1. Speed or Velocity Time Graphs

2. Do In Notes: Sketch a d – t graph for object moving at constant speed
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 v t t

3. Constant Velocity/Speed

4. 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. v d t t

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

6. What is the average speed between 1 – 3 seconds?
1. What is the acceleration of this object?
Slope = accl. Dy/ Dx
(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?

7. 2. What is the acceleration between: 0 – 3 seconds, 5-10 seconds?
Slope = 2 m/s2. 0.

8. 3. What’s going on here?

9. Velocity Time Graphs Vector Nature sign of acceleration

10. Sign of velocity.

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

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

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

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

16. 7. What is the displacement at 10 s?
A1 = 1/2bh = 1/2(4s)(8m/s) = 16 m
A2 = bh = (6 s) (8 m/s) = 48 m
A tot = 64 m.

17. Return to start point

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

19. 9. At what time does the object return to the starting point?
At 5 seconds d = bh =(5s)(1m/s)= + 5 m.
From 5 – 10 seconds d = bh =(5s)(-1m/s)= - 5 m.
At t = 10 s.

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

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

22. d-t: v-t: a-t: slope = velocity area ≠ . slope = accl area = displ
area = D vel
vf – vi.

23. Hwk. handout “Motion Graph Prac”.

24. Objects Falling Under Gravity

25. Freefall Gravity accelerates uniformly masses as they fall and rise.
Earth’s acceleration rate is 9.81 m/s2 – very close to 10 m/s2.

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

27. But with air resistance

28. 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.

29. Apparent Weightlessness Objects in Free-fall Feel Weightless

30. What is the graph of a ball dropped?

32. 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?

33. A ball is thrown upward from the ground level returns to same height.
a is -9.81, the ball is accelerating at constant 9.81 m/s2.
d = ball’s height above the ground
velocity is + when the ball is moving upward
Why is acceleration negative?
Is there ever deceleration?

34. 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 = m/s2. Other planets g is different.

35. 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

36. 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?

37. vi = +25 m/s a = g = -9.81 m/s2. t = 5 s. d = ?
d= vit + ½ at2.
(25m/s)(5s) + 1/2(-9.81 m/s2)(5 s)2.
125 m = +2.4m.
It is 2.4m above the start point.

38. It will be below the start point.
2. If the air time from the previous problem is increased to 5.2 seconds, what will be the displacement?
d = vit + ½ at2.
-2.6 m
It will be below the start point.

39. Ex 3. A 10-kg rock is dropped from a 7- m cliff
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 = m/s2.
vf = ?
Hmmm
vi = 0.
vf 2 = vi2 + 2ad
vf 2 = 2(-9.81m/s2)(7 m)
vf = m/s (down)

40. 4. A ball is thrown straight up into the air with a velocity of 25 m/s
  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) d v (m/s) a (m/s2)

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

42. Mech Universe: The Law of Falling Bodies: