AP Physics, Unit 1, Days 1-2 Objects in Motion

Welcome, Please Find your Seat, Take out your summer homework and a composition notebook.

Summer Homework Check

Classroom Expectations 1) Respect the lab equipment. All lab equipment. 2) Respect each other. 3) Try. And don’t be afraid to ask questions. 4) Work with a sense of urgency. 5) Don’t leave trash in the desk.

About the AP Physics Test VERY heavy emphasis on the conceptual But the mathematical is still necessary VERY heavy emphasis on lab work and exploration

About the AP Physics Class Your lab notebook will be for labs only. All lab information will be recorded in there. All formal lab reports (and many informal ones) will be written in there. Write neatly. Do NOT erase. Homework will contain both readings and practice.

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Objects in Motion Lab

Welcome Back! Homework Check

Constant Velocity Lecture You do NOT need to take notes on everything. Rather, add to your reading notes, or highlight information that seems especially important. For the first couple of lectures, I will be a lot more specific with what you should and should not write down, as you get used to learning from lecture in science.

Chapter 1 Representing Motion Chapter Goal: To introduce the fundamental concepts of motion and to review related basic mathematical principles. © 2015 Pearson Education, Inc.

Chapter 1 Preview Looking Ahead: Describing Motion This series of images of a skier clearly shows his motion. Such visual depictions are a good first step in describing motion. You’ll learn to make motion diagrams that provide a simplified view of the motion of an object. © 2015 Pearson Education, Inc.

Section 1.1 Motion: A First Look © 2015 Pearson Education, Inc.

Types of Motion Motion is the change of an object’s position or orientation with time. The path along which an object moves is called the object’s trajectory. © 2015 Pearson Education, Inc.

Making a Motion Diagram These motion diagrams in one dimension show objects moving at constant speed (skateboarder), speeding up (runner) and slowing down (car). © 2015 Pearson Education, Inc.

Making a Motion Diagram This motion diagram shows motion in two dimensions with changes in both speed and direction. © 2015 Pearson Education, Inc.

QuickCheck 1.1 Car A Car B Motion diagrams are made of two cars. Both have the same time interval between photos. Which car, A or B, is going slower? Answer: A © 2015 Pearson Education, Inc.

QuickCheck 1.1 Car A Car B Motion diagrams are made of two cars. Both have the same time interval between photos. Which car, A or B, is going slower? © 2015 Pearson Education, Inc.

The Particle Model The particle model of motion is a simplification in which we treat a moving object as if all of its mass were concentrated at a single point © 2015 Pearson Education, Inc.

QuickCheck 1.2 Two runners jog along a track. The positions are shown at 1 s intervals. Which runner is moving faster? Answer: A © 2015 Pearson Education, Inc.

QuickCheck 1.2 Two runners jog along a track. The positions are shown at 1 s intervals. Which runner is moving faster? A © 2015 Pearson Education, Inc.

QuickCheck 1.3 Two runners jog along a track. The times at each position are shown. Which runner is moving faster? Runner A Runner B Both runners are moving at the same speed. Answer: C © 2015 Pearson Education, Inc.

QuickCheck 1.3 Two runners jog along a track. The times at each position are shown. Which runner is moving faster? Runner A Runner B Both runners are moving at the same speed. © 2015 Pearson Education, Inc.

Section 1.2 Position and Time: Putting Numbers on Nature © 2015 Pearson Education, Inc.

Position and Coordinate Systems To specify position we need a reference point (the origin), a distance from the origin, and a direction from the origin. The combination of an origin and an axis marked in both the positive and negative directions makes a coordinate system. © 2015 Pearson Education, Inc.

Position and Coordinate Systems The symbol that represents a position along an axis is called a coordinate. © 2015 Pearson Education, Inc.

Time For a complete motion diagram we need to label each frame with its corresponding time (symbol t) as read off a clock. © 2015 Pearson Education, Inc.

Changes in Position and Displacement A change of position is called a displacement. Displacement is the difference between a final position and an initial position: © 2015 Pearson Education, Inc.

Change in Time In order to quantify motion, we’ll need to consider changes in time, which we call time intervals. A time interval Δt measures the elapsed time as an object moves from an initial position xi at time ti to a final position xf at time tf. Δt is always positive. © 2015 Pearson Education, Inc.

