PHYS 201 1st Edition Lecture 4 Outline of Last Lecture I. Adding, subtracting, and multiplying vectors, and graphical representations of vectors.Outline of Current Lecture II. Addition of Vector VelocitiesA. One dimensionB. Two dimensionsIII. Vector displacementIV. Separating x and yA. Applying x and y separation to equationsCurrent LectureRelative Velocity: addition of velocities.Vectors can be used to represent velocity, where speed acts as the magnitude of the two vectors. In this way, we can add or subtract two velocities the same way that we add or subtract two vectors. If both velocities are moving in the same direction (if both velocities are positive or both are negative), we simply add the two velocities together. If the two velocities are moving in opposite directions, (making one velocity positive and one negative) then we will subtract one velocity from another. Example: An escalator moves at a constant speed of 2m/s. If I walk up the moving escalator at a constantspeed of .5m/s, what is my observed speed to someone standing still?E= 2m/s M= .5m/s S= E+M S=2.5m/sSuppose that on this same escalator, I decided to walk at the same speed in the opposite direction that the escalator was moving. In this case, we would use subtraction, because now vector M is negativeS=E-M E= 2m/s M= -.5m/s 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.S=1.5m/sVector addition in two dimensions If our velocities are not moving in either the same direction or in opposite directions, then we solve for them in the same way that we would solve for vectors moving in similar directions. We will plot them graphically, then create a triangle and use trigonometry to solve.Example: I am rowing north at a constant speed of 2m/s across a river that is flowing east at a constant speed of 10m/s. What is the angle of my velocity in relation to the shore and what is my speed? θ Because we only know the magnitude of the sides opposite and adjacent to the unknown angle, we will solve for the angle of my velocity using the equation Tan(θ)=opp/adj.Tan(θ)=2/10=.2θ= Tan-1(.2)=11.3◦My angle of velocity is 11.3◦. To solve for the speed at which I am traveling we need to solve for the hypotenuse. Because we know the other two sides of the triangle, we can solve for the hypotenuse usingthe equation Hyp2=Opp2 + Adj2.|Hyp|=√( Opp2 + Adj2)|Hyp|=√( 22 + 102)|Hyp|=√(104)|Hyp|=10.2Therefore, my speed across the river is 10m/sR 10m/sM 2m/sS= M+RVector displacement We can apply the equations we have used previously on velocity and acceleration to describe displacement and position of vectors. When referring to displacement in vectors, we used Δr to represent the difference between the initial vector(ri) and the final vector(rf). We measure average vector velocity by dividing Δr by Δt(the change in time), essentially the exact same way we find average velocity of a line, except we use the change in vectors instead of the change in position.Separating x and yWhen solving for vectors, the x direction of the vector and y direction of the vector are independent. This means that whatever we do to the vector in the x direction does not affect the vector in the y direction, and vice versa. This means that we can split up our velocity and acceleration of vector measurements into x and y parts.x directiony directionV of vectorVx=Δx/Δt Vy=Δy/Δta of vector ax=ΔVx/Δt ay= ΔVy/ΔtKnowing this, we can apply all of our equations from previous lectures to vectors and separate our equations into x and y directions.Equations:Vx=Vxo + axtVy=Vyo + aytx= xo + Vxot + (1/2)axt2y= yo + Vyot +