2. Problems Chapter 4 6, 20, 23, 45 Chap 2Straight-line motion 4.1-4.4Specifying motion in 2 dimensions 4.5Projectile motion 4.6Range of projectile 4.8Relative motion in 1D 4.9Relative motion in 3-D 4.7Circular motion (study at home) Chap 2Straight-line motion 4.1-4.4Specifying motion in 2 dimensions 4.5Projectile motion 4.6Range of projectile 4.8Relative motion in 1D 4.9Relative motion in 3-D 4.7Circular motion (study at home)

3. Lecture notes on Web Here there is a link to “Lecture resources”, to view individual slides of the lecture The solutions to all problems are available here as well. The lectures are available on the web at: http://www.lms.unimelb.edu.au/login

4. Notices Representative for SSLC (Staff-Student Liaison Cttee) Will select next lecture Representative for SSLC (Staff-Student Liaison Cttee) Will select next lecture

5. Specifying Motion in a straight line in 2 dimensions The importance of

6. Rectilinear Motion Straight line Motion Kinematics HOW things move Dynamics Erect = straight WHY things move

7. x Distance travelled in equal time intervals

8. x Time (t) Dist (x) x x x x

9. Time (t) Dist (x) x x1x1 t1t1 x2x2 x3x3 x4x4 x5x5 t2t2 t3t3 t4t4 t5t5 xx tt Average speed speed = distance. time taken

10. Gradient of the x-t curve Time (t) Dist (x) x x1x1 x2x2 x3x3 x5x5 x6x6 t3t3 x4x4 t4t4 xx tt x4x4 t4t4 xx tt Instantaneous speed

11. If we know x(t), we can find v(t) Position (x) Position as a function of time

12. If we know x (t), we can find v(t) time Speed as a function of time

13. If we know x (t), we can find v(t) Similarly if we know v(t), we can find a(t) since time Acceleration as a function of time

14. If we know a(t), we can find v(t), since: And from v(t) we can find x(t), since From acceleration  speed  distance

15. If acceleration is constant, life is easy! VCE easy t 0 t (t) (accel) (t) speed v 0 Similarly x-x 0 = v 0 t + ½ at 2 v 2 = v 0 2 + 2a(x–x 0 ) etc. if t 0 = 0,

16. 1-D (linear kinematics) with constant acceleration 1-D (linear kinematics) with constant acceleration VCE Physics

17. 2-D Kinematics have magnitude and direction distance  displacement speed  velocity

18. Distance from Melbourne to Bendigo = 1600 km Time for travel = 20 h Displacement of Bendigo rel. to Melbourne = 200 km (NW) Time for travel = 2.5 h Average velocity = displacement/time = 200/2.5 = 80 km/h N.W. Average speed = distance/time = 1600/20 = 80 km/h Average velocity = displacement/time = 200/20 = 10 km/h N.W. Average speed= distance/time = 200/2.5 = 80 km/h

19. To think about: Exercise from last lecture To think about: Exercise from last lecture 1 km 2 km What is her average speed ? What is her average velocity ?What do you mean by velocity? If she rode directly to school, what would be her speed? Velocity? 8 kph 4 kphNorth 8 kph North

20. Y X P 1 (x 1, y 1 ) P 2 (x 2, y 2 ) y1y1 y2y2 x1x1 x2x2 r1r1 r2r2 11 22 r 1 = ix 1 + jy 1 +( kz 1 ) r 2 = ix 2 + jy 2 +( kz 2 ) rr  r = i(x 2 - x 1 ) + j(y 2 -y 1 ) +  r = i  x + j  y + yy xx  r = r 2 - r 1  r = (ix 2 + jy 2 ) -(ix 1 + jy 1 ) r 1 is the vector displacement of P 1 rel to origin r 2 is the displacement vector of P 2 rel to origin  r is the displacement vector of P 2 rel to P 1 The length of r 1 = Unit vectors Z 1 1

21. P 1 (x 1, y 1 ) P2 (x 2, y 2 ) 11 22 rr Y X y1y1 y2y2 x1x1 x2x2 r1r1 r2r2 yy xx velocity V av is in direction of  r InstantaneousAverage

22. if we know the vector displacement r(t), as a function of time, we can find the vector velocity, v(t), at any time, by differentiating r(t) Thus, for motion in 2 or 3 dimensions:

23. Similarly the instantaneous vector acceleration, a(t), is found by differentiating the function that describes the vector velocity v(t).

24. The Kangaroo story with Prof. Gari at Wedderburn, is essentially sample problem 4.3 in your text

25. Displacement of Kangaroo as function of time: = i (-0.31t 2 + 7.2t + 28) + j (0.22t 2 - 9.1t +30) Do I know r(t)? r(t) = i x(t) + j y(t) Remember r can be expressed as components in the x and y directions

26. For example at t = 15 sec r(15) = 66i -57j  87 m tan   x y  r Displacement of Kangaroo as function of time: = i (-0.31t 2 + 7.2t + 28) + j (0.22t 2 - 9.1t +30) do I know r(t)? r(t) = i x(t) + j y(t)

27. = i (-0.62t + 7.2) + j (0.44t -9.1) = i v x (t) + j v y (t) = = Velocity of Kangaroo as a function of time = ( i (-0.31t 2 + 7.2t + 28) + j (0.22t 2 - 9.1t +30)) x y v(t)?  = -130 0 Components of the velocity For example at t = 15 seconds v(t) = i (-2.1) + j (-2.5)  3.3 m s -1 Kangaroo’s velocity at time t = 15 s

28. a(t) ? = Acceleration of Kangaroo? = ( i (-0.62t + 7.2) + j(0.44t -9.1)) = i a x + j a y = 0.76 m s -2 acceleration does not depend on time! = i (-0.62) + j (0.44) x y Kangaroo’s acceleration at time t = 15 s

29. The Kangaroo’s acceleration at t = 15 s

31. Here endeth the lesson lecture No. II

32. Gallileo Parabola Max range 45 o 2 angles for any range Parabola Max range 45 o 2 angles for any range Running out of “impetus”

33. What do we know from experience? The trajectory depends on: Initial velocity Projection angle Anything else? Air resistance Ignore for now

34. Projectile Motion To specify the trajectory we need to specify every point (x,y) on the curve. That is, we need to specify the displacement vector r at any time (t). x y r(t) r(t) = i x(t) + j y(t) x(t) y(t)

35. v 0 = iv ox + j v oy -g v 0 sin  v 0 cos  Usually we know the initial vector velocity v o. We know acceleration is constant = -g. accel is a vector a = ia x + ja y = i 0 - j g = i v 0 cos  jv 0 sin  What do we know? vovo  x y

36. The vertical component of the projectile motion is the same as for a falling object The horizontal component is motion at constant velocity Finding an expression for Projectile motion

37. To define the trajectory, we need r(t) That is:- we need x(t) and y(t) v x (t) = v 0x + a x t x(t) = v 0 cos  t x(t) = v 0x t + ½a x t 2 0 since a x =0 v 0 cos  x y r vovo  Consider Horizontal motion velocity = v 0 cos  displacement -g  const Use x-x 0 = v 0 t + ½at 2

38. a y = -g Use y – y 0 = v 0 t + ½ at 2 y(t) = v 0y t + ½ a y t 2 Recall v 0 sin  x y r vovo  = v 0 sin  t - 1/2gt 2 Re-arranging gives To define the trajectory, we need r(t) That is:- we need x(t) and y(t) Consider Vertical motion

39. y = -k 1 x 2 + k 2 x x y y = kx 2

40. y = -k 1 x 2 + k 2 x x y range vovo

41. Range For what x values does y = 0? i.e. where where Maximum when 2  = 90 0. i.e when  =45 0 x=0 or x=0 R

42. R R’  = y x Gradient of slope Range up a slope Consider omitting

43. True for x = 0 or Consider omitting

44. R  R’ y x x’ y’ If m = 0 (on level ground) Consider omitting

45. y  vovo For collision, x and y of dart and monkey must be the same at instant t Time to travel distance d isd y for dart y-y 0 = v 0 t + ½ at 2 h g

46. y  h vovo d How far has monkey fallen? y for dart at impact (ie at time ) Therefore the height of the monkey is y = h - dist = v 0 monk t + ½ at 2 g

47. Isaac Newton 1642-1727

49. If it is moving with a constant velocity, IT WILL CONTINUE TO DO SO In the absence of a FORCE a body is at rest 1660 AD A body only moves if it is driven. In the absence of a FORCE A body at rest WILL REMAIN AT REST 350 BC

50. 40% Q P. R

51. Dynamics Aristotle For an object to MOVE we need a force. Newton For an object to CHANGE its motion we need a force Newtons mechanics applies for motion in an inertial frame of reference! ???????

52. Both Newton’s and Einstein’s mechanics are relativistic! Values of displacement and velocity made in different inertial reference frames are different but can be simply related. The laws of physics (essentially the forces involved) are always the same in any inertial reference frame. Reference frames that are NOT ACCELERATING Values are relative to the reference frame

53. Ref. Frame P (my seat in Plane) T (Lunch Trolley) x TP Ref. Frame G (ground) x PG x TG x TG = x TP + x PG Speed d/dt() ==> v TG = v TP + v PG Accel d/dt () ==> a TG = a TP + a PG In any inertial frame the laws of physics are the same V PG (const) Gallilean transformations 0

54. N E GROUND N’ E’ AIR r PG r AG r PA a PG = (v PG ) = a PA + 0 v AG r PG = r PA + r AG v PG = (r PG ) = v PA + v AG P Looking from above

55. AG PA AG V PG = V PA + V AG V PA = 215 km/h to East V AG = 65 km/h to North Tan  = 65/215  = 16.8 o Ground Speed of Plane

56. AG PA AG Ground Speed of Plane In order to travel due East, in what direction relative to the air must the plane travel? Do this at home and then do sample problem 4-11

57. A motorboat with its engine running at a constant rate travels across a river from Dock A to Dock B in a straight line, as shown below. Flow N A B Compare the times taken for this crossing when the river is flowing and when it is not. Your answer should include an explanation of what might affect the time taken to go from A to B, and a convincing justification of your conclusion. I will quiz you all on this next lecture