Example 1.1 How long a ride? Carol is enjoying a bicycle ride on a country road that runs east-west past a water tower. Define a coordinate system so that increasing x means moving east. At noon, Carol is 3 miles (mi) east of the water tower. A half-hour later, she is 2 mi west of the water tower. What is her displacement during that half-hour? prepare Although it may seem like overkill for such a simple problem, you should start by making a drawing, like the one in FIGURE 1.15, with the x-axis along the road. © 2015 Pearson Education, Inc.

Example 1.1 How long a ride? (cont.) solve We’ve specified values for Carol’s initial and final positions in our drawing. We can thus compute her displacement: Δx = xf  xi = ( 2 mi)  (3 mi) =  5 mi assess Carol is moving to the west, so we expect her displacement to be negative—and it is. We can see from our drawing in Figure 1.15 that she has moved 5 miles from her starting position, so our answer seems reasonable. Carol travels 5 miles in a half-hour, quite a reasonable pace for a cyclist. © 2015 Pearson Education, Inc.

QuickCheck 1.4 Maria is at position x = 23 m. She then undergoes a displacement x = –50 m. What is her final position? –27 m –50 m 23 m 73 m Answer: A © 2015 Pearson Education, Inc.

QuickCheck 1.4 Maria is at position x = 23 m. She then undergoes a displacement x = –50 m. What is her final position? –27 m –50 m 23 m 73 m © 2015 Pearson Education, Inc.

QuickCheck 1.5 An ant zig-zags back and forth on a picnic table as shown. The ant’s distance traveled and displacement are 50 cm and 50 cm 30 cm and 50 cm 50 cm and 30 cm 50 cm and –50 cm 50 cm and –30 cm Answer: E © 2015 Pearson Education, Inc. 34

QuickCheck 1.5 An ant zig-zags back and forth on a picnic table as shown. The ant’s distance traveled and displacement are 50 cm and 50 cm 30 cm and 50 cm 50 cm and 30 cm 50 cm and –50 cm 50 cm and –30 cm © 2015 Pearson Education, Inc. 35

Section 1.3 Velocity © 2015 Pearson Education, Inc.

Velocity and Speed Motion at a constant speed in a straight line is called uniform motion. © 2015 Pearson Education, Inc.

Example Problem Jane walks to the right at a constant rate, moving 3 m in 3 s. At t = 0 s she passes the x = 1 m mark. Draw her motion diagram from t = –1 s to t = 4 s. © 2015 Pearson Education, Inc.

Velocity and Speed Speed measures only how fast an object moves, but velocity tells us both an object’s speed and its direction. The velocity defined by Equation 1.2 is called the average velocity. © 2015 Pearson Education, Inc.

Example 1.2 Finding the speed of a seabird Albatrosses are seabirds that spend most of their lives flying over the ocean looking for food. With a stiff tailwind, an albatross can fly at high speeds. Satellite data on one particularly speedy albatross showed it 60 miles east of its roost at 3:00 pm and then, at 3:15 pm, 80 miles east of its roost. What was its velocity? © 2015 Pearson Education, Inc.

Example 1.2 Finding the speed of a seabird (cont.) prepare The statement of the problem provides us with a natural coordinate system: We can measure distances with respect to the roost, with distances to the east as positive. With this coordinate system, the motion of the albatross appears as in FIGURE 1.18. The motion takes place between 3:00 and 3:15, a time interval of 15 minutes, or 0.25 hour. solve We know the initial and final positions, and we know the time interval, so we can calculate the velocity: © 2015 Pearson Education, Inc.

Example Problem At t = 12 s, Frank is at x = 25 m. 5 s later, he’s at x = 20 m. What is Frank’s velocity? This example indicates that velocities to the left are negative. The information that t = 12 s is extraneous. Answer: The displacement is 5 m in 5 seconds so his velocity is 5m/5 s = 1 m/s. © 2015 Pearson Education, Inc.

Example Problem Jenny runs 1 mi to the northeast, then 1 mi south. Graphically find her net displacement. © 2015 Pearson Education, Inc.

