1. AP Physics Unit 2 - Vectors

2. Day #3 Relative Motion, Velocity/Position vector relations

3. We’re here on your schedule

12. The relative velocity equation v BS = v BC + v CS v BS v S vBvB v BS C C v BS = + V BS = velocity of the ball relative to the side of the road V BC = velocity of the ball relative to the car V CS = velocity of the car relative to the side of the road Here’s an easy trick to remember the equation

13. * REALLY IMPORTANT NOTE

14. A boy sits in his car with a tennis ball. The car is moving at a speed of 20 mph. If he can throw the ball 30 mph, find the speed with which he will could hit… a)his brother in the front seat of the car. b) A sign on the side of the road that the car is about to pass. c) A sign on the side of the road that the car has already passed v BS = v BC + v CS 50 mph 30 mph20 mph 10 mph 30 mph 20 mph v BS = v BC + v CS 30 mph PROBLEM #1

15. Objects A and B start off the same point. Object A travels 50 km at 40 o N of W. Object B travels at a speed of 30 km/h for 20 minutes in the direction [S 30 o E]. After all the travelling is complete, find the position of object A relative to object B. r AG 40 o r BG 30 o r AG r GB r AB PROBLEM #2 r AB = r AG + r GB

16. r AG = 50 km at 40 o N of W = -50cos40 i + 50 sin40 j r BG = 30 km/h * (1/3)h = 10 km [S 30 o E] = 10sin30 i – 10cos30 j r GB = -10sin30 i + 10cos30 j r AB = (-50cos40 - 10sin30) i + (50 sin40 + 10cos30 ) j = -43.3 i + 40.79 j = 59.5 km [43.3 o N of W] r AG 40 o r BG 30 o r AG r GB r AB PROBLEM #2 r AB = r AG + r GB

17. A golfball is being hit into a gale-force wind, where v W = (-100 j + 50 k) mph. The golfer wants the ball to travel in the + j direction, so he hits the ball accordingly. The ball is hit perfectly, covering the tee to hole distance of 300 yds in 5 seconds. Find the velocity (both magnitude and direction) of the ball relative to the wind directly after being hit. v BW = v BG + v GW PROBLEM #3 300 yds = 900 ft = 900 ft * (1 mi / 5280 ft) =.17 mi.17045 mi / 5 sec =.03409 mi/s * (3600 s / 1h) = 122.727 mph UNIT CONVERSION

18. A golfball is being hit into a gale-force wind, where v W = (-100 j + 50 k) mph. The golfer wants the ball to travel in the + j direction, so he hits the ball accordingly. The ball is hit perfectly, covering the tee to hole distance of 300 yds in 5 seconds. Find the velocity (both magnitude and direction) of the ball relative to the wind directly after being hit. v BW = v BG + v GW PROBLEM #3 V BW = ? V BG 122.727 j V WG = -100 j + 50 k  V GW = 100 j - 50 k V BW = (122.727 j) + (100 j – 50 k) = 222.727 j – 50 k |V BW | = at an angle of 12.65 o off the +j direction in the direction of -k

19. Solve Problems 39, 40, 42, and 49 on pp 67-68. TONIGHTS HW