Section 2.1 Picturing Motion A motion diagram shows the position of an object at successive times. In the particle model, the object in the motion diagram is replaced by a series of single points. Chapter Summary 1

Section 2.2 Where and When? You can define any coordinate system you wish in describing motion, but some are more useful than others. A time interval is the difference between two times. Chapter Summary 4

Section 2.2 Where and When? A vector drawn from the origin of the coordinate system to the object indicates the object’s position. Change in position is displacement, which has both magnitude and direction. Chapter Summary 4

Section 2.2 Where and When? The length of the displacement vector represents how far the object was displaced, and the vector points in the direction of the displacement. Chapter Summary 4

Section 2.2 Position-Time Graphs Position-time graphs can be used to find the velocity and position of an object, as well as where and when two objects meet. Any motion can be described using words, motion diagrams, data tables, and graphs. Chapter Summary 4

Section 2.2 How Fast The slope of an object’s position-time graph is the average velocity of the object’s motion. The average speed is the absolute value of the average velocity. Chapter Summary 4

Section 2.2 How Fast An object’s velocity is how fast it is moving and in what direction it is moving. An object’s initial position, di, its constant average velocity, , its position d, and the time, t, since the object was at its initial position, are related by a simple equation. Chapter Summary 4

Chapter Assessment Questions 1 If an object in the motion diagram is replaced by a series of single points, to which type of motion diagram are you referring? A. Motion diagram B. Particle model C. Coordinate system D. Position-time graph Chapter Assessment Questions 1

Chapter Assessment Questions 2 Reason: A particle model is a simplified version of a motion diagram in which the object in motion is replaced by a series of single points. Chapter Assessment Questions 2

Chapter Assessment Questions 3 Refer to the figure and calculate the distance between the two signals? A. 3 m B. 8 m C. 5 m D. 5 cm Chapter Assessment Questions 3

Chapter Assessment Questions 4 Reason: Distance d = df – di Here, df = 8 m and di = 3 m Therefore, d = 8 m – 3 m = 5 m Chapter Assessment Questions 4

Chapter Assessment Questions 5 From the adjacent position-time graph determine the distance covered by the competitor between 5 s and 10 s? Chapter Assessment Questions 5

Chapter Assessment Questions 6 Answer: The distance covered by the competitor between 5 s and 10 s is d = 20 m – 10 m = 10 m. Chapter Assessment Questions 6

Chapter Assessment Questions 7 A car starting from rest moves with an average speed of 6 m/s. Use the equation of motion for average velocity to calculate the distance the car traveled in 1 minute. A. 6 m B. 10 m C. 60 m D. 360 m Chapter Assessment Questions 7

Chapter Assessment Questions 8 Reason: The equation of motion for average velocity is d = vt + di where d is the object’s position at time t of the object moving with average velocity v, and di is the initial position of the object. In this case di = 0, v = 6 m/s and t = 1 min = 60 s. Therefore d = (6 m/s)(60 s) + 0 m = 360 m. Chapter Assessment Questions 8

Chapter Assessment Questions 9 The position-time graph at right shows three ships moving in a river. Determine which ship has greater average velocity? A. Ship A B. Ship B C. Ship C D. Both ship B and C Chapter Assessment Questions 9

Chapter Assessment Questions 10 Reason: The slope of A is steeper than the slopes of B and C. A steeper slope indicates a greater change in displacement during each time interval, i.e. greater average velocity. Chapter Assessment Questions 10

End of Customs Shows