Chapter 2 Motion in One Dimension Chapter Goal: To describe and analyze linear motion. © 2015 Pearson Education, Inc.

Chapter 2 Preview Looking Ahead: Uniform Motion Successive images of the rider are the same distance apart, so the velocity is constant. This is uniform motion. You’ll learn to describe motion in terms of quantities such as distance and velocity, an important first step in analyzing motion. © 2015 Pearson Education, Inc.

Section 2.1 Describing Motion © 2015 Pearson Education, Inc.

Representing Position We will use an x-axis to analyze horizontal motion and motion on a ramp, with the positive end to the right. We will use a y-axis to analyze vertical motion, with the positive end up. © 2015 Pearson Education, Inc.

Representing Position The motion diagram of a student walking to school and a coordinate axis for making measurements Every dot in the motion diagram of Figure 2.2 represents the student’s position at a particular time. Figure 2.3 shows the student’s motion shows the student’s position as a graph of x versus t. © 2015 Pearson Education, Inc.

From Position to Velocity On a position-versus-time graph, a faster speed corresponds to a steeper slope. The slope of an object’s position- versus-time graph is the object’s velocity at that point in the motion. © 2015 Pearson Education, Inc.

From Position to Velocity We can deduce the velocity-versus- time graph from the position-versus- time graph. The velocity-versus-time graph is yet another way to represent an object’s motion. © 2015 Pearson Education, Inc.

QuickCheck 2.2 Here is a motion diagram of a car moving along a straight road: Which velocity-versus-time graph matches this motion diagram? None of the above. Answer: C © 2015 Pearson Education, Inc. 51

QuickCheck 2.2 Here is a motion diagram of a car moving along a straight road: Which velocity-versus-time graph matches this motion diagram? None of the above. Answer: C C. © 2015 Pearson Education, Inc. 52

QuickCheck 2.3 Here is a motion diagram of a car moving along a straight road: Which velocity-versus-time graph matches this motion diagram? Answer: D © 2015 Pearson Education, Inc. 53

QuickCheck 2.3 Here is a motion diagram of a car moving along a straight road: Which velocity-versus-time graph matches this motion diagram? D. © 2015 Pearson Education, Inc. 54

QuickCheck 2.4 A graph of position versus time for a basketball player moving down the court appears as follows: Which of the following velocity graphs matches the position graph? Answer: C A. B. C. D. © 2015 Pearson Education, Inc.

QuickCheck 2.4 A graph of position versus time for a basketball player moving down the court appears as follows: Which of the following velocity graphs matches the position graph? A. B. C. D. © 2015 Pearson Education, Inc.

Example 2.2 Analyzing a car’s position graph FIGURE 2.11 gives the position-versus-time graph of a car. Draw the car’s velocity- versus-time graph. Describe the car’s motion in words. prepare Figure 2.11 is a graphical representation of the motion. The car’s position- versus-time graph is a sequence of three straight lines. Each of these straight lines represents uniform motion at a constant velocity. We can determine the car’s velocity during each interval of time by measuring the slope of the line. © 2015 Pearson Education, Inc.

Example 2.2 Analyzing a car’s position graph (cont.) solve From t = 0 s to t = 2 s (Δt = 2 s) the car’s displacement is Δx = 4 m  0 m = 4 m. The velocity during this interval is The car’s position does not change from t = 2 s to t = 4 s (Δx = 0 m), so vx = 0 m/s. Finally, the displacement between t = 4 s and t = 6 s (Δt = 2 s) is Δx = 10 m. Thus the velocity during this interval is These velocities are represented graphically in FIGURE 2.12. © 2015 Pearson Education, Inc.

Example 2.2 Analyzing a car’s position graph (cont.) solve The velocity-versus-time graph of Figure 2.12 shows the motion in a way that we can describe in a straightforward manner: The car backs up for 2 s at 2 m/s, sits at rest for 2 s, then drives forward at 5 m/s for 2 s. assess Notice that the velocity graph and the position graph look completely different. They should! The value of the velocity graph at any instant of time equals the slope of the position graph. Since the position graph is made up of segments of constant slope, the velocity graph should be made up of segments of constant value, as it is. This gives us confidence that the graph we have drawn is correct. © 2015 Pearson Education, Inc.

