1. PHYS 20 LESSONS Unit 2: 2-D Kinematics Projectiles Lesson 3: Relative Velocity

2. Reading Segment #1: Relative Velocity To prepare for this section, please read: Unit 2: p.11

3. C. Relative Velocity The speed of an object is always in reference to some other object. For example, when we say an car is moving at 50 km/h, this is in reference to the ground. The ground is assumed to be at rest. 50 km/h ground (at rest)

4. But what if the ground (i.e. the reference frame) is moving? 50 km/h ground 10 km/h In order to determine the velocity of an object relative to a moving reference frame, we need to use vector arithmetic.

5. C1. 1-D Relative Velocity We will first consider motion in one dimension.

6. Consider a person walking on a train: 4.0 m/s10.0 m/s The person is walking 4.0 m/s East (relative to the train), while the train is moving 10.0 m/s (relative to the ground). How fast is the person moving relative to the ground?

7. 4.0 m/s10.0 m/s Since the person and the train are moving in the same direction, the person appears to be moving even faster than the train. v person = (+4.0 m/s) + (+10.0 m/s)Ref: East + = +14.0 m/s West - = 14.0 m/s East

8. What if the person is walking in the opposite direction? 4.0 m/s10.0 m/s The person is walking 4.0 m/s West (relative to the train), while the train is moving 10.0 m/s East. How fast is the person moving relative to the ground?

9. 4.0 m/s10.0 m/s The person and the train are moving in opposite directions. But the train is moving faster. v person = (-4.0 m/s) + (+10.0 m/s)Ref: East + = +6.0 m/s West - = 6.0 m/s East

10. Summary (1-D Relative Velocity) If both objects are moving on the same axis:  State a reference system - choose a positive and a negative direction  Simply add the vectors to get the relative velocity

11. Ex. 1A boat, capable is travelling 9.0 m/s in still water, heads East on a river with a current moving 4.0 m/s West. a) What is its resultant velocity? b) How long would it take to travel 6.0 km upstream?

12. a)current 4.0 m/s 9.0 m/s Ref: East + West -

13. a)current 4.0 m/s 9.0 m/s Ref: East + West -v boat = (+9.0 m/s) + (-4.0 m/s) = +5.0 m/s = 5.0 m/s East

14. b)v = d t d = v t t = d = 6000 m v 5.0 m/s = 1.2  10 3 s

15. Practice Problems Try these problems in the Physics 20 Workbook: Unit 2 p. 13 #1, 2

16. Reading Segment #2: 2-D Relative Velocity To prepare for this section, please read: Unit 2: p.12

17. C2. 2-D Relative Velocity We will now consider relative velocity in two dimensions. This is especially useful for navigation.

18. Terminology: Heading = The direction the plane is aimed i.e. the way it would travel if there is no wind Airspeed = The speed of the plane relative to the air i.e. the speed it would travel if there was no wind Groundspeed = The speed relative to the ground i.e. the actual speed of the plane

19. There are two kinds of problems you will deal with: 1. Crosswind (or cross-current) - the wind will blow you "off course" 2. Heading into the wind (or upstream) - in order to get to a destination, you must aim your craft into the wind

20. Ex. 1(Crosswind question) A plane heads directly North with an airspeed of 300 km/h. However, there is a 50.0 km/h wind blowing from the East (i.e. towards the West). a) What is the resultant velocity of the plane? b) How far would it get blown off-course in 40.0 minutes?

21. a) 300 km/h The plane can travel 300 km/h if there is no wind.

22. a) 50.0 km/h 300 km/h But, what happens to its velocity if there is a 50 km/h crosswind?

23. a) N 50.0 km/h wind R 300 km/h  plane WE Since the velocity vectors are at right angles (i.e. 90  ), add them tail-to-tip. The resultant R represents the plane's actual velocity (with respect to the ground).

24. a) 50.0 km/h R 300 km/h  Pythag: c 2 = a 2 + b 2 R 2 = (50.0 km/h) 2 + (300 km/h) 2 R = 304 km/h Notice that the wind moves the plane even faster.

25. a) 50.0 km/h R 300 km/h  Soh Cah Toa: tan  = 50.0 300  = tan -1 (0.1667) = 9.46 

26. a) N 304 km/h 9.46  WE So, the actual velocity of the plane (relative to the ground) is 304 km/h at 9.46  W of N 80.5  N of W 99.5  rcs

27. b) First, find how far the plane travelled. v = d t d = v t = (304.138 km/h) (40 / 60 hours) = (304.138 km/h) (0.66667 h) = 202.76 km

28. b) N 202.76 km 9.46  WE The plane's displacement is in the same direction as its actual velocity. Thus, it is directed at 9.46  W of N.

29. b) x 202.76 km 9.46  For a right triangle. We are looking for the distance x.

30. b) x 202.76 km 9.46  Soh Cah Toa sin  = x 202.76 km x = (202.76 km) (sin 9.46  ) = 33.3 km

31. Ex. 2(Heading "into the wind") A plane must fly directly West a distance of 1200 km. However, there is a 65.0 km/h wind towards the North. If the plane's airspeed is 340 km/h, a) what heading is needed? b) what is the flight time?

32. a) Destination 340 km/h If there was no wind, the plane could aim directly West and travel at a speed of 340 km/h.

33. a) Destination However, if the plane aims West and there is a wind blowing towards the North, then it will be blown off-course.

34. a) Destination The plane must aim "into the wind". Then, the wind will blow the plane "on-course" and it will arrive at its destination.

35. a) v  60 km/h (wind)340 km/h (still air)

36. a) v  60 km/h (wind)340 km/h Soh Cah Toa: (still air) sin  = 60 340  = sin -1 (0.19118) = 11.0 

37. a) W 11.0  340 km/h S So, the plane must head 11.0  S of W (or 79.0  W of S, 191  )

38. b) v 11.0  60 km/h 340 km/h Find actual speed (groundspeed): c 2 = a 2 + b 2 340 2 = v 2 + 60 2 v 2 = 340 2 - 60 2 v = 333.73 km/hNotice that the wind slowed the plane down.

39. b) Find time to travel 1200 km: v = d t d = v t t = d = 1200 km v333.73 km/h = 3.60 h

40. Animations Boat on Water: http://mysite.verizon.net/vzeoacw1/velocity_composition.html

41. Practice Problems Try these problems in the Physics 20 Workbook: Unit 2 p. 13 #3 - 7