1. Acceleration due to gravity (Earth)

2. Acceleration due to gravity (Earth)
Treated as a constant near the Earth
9.81 m/sec^2 = 981 cm/sec^2

3. Displacement, velocity, and time
Units?
m/sec = m/sec

5. What does the slope mean?

6. Velocity, acceleration, and time
m/sec = m/sec-2 * sec
m/sec-2 = m/sec / sec

7. Assume constant acceleration

8. What is happening here. What does the slope mean
What is happening here? What does the slope mean? What is the linear relationship?

9. Displacement, acceleration, and time

10. Assume constant acceleration

11. What could affect this curve for a falling object?

12. What could affect the curve of displacement vs time?

13. What could affect the curve of displacement vs time?
Initial displacement?
Initial velocity?
Wind resistance?
Mass?
Size?

14. 22. An object shot straight up rises for 7
22. An object shot straight up rises for 7.0 sec before it reaches its maximum height. A second object falling from rest takes 7 sec to reach the ground. Compare the displacements of the objects.
(Start Per 7)

15. 23. Describe the changes in the velocity of a ball thrown straight up into the air. Describe the changes in the acceleration.

16. 24. The value of g on the moon is 1/6 of its value on Earth
24. The value of g on the moon is 1/6 of its value on Earth. Will a ball dropped by an astronaut hit the surface of the moon with a smaller, equal, or larger speed than that of a ball dropped the same height on Earth?

17. 26. One rock is dropped from a cliff, the other thrown upwards from the top of the cliff. They both land at the bottom of the cliff.
Which has a greater velocity at landing?
Which has a greater acceleration?
Which arrives first?

18. Given:
A ball, initially at rest, is dropped. Assuming it is near the surface of the earth, how far does it fall in 2 seconds? 4? 10? 100?
What is its final velocity?

19. d = ½ a t^2, where a = 9.8 m/sec^2 Given:
A ball, initially at rest, is dropped. Assuming it is near the surface of the earth, how far does it fall in 2 seconds? 4? 10? 100?
d = ½ a t^2, where a = 9.8 m/sec^2
What is its final velocity?

20. d = ½ * a * t^2, where a = 9.81 m/sec^2
Given:
A ball, initially at rest, is dropped. Assuming it is near the surface of the earth, how far does it fall in 2 seconds? 4? 10? 100?
d = ½ * a * t^2, where a = 9.81 m/sec^2
What is its final velocity?
V = a * t, where a = 9.81 m/sec^2 and t is found above

21. A ball falls from rest for a distance of 6m
A ball falls from rest for a distance of 6m. How far will it fall in the next 0.1sec?

22. A ball falls from rest for a distance of 6m
A ball falls from rest for a distance of 6m. How far will it fall in the next 0.1sec?
Find t from d = (1/2)at^2
Find d from d = (1/2)a(t+0.1)^2

23. Displacement, velocity, acceleration, and time

24. Optional: Displacement, velocity, and timedf d0 v t
100 10
1000 ? 15 2 12 -3 32 1
0.1 50
500 22 44 6

25. Velocity, acceleration, and time
m/sec = m/sec-2 * sec
m/sec-2 = m/sec * sec v v0 a t
200 20
2000 ? 33
330 3 15
100 2 12 -3 10 32 1
0.1 50
500 22 44 6

26. Displacement, acceleration, and time?
9.8
1000
100
980 15 12 3 32 1 50
500 22 44

27. Displacement, acceleration, velocity, and timedf di v a t ? 10
1000
0.1
333 33 3
200 20 2 5
444 12 -3
100
980

28. Position, velocity and acceleration when t is unknown.
Displacement, acceleration, and velocity
Position, velocity and acceleration when t is unknown.
vf2 = vi2 + 2 * a * d Vf vi a d
100 10 ?
1000 20 15 3 6 12 -3 32 1 50 5
500 11 22 44

29. Example 1: Calculating Distance
A stone is dropped from the top of a tall building. After 3.00 seconds of free-fall, what is the displacement, d of the stone?
Data vf
n/a vi
0 m/s d ?
a = g
-9.81 m/s2 t
3.00 s

