1. PHY 101: Lecture 2 2.1 Displacement 2.2 Speed and Velocity
2.3 Acceleration
2.4 Equations of Kinematics for Constant Acceleration
2.5 Applications of the Equations of Kinematics
2.6 Freely Falling Bodies
2.7 Graphical Analysis of Velocity and Acceleration

2. PHY 101: Lecture 2 Kinematics in One Dimension
2.1
Displacement

3. Kinematics Definition of Distance
Type of Quantity: Scalar
Total path length traversed in moving from one position to another
Distance depends on path and not on starting and ending points
Symbol: d
SI Unit: meter (m)

4. Kinematics Distance - Example
What is the distance for each of the following trips?
1. 3 m east, 4 m east
2. 3 m east, 4 m north
3. 3 m east, 4 m west
For each trip, the distance is 7 m

5. Definition of Initial PositionXi
Definition of Initial Position
Type of Quantity: Vector
Initial position of an object is indicted by a position vector, xi, from the origin to the position xi
Magnitude of xi is distance from the origin to position xi
Direction of xi is either + or -
Symbol: xi
SI Unit: meter (m)
Note: Subscript f means the final position, xf xi xi

6. 2-Dimensional Example Position / Free Vectors

7. Definition of Displacement
Type of Quantity: Vector
Vector drawn from initial position to final position
xi + Dx = xf
Dx = xf - xi
Magnitude equals shortest distance between initial and final positions
Direction points from initial to final position
Displacement depends on the initial and final positions and not on path length
Symbol: Dx
SI Unit: meter (m)
D is always final value minus initial value Xi DX Xf

8. Distance / Displacement

9. Direction of Displacement
Vector nature of displacement is giving by
+ sign to indicate displacement in the +x direction, to the right, or east
- sign to indicate displacement in the –x direction, to the left, or west
Note: The same is true for position, velocity, and acceleration

10. Example of + / - Sign 500 meters in the +x direction is +500 meters

11. Distance vs. Displacement
An object can move so that the path length (distance) is large but the displacement is zero or small

12. Displacement – Example 1
What is the displacement for each of the following situations?
1. 3 m east, 4 m east
Displacement is 7 m east
2. 3 m east, 4 m north
Displacement is 5 m at 530 above +x axis
3. 3 m east, 4 m west
Displacement is 1 m west

13. Displacement – Example 2
A student throws a rock straight upward from shoulder level, which is 1.65 m above the ground
What is the displacement of the rock when it hits the ground?
Displacement is a vector that points from the initial position to the final position
Initial position is 1.65 m above the ground
Final position is the ground, height 0 m
Magnitude of the displacement vector is then 1.65 m
Direction of the displacement vector is downward

14. PHY 101: Lecture 2 Kinematics in One Dimension
2.2
Speed and Average Velocity

15. Definition of Average Speed
Type of Quantity: Scalar
Average Speed = Distance / Time
savg = d / t
Symbol: savg
SI Unit: meter/s (m/s)

16. What Does Constant Speed Mean?
Ball has initial position xi = 0 m at time ti = 0
Ball has constant velocity t = 5 m/s
This means that during:
1st second ball moves 5 m
2nd second ball moves 5 m
3rd second ball moves 5 m, and so on …
Position at time t is:
t = 0 s, x = 0 m
t = 1 s, x = 5 m
t = 2 s, x = 10 m
t = 3 s, x = 15 m

17. Average Speed – Example 1
A race car circles 10 times around an 8-km track in 1200 s
What is its average speed in m/s?
distance traveled is d = 10 x 8 km = 80 km = 80,000 m
t = 1200 s
savg = d/t = / 1200 s = 66.7 m/s

18. Average Speed – Example 2
A motorist drives 150 km from one city to another in 2.5 h, but makes the return trip in only 2.0 h
What are the average speeds for
(a) each half of the round-trip, and
(b) the total trip?
(a)
First half of trip
average speed = d/t = 150 km/2.5 h = 60 km/h
Second half of trip
average speed = d/t = 150 km/2 h = 75 km/h
(b)
Entire trip
average speed is d/t = ( ) / ( ) = km/h

19. Definition of Average Velocity
Type of Quantity: Vector
Average Velocity = Displacement / Time
vavg = Dx/t = (xf – xi)/t
Direction is the same as the direction of displacement
Symbol: vavg
SI Unit: meter/second (m/s)

20. Average Velocity – Example 1
A race car circles 10 times around an 8-km track in 1200 s
What is its average velocity in m/s
After 10 laps, the ending position is the same as the starting position
Therefore, displacement is zero, even though distance is 80,000 m
Since displacement is zero, average velocity is zero.

