1. KINEMATICS in One Dimension
REPRESENTING MOTION
KINEMATICS in One Dimension
“To understand motion is to understand nature.”
Leonardo da Vinci

2. Study of motion, force and energy
MECHANICS
Study of motion, force and energy
Kinematics
How
objects move
Dynamics
Why
objects move

3. Kinematics Objectives
Represent motion through the use of words, motion diagrams, graphs, and mathematical models.
Use the terms position, distance, displacement, and time interval in a scientific manner to describe motion. 3

4. Motion Motion is instinctive
Eyes will notice moving objects more readily than stationary ones
Object changes position
Motion can occur in many directions and paths

5. Representing Motion A description of motion relates PLACE and TIME.
Answers the questions WHERE? and WHEN?
PLACE
TIME

6. Motion Diagram & Particle Model
Simplified version of a motion diagram in which the object in motion is replaced by a series of single points
Size of object must be much less than the distance it moves

7. Describe motion of the car…
Draw a particle model…

8. How are the two particle models different? Describe the motion of each.

9. Reference Frames
Any measurement of position, distance or speed must be made with respect to a frame of reference
80 km/h

10. Coordinate System
Tells you the location of the zero point of the variable you are studying and the direction in which the values of the variable increase.
ORIGIN
The point at which both variables have the value zero

11. Distance and Displacement
Distance, d – total ground covered
Displacement, Dx –
change in position of an object (position is measured from the origin of a chosen coordinate system)

12. Distance and Displacement
Example - A car travels 400 km from Livingston to Philadelphia and then back 200 km to Trenton. What is the displacement of the car? What distance did the car travel?
X =
100
200
300
400 km
Livingston
Trenton
Philadelphia
displacement
distance

13. Distance and Displacement
Example - A car travels 400 km from Livingston to Philadelphia and then back 200 km to Trenton. What is the displacement of the car? What distance did the car travel?
X =
100
200
300
400 km
Livingston
Trenton
Philadelphia
displacement
distance

14. Displacement Has magnitude (size) and direction. It is a VECTOR
Vectors are represented by arrows
x(m) 1 2 3 -1

15. Has magnitude and direction. It is a VECTOR
Displacement
Has magnitude and direction. It is a VECTOR
x(m) 1 2 3 -1 15

16. Average Speed and Average Velocity
Physics C 1-D Motion
Average Speed and Average Velocity
Average speed describes how fast a particle is moving. It is calculated by:
Average velocity describes how fast the displacement is changing with respect to time:
always positive
Scalar (has magnitude
only)
sign gives direction in 1 Dimension
Vector (has magnitude
and direction)
Average speed and average velocity often have the same magnitude, but not always
Bertrand

17. Example - A car travels 400 km from Philadelphia to Livingston in 2 hours and then back 200 km to Trenton in 1 hr. What is the car’s speed and velocity? Dx
X =
100
200
300
400 km
Livingston
Trenton
Philadelphia
Average speed
Average velocity

18. Example
The position-time graph shows the progress of two runners, A and B.
When does runner B pass runner A?
Where does runner B pass runner A?
What is the starting position for runner A? runner B?
After 10 hrs, what is the average velocity of runner A? After 10 hrs, what is the average velocity of runner B?
If the finish line is at 40 km, who won the race? 18

19. Analyzing Graphs UNITS are the key to analyzing graphs
When analyzing graphs always check for the following two things:
Slope: Look at the units of the slope to see if it corresponds to a physically meaningful measurement.
Area under the curve: look at the units for the area under the curve to see if it corresponds to a measurement.

