Phys 201 1nd Edition Lecture 3 Outline of Last Lecture I. Applied equations in solving for velocity, acceleration or gravity(freefall), time, and distance. Outline of Current LectureII. VectorsA. DescriptionB. Two-dimensional motion III. Graphing VectorsA. AdditionB. SubtractionC. MultiplicationIV. Solving For VectorsA. Trigonometry equationsB. Solving for Componentsa. Single Vector Problemsb. Vector Addition ProblemsCurrent LectureVectors:Vectors are used to describe motion. The two factors necessary to measure a vector are magnitude, or length, and direction. Comparing vectors allows us to figure out displacement of an object in two-dimensional motion problems.These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.Two-dimensional motionTwo-dimensional motion refers to movement in more than one direction. For example, if a ball is rolling down an incline, it is moving both forward, and down, or if it is thrown in an arc, it is moving first forward and up, then forward and down. It is best to represent vectors on a graph as an arrow. Because direction is one of the main factors effecting a vector, it is important to understand where a vector in a certain direction will fall on a graph.Graphing VectorsThink of the x and y intercept of a graph as the center point on a circle or compass. Region on a Graph Angles in a circleDirections on a compassRegion 1(+x, +y) 0◦-90◦ E, NE, NRegion 2(-x, +y) 90◦-180◦ N, NW, WRegion 3(-x, -y) 180◦-270◦ W, SW, SRegion 4(+x, -y) 270◦-360◦ S, SE, ETo compare vectors, they can be added, subtracted, or multiplied.AdditionA= B= And C = A + BTo find C in this case, we can graphically represent A + B. In the case above, A1 and B1 represent the equation written as A+B. This means that A is graphed first and the tail of B is attached to the head of A. A2 and B2 represent the equation written as B+A. The vector labeled C represents the vectors A and B together as it attached from the tail of the first vector to the head of the second vector. SubtractionB1B2A2A1CWe can do subtraction with vectors in a very similar way. However, in the case of the equation C=A-B, thedirection of the B vector must be reversed in order to make it negative. This means that the B vector on the graph will be drawn as -B= instead of B= MultiplicationMultiplication with vectors works just like multiplication. For example, if D = 4B, then D=B+B+B+B. If we were show this graphically, it would look like this; Remember, if we are multiplying a single vector, you are affecting the size of the vector, but not the direction. The only way to alter the direction of a vector represented this way is to add or subtract a different vector to it. Solving for Vectors:To accurately compare and/or solve vector problems, we need to first picture our vector as part of a triangle and then apply trigonometry. The trigonometric rules to remember are;Sin(θ)= opp/hypCos(θ)= adj/hypTan(θ)=opp/adjHyp2=adj2+opp2ComponentsThese equations will allow us to find the components of the vector. The components are corresponding xand y lines that form the right angle of our hypothetical triangle. Typically, components are represented by the variable representation of your vector with a corresponding x or y coefficient. If you have a vector labeled C, the components of the vector will be Cx and Cy.Example: If a bird flies as 30◦ North East for 2 km, how far East does the bird end up? BBBBD2kmCxCyᶿBecause we are trying to determine how far east the bird traveled, we are looking for the value of Cx. To find this, we plug in our known values into our trigonometric rules. Because we know the length of the hypotenuse and we are trying to find the length of the side adjacent to θ we will use the equation; Cos(θ)=adj/hypTo solve for the adjacent side, rearrange the equation so that is reads; Cos(θ)hyp=adjIf we plug in our values we get;Cos(30)2=adj= 1.7Therefore, the bird has traveled 1.7km east.These steps work the same way when dealing with vector addition. Suppose the same bird from the first problem, after his initial flight, changes direction and flies 60◦ North West until he is directly above his starting point. We want to find out how far the bird traveled all together.To solve this, we will add the new vector (we can label it D) to our original vector (labeled C) on a graph. CCxCyDDxDyIn this case we are trying to determine the length of the hypotenuse D. Looking at the graph, we can see Cx and Dx are of equal length. Therefore, we know the measurement of the angle and the correspondingadjacent side. Therefore, we will use the same equation from the first problem, only this time we will rearrange it to read;D= Dx/Cos(θ)D=1.7/Cos(60)= 3.4Total length of flight= C + D = 3.4 + 2 =5.4Therefore, the bird flew a total of