21. Average Velocity – Example 2
Car travels half a lap in 3 seconds along a circle with radius = 150 m
(a) What is the distance traveled by the car?
Distance is the path length = half the circumference of the circle
d = 2pr/2 = p(150) = 471 m
(b) What is the magnitude of the car’s displacement?
Displacement is the vector from the initial position to the final position
Vector is across the diameter of the circle
Magnitude of displacement is 2 x 150 = 300 m
(c) What is the average speed of the car?
Average speed s = distance / time = m / 3 s = 157 m/s
(d) What is the magnitude of the average velocity of the car?
Average velocity = displacement / time = 300 m / 3 s = 100 m/s

22. Average Velocity – Example 3
A car travels a full lap in 5 seconds along a circle that has a radius of 150 m?
(a) What is the distance traveled by the car?
Distance is the path length = circumference of the circle
d = 2pr = 2p(150) = 942 m
(b) What is the magnitude of the car’s displacement?
Displacement is vector from the initial position to the final position
The initial and final positions are at the same location
Magnitude of displacement vector is zero.
(c) What is the average speed of the car?
Average speed s = distance / time = m / 5 s = 189 m/s
(d) What is magnitude of the average velocity of car?
Average velocity = displacement / time = 0 m / 5 s = 0 m/s

23. PHY 101: Lecture 2 Kinematics in One Dimension
2.3
Acceleration

24. Definition of Average Acceleration
Type of Quantity: Vector
Acceleration = Change in Velocity / Time
aavg = Dv/t = (vf – vi)/t
Symbol: aavg
SI Unit: meter/second2 (m/s2)
Note: When velocity and acceleration have opposite direction (signs), the object is decelerating

25. Average Acceleration – Example 1
An automobile traveling at 8 m/s along a straight, level road accelerates to 20 m/s in 6.00 s
What is the magnitude of the auto’s average acceleration?
average acceleration =
(vf – vi)/t = (20 – 8) / 6 = 2 m/s2

26. Average Acceleration – Example 2
Motorcycle has a constant acceleration of 2.5 m/s2
Both the velocity and acceleration of the motorcycle point in the same direction
How much time is required for the motorcycle to change its velocity from 21 to 31 m/s.
aavg = (vf – vi) / t
2.5 m/s2 = (31 m/s – 21 m/s) / t
t = 10 / 2.5 = 4 s

27. PHY 101: Lecture 2 Kinematics in One Dimension
2.4
Equations of Kinematics for Constant Acceleration

28. Equation of Motion Equations that give x, v, and a as functions of t
x = f(t)
v = g(t)
a = h(t)
We will look at a special case where a is constant
a = aavg = constant independent of time
We can also eliminate t from x and v to obtain
v = j(x)

29. Development of Equations of Motion
a = aavg = (vf – vi) / t
vf = vi + at (constant acceleration)
vavg = (xf – xi) / t
xf = xi + vavgt
vavg = ½ (vf + vi) (constant acceleration)

30. Equations of Motion vf = vi + at xf = xi + vit + (1/2)at2
vf2 = vi2 + 2a(xf – xi)
These are vector equations
xi, xf, vi, vf, a, and t can be + or -

31. Scalar vs. Vector Magnitude
Scalar can be negative or positive
For example, temperature, work, and energy can be negative or positive
Magnitude of a vector is always positive
For example, displacement, velocity, and acceleration magnitude is always positive
Negative sign on a magnitude is an indication that the vector points in the negative (opposite) direction

32. PHY 101: Lecture 2 Kinematics in One Dimension
2.5
Applications of the Equations of Kinematics

33. Equations of Motion – Example 1
Basketball player starts from rest and accelerates uniformly to speed 6.0 m/s in 1.5 s
What distance does the player run?
vf = vi + at
xf = xi + vit + ½at2
vf2 = vi2 + 2a(xf – xi)
xi = 0 xf = ?
vi = 0 vf = 6.0
a = ? t = 1.5
6=0+a(1.5) Equation has a (1st)
xf=0+0(1.5)+½a(1.5)2 Equation has 2 unknowns, xf, a (2nd)
62=02+2a(xf–0) Equation has two unknowns, xf, a
6=0+a(1.5); a = 6/1.5 = 4 m/s2
xf=0+0(1.5)+½(4)(1.5)2 = 4.5 m/s