20. Graphical Representation of Motion
Physics C 1-D Motion
Graphical Representation of Motion
Position-Time Graph A B x Dx Dt
slope t
Steepness = speed
slope
Sign = direction
Velocity = speed + direction
Bertrand

21. Average Velocity from a Graph
Physics C 1-D Motion
Average Velocity from a Graph t x A B Dx Dt
Mathematical Model
x = position
x0 = initial position
v = average velocity
t = time
Bertrand

22. Graphs of Motion Mathematical Model x0 = 20 m Slope v = 2 m/s
Physics C 1-D Motion
Graphs of Motion
Mathematical Model
Dx =20m
Dx =20m
x0 = 20 m
v = 2 m/s
Slope
SLOPE
AREA
Mathematical Model
(UNIFORM VELOCITY)
Area=20m = Dx
Area
(init and final positions unknown. ONLY KNOW DISPLACEMENT)
Bertrand 22

23. What is happening in this graph?
Mathematical
Graphical
Verbal: For every second that passes the buggy
travels +0.4 meter every second from an initial
position of 0.5 meter in the positive x direction.
NEGATIVE SPEED???
End t=6 s 5s 4s 3s 2s 1s
t=0 s START
X =
-10 -5 5 10 15
20 m
Diagrammatic
v = 23

24. Plot the corresponding v-t graph
x0 = 20 m
v = -5 m/s
Slope
SLOPE 24

25. What is happening?
Follow pos – goes away from starting point then at C comes closer
Always at same speed?
Moving away from starting point to A
Slows down to B
Speeds up to C
Reverses direction and comes closer to start with

26. What is happening in each?D. B.

27. Draw the corresponding v-t graph v-t
Physics C 1-D Motion
Draw the corresponding v-t graph
SLOPE
v-t
Bertrand

28. F=D=E>A>B>C C
Physics C 1-D Motion B A
Which regions shows positive displacement? negative? When is the object moving in the + direction? When is the object moving in the – direction? Which region is the object moving with maximum + velocity? Rank speeds from greatest to least When is the object at rest When does the object change direction?
A,F
B,D,E
A,F
B,D,E F
F=D=E>A>B>C C
At 2s, at 7 s
Bertrand

29. Physics C 1-D Motion B A
What is the total displacement for the 8 sec? What is the average velocity for the entire 8 sec trip? What is the distance traveled in the 8 sec trip? What is the average speed for the 8 sec trip? 0m
0m/s
32m
4m/s
Bertrand

30. Problem: A car starting from rest moves with an average speed of 6 m/s
Problem: A car starting from rest moves with an average speed of 6 m/s. Calculate the distance the car traveled in 1 minute.
GIVEN: s = 6 m/s
t = 1 min =60 s
UNKNOWN: d = ? m
FORMULA: s = d / t
SUBSTITUTION: 6 m/s = d / 60 s
SOLUTION d = 360 m

31. Problem: An object moves from the position +47 m to the position +15 m in 8 s. What is its total displacement? What is its average velocity?
GIVEN: xo = 47m xf = 15m
t = 8 s
UNKNOWN: Dx = ? m
FORMULA: Dx=xf-xo
SUBSTITUTION: Dx= 15-47
SOLUTION Dx = -32 m
GIVEN: xo = 47m xf = 15m Dx = -32 m t = 8 s
UNKNOWN: vav = ? m/s
FORMULA:
SUBSTITUTION:
SOLUTION

32. v-t Draw the corresponding x-t graph
Physics C 1-D Motion
AREA
v-t
Dx=
+12m
Draw the corresponding x-t graph 3m
Area= Dx
-6m
Dx=
-16m
Bertrand

33. The Meaning of Shape of a Position-Time graph
Contrast a constant and changing velocity
Contrast a slow and fast moving object

34. Average Velocity and Instantaneous Velocity
Average velocity only depends on the initial and final positions. These 2 cars have the same average velocities but different velocities at each instant. When the velocity is not uniform, the instantaneous velocity is not the same as the average velocity.
AVERAGE VELOCITY:
X =
-40
-20 20 40 60
80 km
X =
-40
-20 20 40 60
80 km
Start t=0
End t=2 hr

