PHY 107 1st Edition Lecture 9 Outline of Last Lecture I. Galilean TransformationII. Relative Motion in One DimensionIII. Relative Motion in Two DimensionsOutline of Current Lecture IV. Relative Motion (review)V. Practice Questions (up to chapter 4)VI. Force & Motion IntroductionCurrent LectureRelative Motion:- The velocity of a particle, P, determined by 2 different observers, A and B, varies from observer to observer.- In one dimension – o xA = xB + xBA → ddt(xA) = ddt(xB) + ddt(xBA) → vA = vB + vBAod vBAdt = 0 → aA = aB (even though the observers measure different velocities for P, they measure the same acceleration)- In two dimensions – o´rA = ´rB + ´rBA → ddt´rA = ddt´rB + ddt´rBA → ´vA = ´vB +´vBA oddt´vA = ddt´vB + ddt´vBA → d vBAdt = 0 → ´aA = ´aB (even though the observers measure different velocities for P, they measure the same acceleration)Practice Questions:A particle moves horizontally in uniform circular motion, over a horizontal xy plane. At one instant, it moves through the point at coordinates (4.00 m, 4.00 m) with a velocity of – 5.00´i 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.m/s and an acceleration of + 12.5´j m/s2. What are the (a) x and (b) y coordinates of the center of the circular path?o Since the acceleration is + 12.5´j, the acceleration is pointedstraight up. We know that the acceleration in centripetalmotion is pointed towards the center of the circle, so the center is straight above the point (4.00 m, 4.00 m)o The x coordinate of the center is 4.00 mo For the y coordinate, we must find the radius:a = v2r → r = v2a = (5.00ms)212.5m/ s2 = 2.00 mo So, the y coordinate of the center is 2.00 m + 4.00 m = 6.00 m(a) 4.00 m (b) 6.00 mA boat is traveling upstream in the positive direction of an x axis at 14 km/h with respect to the water of a river. The water is flowing at 9.0 km/h with respect to the ground. What are the (a) magnitude and (b) direction of the boat’s velocity with respect to the ground? A child on the boat walks from front to rear at 6.0 km/h with respect to the boat. What are the (c) magnitude and (d) direction of the child’s velocity with respect to the ground?*Upstream corresponds to the +x direction*o´vA = ´vB + ´vBA → ´vBG = ´vBW + ´vWG and ´vCG = ´vCB + ´vBGo´vBG = ´vBW + ´vWG = (14 km/h)´i + (-9 km/h)´i = (5 km/h)´i → |´vBG| = (a) 5 km/ho The direction of ´vBG is (b) +x, or upstream.o´vCG = ´vCB + ´vBG = (-6 km/h)´i + (5 km/h)´i = (-1 km/h)´i → |´vCG| = (c)1 km/ho The direction of ´vCG is (d) –x, or downstream. A particle starts from the origin at t = 0 with a velocity of 8.0´j m/s and moves in the xy planewith constant acceleration (4.0´i + 2.0´j) m/s2. When the particle’s x coordinate is 29 m, what are its (a) y coordinate and (b) speed?o x = voxt + axt22, y = voyt + ayt22 o vx = vox + axt, vy = voy + ayto´r = ´v0t + ´a t22 = (8.0´j) + (4.0´i+2.0´j)t22 = (2.0t2)´i + (8.0t + 1.0t2)´j(a) x = 29 m; 2.0t2 = 29, t = 3.8 so y = (8.0 m/s)(3.8 s) + (1.0 m/s2)(3.8 s)2 = 45´v = ´v0 + ´at at t = 3.8 s, v = √vx2+vy2 = √(15.2ms)2+(15.6ms)2 = (b) 22 m/sForce & Motion:- “Dynamics” - Newton’s Three Laws of Motion: o Describe physical phenomena of a vast range including motion of objects from proteins and cells to stars and planetso Fails in 2 circumstances:1. When the speed of the object is near (1% or more) the speed of light in a vacuum(3 x 108 m/s)*In this case, must use theory of relativity*2. When objects are very small *In this case, must use quantum