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

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Presentation on theme: "REPRESENTING MOTION KINEMATICS in One Dimension “To understand motion is to understand nature.” Leonardo da Vinci."— Presentation transcript:

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

2 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.

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 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 Motion Diagram & Particle Model

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

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

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,  x – change in position of an object (position is measured from the origin of a chosen coordinate system)

12 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? kmX = LivingstonPhiladelphiaTrenton displacement distance 0 Distance and Displacement

13 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? kmX = LivingstonPhiladelphiaTrenton displacement distance 0 Distance and Displacement

14 Displacement Has magnitude (size) and direction. It is a VECTOR 123 x(m) Vectors are represented by arrows

15 Displacement Has magnitude and direction. It is a VECTOR 123 x(m)

16 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 sign gives direction in 1 Dimension Scalar (has magnitude only) Vector (has magnitude and direction) Average speed and average velocity often have the same magnitude, but not always

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? kmX = LivingstonPhiladelphiaTrenton Average speed Average velocity xx

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

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 t x A B xx tt slope Steepness = speed Sign = direction Velocity = speed + direction Position-Time Graph slope

21 Average Velocity from a Graph t x A B xx tt Mathematical Model x = position x 0 = initial position v = average velocity t = time

22 Graphs of Motion Mathematical Model x 0 = 20 m v = 2 m/s Slope Mathematical Model (UNIFORM VELOCITY) Area Area=20m =  x  x =20m SLOPE AREA  x =20m (init and final positions unknown. ONLY KNOW DISPLACEMENT)

23 What is happening in this graph? mX = t=0 s START End t=6 s v = 1s 2s3s4s5s Mathematical Diagrammatic Graphical

24 Plot the corresponding v-t graph x 0 = 20 m v = -5 m/s Slope SLOPE

25 What is happening?

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

27 Draw the corresponding v-t graph v-t SLOPE

28 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 B A,F B,D,E A,F B,D,E F F=D=E>A>B>C C At 2s, at 7 s

29 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? A B 0m 0m/s 32m 4m/s

30 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 Problem: A car starting from rest moves with an average speed of 6 m/s. Calculate the distance the car traveled in 1 minute.

31 GIVEN: x o = 47m x f = 15m t = 8 s UNKNOWN:  x = ? m FORMULA:  x=x f -x o SUBSTITUTION:  x= SOLUTION  x = -32 m 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: x o = 47m x f = 15m  x = -32 m t = 8 s UNKNOWN: v av = ? m/s FORMULA: SUBSTITUTION: SOLUTION

32 Draw the corresponding x-t graph Area=  x -6m  x= +12m 3m  x= -16m AREA v-t

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: kmX = kmX = Start t=0End t=2 hr 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.

35 Acceleration Average acceleration describes how quickly or slowly the velocity changes. It is calculated by: Vector 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. SI units: m/s 2

36 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 v 1s2s3s4s5s a x m/s 2 20m/s 0 0

37 START 1s2s3s4s5s x t=0 t=5 t=4 t=3 t=2 t=1 x va Position-Time Graph

38 START v 1s2s3s4s5s a x START v 1s2s3s4s5s a x Speeding up in + direction Slowing down in + direction a and v SAME direction a and v OPP direction 0 0

39 START v 4s3s2s1s0 a x Slowing up in - direction a and v OPP direction 5s START v 4s3s2s1s 0 a x Speeding up in - direction a and v SAME direction 5s

40 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

41 START 1s2s3s4s5s x t=0 t=5 t=4 t=3 t=2 t=1 x va Position-Time Graph Speeding up in + direction 0

42 START 1s2s3s4s5s x t=0 t=5 t=3 t=2 t=1 t=4 xva Position-Time Graph Slowing down in + direction 0

43 t=5 t=0 t=1 t=2 t=3 t=4 x va Speeding up in - direction START 4s3s2s1s5s x 0

44 START 4s3s2s1s5s x t=5 t=0 t=2 t=3 t=4 t=1 x va Slowing down in - dire ct ion 0

45 START 4s3s2s1s5s x t=5 t=0 t=2 t=3 t=4 t=1 x va Slowing down in - dire ct ion What is this object doing? 0

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

47 t=5 t=0 t=1 t=2 t=3 t=4 x va Speeding up in - direction START 4s3s2s1s5s x 0 What is this object doing?

48 SLOPE AREA PHYSICS DEPARTMENT STORE

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

50 SLOPE = AREA vv slope SLOPE = AREA slope AREA constant acceleration Velocity NOT constant xx

51 Estimate the net displacement from 0 s to 5.0 s  x = = 10.5 m Area under v-t curve

52 Construct the corresponding x-t and a-t curves AREA Curved (acceleration) Straight (constant v) slope

53 Estimate the displacement from 0 s to 4.0 s  x = 2 -2 = 0 m Area under v-t curve

54 AREA Construct the corresponding x-t and a-t curves slope All Curved (acceleration)

55 Construct the corresponding x-t and a-t curves curved straight slope AREA A B C D E F

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

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

58 Mathematical Equations to Represent Constant Accelerated Motion and relationship to graphs 4 1

59 KINEMATIC EQUATIONS (Constant Acceleration)  x = v  t (definition of average velocity) v v 0 + v f  = (average velocity for constant acceleration)  v = a  t (definition of avr a)  t = (v f – v 0 )/a  x = (v 0 + v f )(v f – v 0 )/2a  v f 2 = v a  x (time independent)

60 KINEMATIC EQUATIONS (Constant Acceleration)

61 x-t graph is quadratic: v-t graph is linear: a-t graph a = constant Representations of Accelerated Motion t Graphical Mathematical x = At 2 +Bt + x 0 x = ½at 2 +v 0 t + x 0

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? a = -2 m/s 2 t stop Area  x=36m Slope=a 12 v-t

63 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? v 0 = 12 m/s v f = 0 a = ? m/s 2  x = 36 m t a = -2 m/s 2

64 5-Step Problem Solving GIVEN – 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 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/s 2. 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? v i = 70 m/s v f = 2 a = -3 m/s 2  x = ? m t  x = 816 m xx vivi a vfvf a

66 KINEMATICS PROBLEMS A runner goes 12 m in 3 s at a constant acceleration of 1.5 m/s 2. What is her velocity at the end of the 12 m? v i = v f = ? m/s a = 1.5 m/s 2  x = 12 m t = 3 s v i = 1.75 m/s v f = 6.25 m/s

67 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/s 2 while rolling across the grass, find its velocity at the time of interception. KINEMATICS PROBLEMS v i = 4 m/s v f = ? m/s a = m/s 2  x = 7 m t = v f = 3 m/s


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