35. Physics C 1-D Motion
Acceleration
Average acceleration describes how quickly or slowly the velocity changes. It is calculated by:
Vector
SI units: m/s2
Instantaneous acceleration describes how the velocity changes over a very short time interval:
Acceleration tells us how fast the velocity changes.
Velocity tells us how fast the position changes.
Bertrand

36. x v a Average acceleration
Example - A car accelerates along a straight road from rest to 24 m/s in 6.0 s. What is the average acceleration?
Average acceleration
START 1s 2s 3s 4s 5s x v 4 8 12 16
20m/s a
4m/s2

37. x x v a Position-Time Graph t=0 t=5 t=4 t=3 t=2 t=1 1s 2s 3s 4s 5s
START 1s 2s 3s 4s 5s x

38. x v a x v a Speeding up in + direction a and v SAME direction
START 1s 2s 3s 4s 5s x v a
Slowing down in + direction
a and v OPP direction
START 1s 2s 3s 4s 5s x v a

39. x v a x v a Speeding up in - direction a and v SAME direction
START 5s 4s 3s 2s 1s x v a
Slowing up in - direction
a and v OPP direction
START 5s 4s 3s 2s 1s x v a

40. Displacement and velocity are in the direction of motion
Physics C 1-D Motion
Displacement and velocity are in the direction of motion
When acceleration is in the SAME direction as velocity, the object is speeding up
When acceleration is in the OPPOSITE direction to velocity, the object is slowing down
Bertrand

41. x x v a Speeding up in + direction t=0 t=5 t=4 t=3 t=2 t=1 1s 2s 3s 4s
Position-Time Graph
START 1s 2s 3s 4s 5s x

42. x Slowing down in + direction x v a t=0 t=5 t=3 t=2 t=1 t=4 1s 2s 3s
Position-Time Graph
START 1s 2s 3s 4s 5s x

43. x Speeding up in - direction x v a t=5 t=0 t=1 t=2 t=3 t=4 5s 4s 3s 2s
START 5s 4s 3s 2s 1s x

44. x Slowing down in - direction x v a t=0 t=1 t=2 t=3 t=4 t=5 5s 4s 3s
START 5s 4s 3s 2s 1s x

45. x What is this object doing? Slowing down in - direction x v a t=5 t=0
START 5s 4s 3s 2s 1s x

46. Draw the corresponding v-t and a-t graphs x-t graph v-t graph

47. x What is this object doing? Speeding up in - direction x v a t=5 t=0
START 5s 4s 3s 2s 1s x

48. PHYSICS DEPARTMENT STORE
SLOPE
AREA
SLOPE
AREA

49. Constant Acceleration Motion
AVERAGE VELOCITY
SLOPE =
acceleration
Connect with curved line
No physical meaning
AREA
slope
AVERAGE acceleration
SLOPE =
AREA
AREA
slope
AREA
SLOPE =
Dv =-9
AREA
RUNNING TOTAL

50. constant acceleration
SLOPE =
Velocity NOT constant
slope
AREA
SLOPE = Dx
AREA
slope
constant acceleration
AREA
AREA Dv

51. Estimate the net displacement from 0 s to 5.0 s
Physics C 1-D Motion
Estimate the net displacement from 0 s to 5.0 s
Area under v-t curve
Dx = = 10.5 m
Bertrand

52. Construct the corresponding x-t and a-t curves
Physics C 1-D Motion
Construct the corresponding x-t and a-t curves
Straight (constant v)
Curved (acceleration)
AREA
slope
Bertrand 52

53. Estimate the displacement from 0 s to 4.0 s
Physics C 1-D Motion
Estimate the displacement from 0 s to 4.0 s
Area under v-t curve
Dx = 2 -2 = 0 m
Bertrand