30. Example 1: Calculating Distance
Since vi = 0 we will substitute g for a and get:
d = ½ gt2
d = ½ (-9.81 m/s2)(3.00 s)2
d = m

31. Example 2: Calculating Final Velocity
What will the final velocity of the stone be?
Data vf ? vi
0 m/s d
-44.1 m
a = g
-9.81 m/s2 t
3.00 s

32. Example 2: Calculating Final Velocity
Again, since vi = 0 we will substitute g for a and get:
vf = gt
vf = (-9.81 m/s2)(3.00 s)
vf = m/s
Or, we can also solve the problem with:
vf2 = vi2 + 2ad, where vi = 0
vf = [(2(-9.81 m/s2)(44.1 m)]1/2

33. Example 3: Determining the Maximum Height (per 6)
How high will the coin go?
Data vf
0 m/s vi
5.00 m/s d ?
a = g
-9.81 m/s2 t

34. Example 3: Determining the Maximum Height
Since we know the initial and final velocity as well as the rate of acceleration we can use:
vf2 = vi2 + 2ad
Since Δd = Δy we can algebraically rearrange the terms to solve for Δy.

35. Example 4: Determining the Total Time in the Air
How long will the coin be in the air?
Data vf
0 m/s vi
5.00 m/s d
1.28 m
a = g
-9.81 m/s2 t ?

36. Example 4: Determining the Total Time in the Air
Since we know the initial and final velocity as well as the rate of acceleration we can use:
vf = vi + aΔt, where a = g
Solving for t gives us:
Since the coin travels both up and down, this value must be doubled to get a total time of 1.02s

37. What information does the shape of the curve provide?
Straight curve = constant velocity.
Changing curve = changing velocity (i.e. acceleration).
Time
Position
Constant Velocity
Position
Time
Changing Velocity

38. Vf (m/s) Vi (m/s) d (m) a (m/s^2) t(s)
Sample Problems
A horse rounds the curve at 11m/s and accelerates to 17.3m/s. His acceleration is 1.8m/sec^2. How long does it take him to round the curve and what distance does he travel?
Vf (m/s)
Vi (m/s)
d (m)
a (m/s^2)
t(s)

39. Vf (m/s) Vi (m/s) d (m) a (m/s^2) t(s)
Sample Problems
Vf (m/s)
Vi (m/s)
d (m)
a (m/s^2)
t(s)
A car slows from 15.6m/sec to 0.9m/sec over a distance of 29m. How long does this take and at what acceleration?

40. Vf (m/s) Vi (m/s) d (m) a (m/s^2) t(s)
Sample Problems
Vf (m/s)
Vi (m/s)
d (m)
a (m/s^2)
t(s)
Natara is running a 1km race, and during the second half of the race suddenly increases her speed from 9.3m/s to 10.7m/s over a 5.3second interval. What was her acceleration and how far did she run while accelerating?

41. Vf (m/s) Vi (m/s) d (m) a (m/s^2) t(s)
Sample Problems
Vf (m/s)
Vi (m/s)
d (m)
a (m/s^2)
t(s)
Alex, while driving through the parking lot, brakes at 3m/sec^2 over a distance of 47m. What is his final velocity? For how long did he brake?

42. Vf (m/s) Vi (m/s) d (m) a (m/s^2) t(s)
Sample Problems
Vf (m/s)
Vi (m/s)
d (m)
a (m/s^2)
t(s)
A train traveling at 5.2m/sec accelerates at 2.3m/sec^2 over a 4.2sec period. What is its final velocity? How far does it travel while accelerating?

43. Vf (m/s) Vi (m/s) d (m) a (m/s^2) t(s)
Sample Problems
Vf (m/s)
Vi (m/s)
d (m)
a (m/s^2)
t(s)
Connor throws a bowling ball out the window of Planet WeirdPicture with a downward velocity of 14.9m/s. The ball falls 32m in 9.3s. What is its final velocity and the acceleration due to gravity on planet WeirdPicture?