From Velocity to Position We can deduce the position-versus- time graph from the velocity-versus- time graph. The sign of the velocity tells us whether the slope of the position graph is positive or negative. The magnitude of the velocity tells us how steep the slope is. © 2015 Pearson Education, Inc.

QuickCheck 2.1 Here is a motion diagram of a car moving along a straight road: Which position-versus-time graph matches this motion diagram? Answer: E © 2015 Pearson Education, Inc. 61

QuickCheck 2.1 Here is a motion diagram of a car moving along a straight road: Which position-versus-time graph matches this motion diagram? E. © 2015 Pearson Education, Inc. 62

QuickCheck 2.6 A graph of velocity versus time for a hockey puck shot into a goal appears as follows: Which of the following position graphs matches the velocity graph? Answer: D A. B. C. D. © 2015 Pearson Education, Inc.

QuickCheck 2.6 A graph of velocity versus time for a hockey puck shot into a goal appears as follows: Which of the following position graphs matches the velocity graph? A. B. C. (d) D. © 2015 Pearson Education, Inc.

Section 2.2 Uniform Motion © 2015 Pearson Education, Inc.

Uniform Motion Straight-line motion in which equal displacements occur during any successive equal-time intervals is called uniform motion or constant- velocity motion. An object’s motion is uniform if and only if its position-versus-time graph is a straight line. © 2015 Pearson Education, Inc.

Equations of Uniform Motion The velocity of an object in uniform motion tells us the amount by which its position changes during each second. The displacement Δx is proportional to the time interval Δt. © 2015 Pearson Education, Inc.

Equations of Uniform Motion © 2015 Pearson Education, Inc. Text: p. 34

QuickCheck 2.8 Here is a position graph of an object: At t = 1.5 s, the object’s velocity is 40 m/s 20 m/s 10 m/s –10 m/s None of the above Answer: B © 2015 Pearson Education, Inc. 69

QuickCheck 2.8 Here is a position graph of an object: At t = 1.5 s, the object’s velocity is 40 m/s 20 m/s 10 m/s –10 m/s None of the above © 2015 Pearson Education, Inc. 70

Example 2.3 If a train leaves Cleveland at 2:00… A train is moving due west at a constant speed. A passenger notes that it takes 10 minutes to travel 12 km. How long will it take the train to travel 60 km? prepare For an object in uniform motion, Equation 2.5 shows that the distance traveled Δx is proportional to the time interval Δt, so this is a good problem to solve using ratio reasoning. © 2015 Pearson Education, Inc.

Example 2.3 If a train leaves Cleveland at 2:00…(cont.) solve We are comparing two cases: the time to travel 12 km and the time to travel 60 km. Because Δx is proportional to Δt, the ratio of the times will be equal to the ratio of the distances. The ratio of the distances is This is equal to the ratio of the times: It takes 10 minutes to travel 12 km, so it will take 50 minutes—5 times as long— to travel 60 km. © 2015 Pearson Education, Inc.

Example Problem A soccer player is 15 m from her opponent’s goal. She kicks the ball hard; after 0.50 s, it flies past a defender who stands 5 m away, and continues toward the goal. How much time does the goalie have to move into position to block the kick from the moment the ball leaves the kicker’s foot? Answer: calculate the speed of the ball which is 30 m/s. The time to travel 5 m at that speed is 0.17 s. This is less than human reaction time. © 2015 Pearson Education, Inc.

From Velocity to Position, One More Time The displacement Δx is equal to the area under the velocity graph during the time interval Δt. © 2015 Pearson Education, Inc.

QuickCheck 2.11 Here is the velocity graph of an object that is at the origin (x  0 m) at t  0 s. At t  4.0 s, the object’s position is 20 m 16 m 12 m 8 m 4 m Answer: C © 2015 Pearson Education, Inc. 75

QuickCheck 2.11 Here is the velocity graph of an object that is at the origin (x  0 m) at t  0 s. At t  4.0 s, the object’s position is 20 m 16 m 12 m 8 m 4 m Displacement  area under the curve © 2015 Pearson Education, Inc. 76

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