 cos adj V.C. =? hyp adjacent hypotenuse cos sine opp opposite hypotenuse hyp sin 63.5 m/s adj H. C. 63.5cos 3 53.85m / s, E opp V. C. 63.5sin 3 33.64m / s, S
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. 5000 km, E 1500 km 40 H.C. cosine adj hyp cos V.C. adj adj hypotenuse sine opp V. C. 1500sin 40 964. opp hypotenuse opp hyp sin H. C. 1500cos 40 1149.1 km, E km, N 1500 km + 1149.1 km = 649.1 km R 649.1 km 964. km R 649.1 964. 964. Tan 0.364 649.1 1 Tan (0.364) 0.0 819.1km The Final Answer: 819.1 km @ 0 degrees, East of North
We use the term VECTOR RESOLUTION to suggest that any vector which IS NOT on an axis MUST be broken down into horizontal and vertical components. BUT --- the ultimate and recurring themes in physics is take any and all vectors and turn them all into ONE BIG RIGHT TRIANGLE.
1. Make a drawing showing all the vectors, angles, and given directions.. Make a chart with all the horizontal components in one column and all the vertical components on the other. 3. Make sure you assign a negative sign to any vector which is moving WEST or SOUTH. 4. Add all the horizontal components to get ONE value for the horizontal. Do the same for the vertical. 5. Use the Pythagorean Theorem to find the resultant and Tangent to find the direction.
A search and rescue operation produced the following search patterns in order: 1: 30 meters, west : 65 meters, 3 degrees East of South 3: 130 meters, east 4: 4 meters, degrees West of North
1: 30 meters, west : 65 meters, 3 degrees East of South 3: 130 meters, east 4: 4 meters, degrees West of North 30 m, W 3 ο 4 m 65 m ο 130 m, E
1: 30 meters, west : 65 meters, 3 degrees East of South 3: 130 meters, east 4: 4 meters, degrees West of North Leg Horizontal Vertical 1 30 m 0 m 3 130 m 0 m 4
65 meters, 3 degrees East of South Leg Horizontal Vertical 1 30 m 0 m 65cos3 = 55.1 m v.c 65 m 3 h.c. 65sin3 =34.44 m 34.44 m 55.1 m 3 130 m 0 m 4
4 meters, degrees West of North h.c. 4sin =15.73 m Leg Horizontal Vertical 1 30 m 0 m 4 m v.c 4cos=38.94 m 34.44 m 55.1 m 3 130 m 0 m 4 15.73 m 38.94 m
Leg Horizontal Vertical 1: 30 meters, west : 65 meters, 3 degrees East of South 3: 130 meters, east 4: 4 meters, degrees West of North 1-30 m 0 m 34.44 m - 55.1 m 3 130 m 0 m 4-15.73 m 38.94 m
1: 30 meters, west : 65 meters, 3 degrees East of South 3: 130 meters, east 4: 4 meters, degrees West of North Leg Horizontal Vertical 1-30 m 0 m 34.44 m -55.1 m 3 130 m 0 m 4-15.73 m 38.94 m 118.71 m -16.18 m Total 118.71 m -16.18 m What does this mean???
30 m, W 118.71 m 3 65 m -16.18 m 4 m 130 m, E R 118.71 ( 16.18) 119.81 m 16.18 Tan 0.136 118.71 1 Tan (0.136) 7.76 Final Answer: 119.81 m, 7.76 degrees, South of East