34. Equations of Motion–Example 2a
Car accelerates from rest at rate of 2.0 m/s2 for 5.0 s
(a) What is the speed of the car at the end of that time?
vf = vi + at
xf = xi + vit + ½at2
vf2 = vi2 + 2a(xf – xi)
xi = 0 xf = ?
vi = 0 vf = ?
a = 2.0 t = 5.0
vf=0+2(5) Equation has vf
xf=0+0(5)+½(2)(5)2 Equation has xf
vf2=02+2(2)(xf–0) Equation has two unknowns, xf, vf
vf=0+2(5)=10 m/s

35. Equation of Motion – Example 2b
Car accelerates from rest at rate of 2.0 m/s2 for 5.0 s
(b) How far does the car travel in this time?
vf = vi + at
xf = xi + vit + ½at2
vf2 = vi2 + 2a(xf – xi)
xi = 0 xf = ?
vi = 0 vf = 10
a = 2.0 t = 5.0
10=0+2(5) Can’t use Equation
xf=0+0(5)+½(2)(5)2 Equation has xf
102=02+2(2)(xf–0) Equation has xf
xf=0+0(5)+½(2)(5)2 = 25 m

36. Equation of Motion – Example 3a
A car traveling at 15 m/s stops in 35 m
(a) What is the acceleration?
vf = vi + at
xf = xi + vit + ½at2
vf2 = vi2 + 2a(xf – xi)
xi = 0 xf = 35
vi = 15 vf = 0
a = ? t = ?
0=15+at Equation has a, t
35=0+15t+½(a)(t)2 Equation has a, t
02=152+2a(35–0) Equation has a
02=152+2a(35–0)
a = -225/70 = -3.2 m/s2

37. Motion – Example 3b A car traveling at 15 m/s stops in 35 m
(b) What is time during this deceleration until car stops?
vf = vi + at
xf = xi + vit + ½at2
vf2 = vi2 + 2a(xf – xi)
xi = 0 xf = 35
vi = 15 vf = 0
a = -3.2 t = ?
0=15-3.2t Equation has t
35=0+15t+½(-3.2)(t)2 Equation has t
02=152+2(-3.2)(35–0) Can’t use equation
0=15-3.2t
t = -15/-3.2 = 4.67 s

38. Equation of Motion – Example 4
A plane accelerates at 8 m/s2 on a runway that is 500 m long
The take off speed of the plane is 80 m/sec
Can the plane takeoff?
vf = vi + at
xf = xi + vit + ½at2
vf2 = vi2 + 2a(xf – xi)
xi = 0 xf = 500
vi = 0 vf = ?
a = 8 t = ?
vf=0+8t Equation has vf, t
500=0+0t+½(8)(t)2 Equation has t
vf2=02+2(8)(500–0) Equation has vf
vf2=02+2(8)(500–0) = 500(16)
vf=sqrt(500(16) = 89.4 m/s
The plane can takeoff

39. Equation of Motion – Example 5
A plane accelerates at 8 m/s2
The take off speed of the plane is 80 m/sec
What is minimum length of run way for plane to reach take off speed?
vf = vi + at
xf = xi + vit + ½at2
vf2 = vi2 + 2a(xf – xi)
xi = 0 xf = ?
vi = 0 vf = 80
a = 8 t = ?
80=0+8t Equation has t
xf=0+0t+½(8)(t)2 Equation has xf, t
802=02+2(8)(xf–0) Equation has xf
802=02+2(8)(xf–0)
xf=6400/16 = 400 m

40. Eq. of Motion – Example 6 Bull
Bull runs 8 m/sec
A boy at rest has a head start of 12 m
When he sees the bull he accelerates at 2 m/s2
Does the bull catch the boy?
vf = vi + at
xf = xi + vit + ½at2
vf2 = vi2 + 2a(xf – xi)
xi = 0 xf = ?
vi = 8 vf = 8
a = 0 t = ?
8=8+0t Can’t use this equation
xf=0+8t+½(0)(t)2 Equation has xf, t
82=82+2(0)(xf–0) Can’t use this equation

41. Eq. of Motion – Example 6 Boy
Bull runs 8 m/sec
A boy at rest has a head start of 12 m
When he sees the bull he accelerates at 2 m/s2
Does the bull catch the boy?
vf = vi + at
xf = xi + vit + ½at2
vf2 = vi2 + 2a(xf – xi)
xi = 12 xf = ?
vi = 0 vf = ?
a = 2 t = ?
vf=0+2t Equation has vf, t
xf=12+0t+½(2)(t)2 Equation has xf, t
vf2=02+2(2)(xf–12) Equation has xf, vf