44. Vf (m/s) 12.9 Vi (m/s) d (m) 97 a (m/s^2) 4.8 t(s)
Sample Problems
Vf (m/s)
12.9
Vi (m/s)
d (m) 97
a (m/s^2)
4.8
t(s)

45. Vf (m/s) 4.8 Vi (m/s) d (m) a (m/s^2) 4.4 t(s) 0.6
Sample Problems
Vf (m/s)
4.8
Vi (m/s)
d (m)
a (m/s^2)
4.4
t(s)
0.6

46. Vf (m/s) 4 Vi (m/s) d (m) 43 a (m/s^2) t(s) 4.5
Sample Problems
Vf (m/s) 4
Vi (m/s)
d (m) 43
a (m/s^2)
t(s)
4.5

47. Sample Problems
Vf (m/s)
17.3
15.6
10.7
12.9
4.8 4
Vi (m/s) 11
0.9
9.3
14.5
5.2
14.9
d (m) 29 47 32 97 43 54
a (m/s^2)
1.8 -3
2.3
4.4
-4.2
t(s)
5.3
4.2
0.6
4.5
7.3

48. Vi (m/s) Vf (m/s) a (m/s^2) d (m) t(s)
Sample Problems
Vi (m/s)
Vf (m/s)
a (m/s^2)
d (m)
t(s)
-4.2 54
7.3

49. What information does the shape of the curve provide?
Positive Acceleration
Negative Acceleration
Time
Position
Time
Position
Decreasing Velocity
Position
Time
Position
Time
Increasing Velocity

50. Characterize the motion of the object from A to E.
Constant velocity in the positive direction.
Decreasing velocity.
Stationary.
Increasing velocity.
Time
Position A B C D E

51. What information does the slope of the curve provide?
Positively sloped curve = movement in the positive direction.
Negatively sloped curve = movement in the negative direction.
Time
Position
Positive Direction
Time
Position
Negative Direction

52. What else does the slope tell us?
What is the motion described by each curve?
How does the velocity of each curve compare to one another?
A is stationary
B is moving at the same
speed as C, but their starting
position is different.
D is moving slower than
B or C.
Time
Position A C B D

53. What can be inferred from the following graph?

54. How do you determine the instantaneous velocity?
What is the runners velocity at t = 1.5s?
Instantaneous velocity = slope of line tangent to curve.

55. Determining the instantaneous velocity from the slope of the curve.
m = rise/run
m = 25m – 5 m
3.75s – 1.0s
m = 7.3 m/s
v = s

56. Instead of position vs. time, consider velocity vs. time.
Relatively constant
velocity
High acceleration

57. How can displacement be determined from a v vs. t graph?
Measure the area under the curve.
d = v*t
Where
t is the x component
v is the y component
Time
Velocity A2 A1
A1 = d1 = ½ v1*t1
A2 = d2 = v2*t2
dtotal = d1 + d2

58. Measuring displacement from a velocity vs. time graph.
A = b x h
A = (7.37)(11.7)
A = 86.2 m
A = ½ b x h
A = ½ (2.36)(11.7)
A = 13.8 m

59. What information does the slope of the velocity vs. time curve provide?
Positively sloped curve = increasing velocity (Speeding up).
Negatively sloped curve = decreasing velocity (Slowing down).
Horizontally sloped curve = constant velocity.
Time
Velocity
Positive Acceleration A
Time
Velocity
Negative Acceleration B
Time
Velocity
Zero Acceleration C
Velocity
Velocity
Velocity

60. Determining velocity from acceleration
If acceleration is considered constant:
a = v/t = (vf – vi)/(tf – ti)
Since ti is normally set to 0, this term can be eliminated.
Rearranging terms to solve for vf results in:
vf = vi + at
Time
Velocity
Positive Acceleration
Velocity

61. Displacement when acceleration is constant.
Displacement = area under the curve.
Δd = vit + ½ (vf – vi)*t
Simplifying:
Δd = ½ (vf + vi)*t
If the initial position, di, is not 0, then:
df = di + ½ (vf + vi)*t
By substituting vf = vi + at
df = di + ½ (vi + at + vi)*t
df = di + vit + ½ at2 vf
d = ½ (vf-vi)t
d = vit vi t