1. Speed or Velocity Time Graphs

2. Do In Notes: Sketch a d – t graph for object moving at constant speed. Now sketch a speed time graph showing the same motion. d t v t

3. Constant Velocity/Speed

4. What is an object is uniformly speeding up?

5. Sketch a distance – t graph for object starting from rest and speeding up with constant acceleration. Now sketch a speed time graph showing the same motion. d t v t

6. Constant / Uniform Acceleration. Slope = Constant Acceleration of straight line on v-t graph. Average Velocity is the midpoint between 2 speeds. v f + v i 2

7. What is the acceleration of this object? Slope = accl.  y/  x (50 – 10) m/s (5 – 1) s What is the sign of accl? What is the average speed between 1 – 3 seconds? What is the average speed between 3 – 5 seconds?

8. . What is the acceleration between: 0 – 3 seconds, 5-10 seconds? Slope = 2 m/s 2. 0.

9. What’s going on here?

10. Velocity Time Graphs Vector Nature sign of acceleration

11. Sign of velocity.

13. Finding Distance or Displacement on V-T graphs Speed Time = distance Velocity Time = displacement

14. Displacement = Area Under Curve at specific time. Drop a vertical to the X axis. A = bh = (20s)(30m/s) What is the displacement at 20 sec?

15. 2. What is the displacement at 5 s? Area = bh = (5 s) x (1 m/s) = 5 m.

16. 3: What is the displacement at 4 seconds? A = ½ bh = ½ (4s)(40 m/s) = 80 m.

17. 4. What is the displacement at 10 s? A 1 = 1/2bh = 1/2(4s)(8m/s) = 16 m A 2 = bh = (6 s) (8 m/s) = 48 m A tot = 64 m.

18. Return to start point

19. How can you tell when object is back to starting point? Positive displacement = negative displacement. Tot displacement = 0.

20. 5. At what time does the object return to the starting point? At 5 seconds d = 5 m. From 5 – 10 seconds d = - 5 m. At t = 10 s.

22. Given the v – t graph below, sketch the acceleration – t graph for the same motion.

23. Acceleration – time Graphs What is the physical behavior of the object? Slowing down pos direction, constant vel neg accel.

24. d-t: –slope = velocity – area ≠. v-t: –slope = accl – area = displ a-t: – slope ≠. – area =  vel –v f – v i.

25. Hwk. Text do pg 70 #17, 30 and packet “Motion Graph Prac”. In class pg 59 #5.

27. Objects Falling Under Gravity

28. Freefall Gravity accelerates uniformly masses as they fall and rise. Earth’s acceleration rate is 9.81 m/s 2 – very close to 10 m/s 2.

29. Falling objects accelerate at the same rate in absence of air resistance

30. But with air resistance

31. Fortunately there is a “terminal fall velocity.” After a while, the diver falls with constant velocity due to air resistance. Unfortunately terminal fall velocity is too large to live through the drop.

32. Apparent Weightlessness Objects in Free-fall Feel Weightless

33. What is the graph of a ball dropped?

35. What do the d-t, v-t, and a – t, graphs of a ball thrown into the air look like if it is caught at the same height?

36. A ball is thrown upward from the ground level returns to same height. d = ball’s height above the ground velocity is + when the ball is moving upward Why is acceleration negative? a is -9.81, the ball is accelerating at constant 9.81 m/s 2. Is there ever deceleration?

37. Free-fall Assumptions Trip only in the air. Trip ends before ball caught. -Symmetrical Trip time up = time down -Top of arc: v = 0, a = ?? -On Earth g = -9.81 m/s 2. Other planets g is different.

38. Solving: Use accl equations replace a with -g. List given quantities & unknown quantity. Choose accl equation that includes known & 1 unknown quantity. Be consistent with units & signs. Check that the answer seems reasonable Remain calm

39. Practice Problem. 1. A ball is tossed upward into the air from the edge of a cliff with a velocity of 25 m/s. It stays airborne for 5 seconds. What is its total displacement?

40. v i = +25 m/s a = g = -9.81 m/s 2. t = 5 s. d = ? d= v i t + ½ at 2. (25m/s)(5s) + 1/2(-9.81 m/s 2 )(5 s) 2. 125 m- 122. 6 = +2.4m. It is 2.4m above the start point.

41. 2. If the air time from the previous problem is increased to 5.2 seconds, what will be the displacement? d = v i t + ½ at 2. -2.6 m It will be below the start point.

42. Ex 3. A 10-kg rock is dropped from a 7- m cliff. What is its velocity just before hitting the ground? d = 7m a = -9.81 m/s 2. v f = ? Hmmm v i = 0. v f 2 = v i 2 + 2ad v f 2 = 2(-9.81m/s 2 )(7 m) v f = -11.7 m/s (down)

43. 4. A ball is thrown straight up into the air with a velocity of 25 m/s. Create a table showing the balls position, velocity, and acceleration for each second for the first 5 seconds of its motion. T(s)dv (m/s)a (m/s 2 ) 0025-9.81 12015.2-9.81 2305.4-9.81 331-4.43-9.81 52.4-24-9.91

44. Read Text pg 60-64 Do prb’s pg 64# 1-5 show work.

45. Mech Universe: The Law of Falling Bodies: http://www.learner.org/resources/series42.html?pop=yes&pid=549#