42. Example 6 Bull / Boy Bull runs 8 m/sec
A boy at rest has a head start of 12m
When he sees the bull he accelerates at 2 m/s2
Does the bull catch the boy?
xf=0+8t+½(0)(t)2 (from Bull equations)
xf=12+0t+½(2)(t)2 (from Boy equations)
They catch up to each other when the boy xf = bull xf
8t=12+t2
t2-8t+12=0
Factor
(t-2)(t-6) = 0
Bull catches boy at 2 s and 6 s
2 s is bull catching up to the boy
6 s is the accelerating boy catching up to the bull

43. PHY 101: Lecture 2 Kinematics in One Dimension
2.6
Freely Falling Bodies

44. Free Fall Objects falling vertically Near the surface of the earth
Acceleration due to gravity
a = -9.8 m/s2

45. Free Fall – Example 1 A bomb is dropped from 6000m
With what speed does it hit the ground?
vf = vi + at
yf = yi + vit + ½at2
vf2 = vi2 + 2a(yf – yi)
yi = 6000 yf = 0
vi = 0 vf = ?
a = t = ?
vf = 0 – 9.8t Equation has vf, t
0 = t + ½ (-9.8)t2 Equation has t
vf2 = (-9.8) (0 – 6000) Equation has vf
vf2 = (-9.8)(-6000)
vf = sqrt(117600) = -343 m/s

46. Free Fall – Example 2a
A student drops a ball from the top of a tall building
It takes 2.8 s for the ball to reach the ground
(a) What is the height of the building?
vf = vi + at
yf = yi + vit + ½at2
vf2 = vi2 + 2a(yf – yi)
yi = ? yf = 0
vi = 0 vf = ?
a = -9.8 t = 2.8
vf = 0 – 9.8(2.8) Equation has vf
0 = yi + 0(2.8) + ½(-9.8)(2.8)2 Equation has yi
vf2 = (-9.8) (0 – yi) Equation has yi, vf
0 = yi + (1/2)(-9.8)(2.8)2
yi = 38.4 m

47. Free Fall – Example 2b
A student drops a ball from the top of a tall building
It takes 2.8 s for the ball to reach the ground
(b) What was the ball’s velocity just before hitting the ground?
vf = vi + at
yf = yi + vit + ½at2
vf2 = vi2 + 2a(yf – yi)
yi = yf = 0
vi = 0 vf = ?
a = t = 2.8
vf = 0 – 9.8(2.8) Equation has vf
0 = (2.8) + ½(-9.8)(2.8)2 Can’t use equation
vf2 = (-9.8) (0 – 38.4) Equation has vf
vf = 0 – 9.8(2.8) = m/s (- means downward)

48. Free Fall – Example 3
Boy throws stone straight upward with an initial velocity of 15 m/s
What maximum height will stone reach before falling back down?
At the maximum height vy = 0
vf = vi + at
yf = yi + vit + ½at2
vf2 = vi2 + 2a(yf – yi)
yi = 0 yf = ?
vi = 15 vf = 0
a = t = ?
0 = 15 – 9.8t Equation has t
yf = t + ½(-9.8)t2 Equation has yf, t
02 = (-9.8) (yf – 0) Equation has yf
02 = (-9.8) (yf – 0)
yf = -152/2/-9.8 = m

49. Free Fall – Example 4
Ball thrown upwards at 40m/s. Calculate time to reach 35m
vf = vi + at
yf = yi + vit + ½at2
vf2 = vi2 + 2a(yf – yi)
yi = 0 yf = 35
vi = 40 vf = ?
a = t = ?
vf = 40 – 9.8t Equation has vf, t (2nd)
35 = t + ½(-9.8)t2 Equation has t
vf2 = (-9.8) (35 – 0) Equation has vf (1st)
vf2 = (-9.8) (35 – 0) = 914
vf = or -30.2
30.2 = 40 – 9.8t
t = (30.2 – 40) / (-9.8) = 1.0 s (time on the way up)

50. Free Fall – Example 5 Ball 1
A ball is dropped from rest
Four seconds later, a second ball is thrown down at 50m/s Calculate when and where the two balls meet
vf = vi + at
yf = yi + vit + ½at2
vf2 = vi2 + 2a(yf – yi)
yi = 0 yf = ?
vi = 0 vf = ?
a = t = ?
vf = 0 – 9.8t Equation has vf, t
yf = 0 + 0t + ½(-9.8)t2 Equation has yf, t
vf2 = (-9.8) (yf – 0) Equation has yf, vf