54. Construct the corresponding x-t and a-t curves
Physics C 1-D Motion
All Curved (acceleration)
AREA
slope
Bertrand 54

55. Construct the corresponding x-t and a-t curves
curved
Construct the corresponding x-t and a-t curves
straight
straight
curved
curved B
AREA A C
What region moving + direction, -direction, reverse?
What region speeding up, slowing down, constant v, stopped D F E
slope 55

56. Representations of Accelerated Motion Graphical Mathematical
x-t graph is quadratic:
A and B have physical meaning which we will derive
v-t graph is linear:
a-t graph a = constant

57. (Equation of v-t graph)
Mathematical Equations to Represent Constant Accelerated Motion and relationship to graphs
Definition of average velocity:
(Slope of x-t graph) 1
Definition of average acceleration: 2
(Equation of v-t graph)
(Slope of v-t graph)
constant
acceleration
For constant acceleration: 3

58. Mathematical Equations to Represent Constant Accelerated Motion and relationship to graphs1 4

59. (Constant Acceleration)
KINEMATIC EQUATIONS
(Constant Acceleration)
Dx = vDt (definition of average velocity)
Dv = aDt (definition of avr a)
Dt = (vf – v0)/a v
v0 + vf 2 =
(average velocity for constant acceleration)
Dx = (v0 + vf)(vf – v0)/2a
5. vf2 = v02 + 2aDx
(time independent)

60. (Constant Acceleration)
KINEMATIC EQUATIONS
(Constant Acceleration) 2. 3. 4. 5.

61. Representations of Accelerated Motion Graphical Mathematical
x-t graph is quadratic: t
x = At2 +Bt + x0
x = ½at2 +v0t + x0
v-t graph is linear:
a-t graph a = constant

62. Do Now (solve graphically): A car moving at 12 m/s is 36 m away from a stop sign. What acceleration will stop the car exactly at the stop sign?
v-t 12
Slope=a
a = -2 m/s2
Solve graphically
Area
Dx=36m
tstop

63. Example: A car moving at 12 m/s is 36 m away from a stop sign
Example: A car moving at 12 m/s is 36 m away from a stop sign. What acceleration will stop the car exactly at the stop sign?
v0= 12 m/s
vf = 0
a = ? m/s2
Dx = 36 m t
a = -2 m/s2

64. 5-Step Problem Solving GIVEN UNKNOWN FORMULA SUBSTITUTION SOLUTION
Read & Draw Diagram (when needed)
List all the given information using variables and units
NO WORDS
UNKNOWN
What variable are you looking to solve
Variable = ? units
FORMULA
Write formula
SUBSTITUTION
Substitute givens into formula (include units)
SOLUTION
Box your final answer: make sure units are expressed

65. Dx = 816 m KINEMATICS PROBLEMS
A passenger jet lands on a runway with a velocity of 70 m/s. Once it touches down, it accelerates at a constant rate of -3 m/s2. How far does the plane travel down the runway before its velocity is decreased to 2 m/s, its taxi speed to the landing gate?
vi = 70 m/s
vf = 2
a = -3 m/s2
Dx = ? m t
Dx = 816 m a a vi vf Dx

66. vi = 1.75 m/s vf = 6.25 m/s KINEMATICS PROBLEMS
A runner goes 12 m in 3 s at a constant acceleration of 1.5 m/s2. What is her velocity at the end of the 12 m?
vi =
vf = ? m/s
a = 1.5 m/s2
Dx = 12 m
t = 3 s
vi = 1.75 m/s
vf = 6.25 m/s

67. vf = 3 m/s KINEMATICS PROBLEMS
The U.S. and South Korean soccer teams are playing in the first round of the world cup. An American kicks the ball giving it an initial velocity of 4 m/s. The ball rolls a distance of 7 m and is then intercepted by a South Korean player. If the ball accelerated at m/s2 while rolling across the grass, find its velocity at the time of interception.
vi = 4 m/s
vf = ? m/s
a = m/s2
Dx = 7 m
t =
Signs and root issue
vf = 3 m/s