51. Free Fall – Example 5 Ball 2
A ball is dropped from rest
Four seconds later, a second ball is thrown down at 50m/s Calculate when and where the two balls meet
vf = vi + at
yf = yi + vit + ½at2
vf2 = vi2 + 2a(yf – yi)
yi = 0 yf = ?
vi = -50 vf = ?
a = t = t – 4 ?
vf = -50–9.8(t-4) Equation has vf, t
yf = 0-50(t-4)+½(-9.8)(t-4)2 Equation has yf, t
vf2 = 502+2(-9.8) (yf – 0) Equation has yf, vf

52. Free Fall – Example 5 Balls 1 and 2
A ball is dropped from rest
Four seconds later, a second ball is thrown down at 50m/s
Calculate when and where two balls meet
yf = 0 + 0t + ½(-9.8)t2 (from ball 1)
yf = 0+50(t-4)+½(-9.8)(t-4)2 (from ball 2)
Substitute equation 1 into equation 2
-4.9t2 = -50(t-4) – 4.9(t-4)2
-4.9t2 = -50t – 4.9t t – 78.4
-10.6t = -0
t = / = 11.4 s

53. Free Fall – Example 6a
From tower 100 m high, ball is thrown up with speed of 40 m/s
(a) How high does it rise?
vf = vi + at
yf = yi + vit + ½at2
vf2 = vi2 + 2a(yf – yi)
yi = 100 yf = ?
vi = 40 vf = 0
a = t = ?
0 = 40 – 9.8t Equation has t
yf = t + ½(-9.8)t2 Equation has yf, t
02 = 402+2(-9.8)(yf–100) Equation has yf
02 = 402+2(-9.8)(yf–100)
yf = (-1600/2/-9.8) = m

54. Free Fall – Example 6b
From tower 100 m high, ball is thrown up with speed of 40 m/s
(b) When does it hit ground?
vf = vi + at
yf = yi + vit + ½at2
vf2 = vi2 + 2a(yf – yi)
yi = 100 yf = 0
vi = 40 vf = ?
a = t = ?
vf = 40 – 9.8t Equation has vf, t
0 = t + ½(-9.8)t2 Equation has t
vf2 = 402+2(-9.8)(0–100) Equation has vf
0 = t + ½(-9.8)t2
-4.9t2 + 40t = 0 (Use quadratic equation)
t = 10.2 s

55. Free Fall – Example 6c
From tower 100 m high, ball is thrown up with speed of 40 m/s
(c) How fast is it moving at the ground ?
vf = vi + at
yf = yi + vit + ½at2
vf2 = vi2 + 2a(yf – yi)
yi = 100 yf = 0
vi = 40 vf = ?
a = t = 10.2
vf = 40 – 9.8(10.2) Equation has vf
0 = (10.2)+½(-9.8)(10.2)2 Can’t use equation
vf2 = 402+2(-9.8)(0–100) Equation has vf
vf2 = 402+2(-9.8)(0–100) = = 3560
vf = m/s

56. PHY 101: Lecture 2 Kinematics in One Dimension
2.7
Graphical Analysis of Velocity and Acceleration

57. Graphical Analysis 1 Velocity = 0 Acceleration = 0
Horizontal line on P-T graph: velocity = 0
Time 1 2 3 4 5
Pos

58. Graphical Analysis 2 Velocity = 1 Acceleration = 0
Sloping line on P-T graph: velocity <> 0
Time 1 2 3 4 5
Pos

59. Graphical Analysis 3 Velocity = 2 Acceleration = 0
Steeper slope means greater velocity
Time 1 2 3 4 5
Pos 6 8 10

60. Graphical Analysis 4 Velocity = -1 Acceleration = 0
Negative slope means velocity in opposite direction
Time 1 2 3 4 5
Pos -1 -2 -3 -4 -5

61. Graphical Analysis 5 Velocity = 1 Acceleration = 1
Curving line on P-T graph: acceleration <> 0
Time 1 2 3 4 5
Pos
1.5
7.5 12
17.5

62. Speed vs Average Speed (Charts)
Time (min)
Position (m)
Speed (m/min) 1
400
0.5 2
1200
1.5
800 3
2700
2.5
1500 4
2900
3.5
200 5
4.5 6
4000
5.5
1100 7
5500
6.5 8
7200
7.5
1700 9
7300
8.5
100

63. Speed vs. Avg. Speed (Graphs)

64. Speed vs. Average Speed Prior charts & graphs show trip of car
During trip the velocity changes
Average velocity remains constant during the trip
Average velocity produces same final position as does the varying velocity