Chapter 2,3,7 MASTER CLASS NOTES FOR MOTION Chapter 2,3 & 7

Chapter 2,3,7 MASTER CLASS NOTES FOR MOTION Chapter 2,3 & 7
MASTER NOTES MASTER CLASS NOTES FOR MOTION Chapter 2,3 & 7

Chapter 2 Table of Contents Section 1 Displacement and Velocity
Motion in One Dimension Table of Contents Section 1 Displacement and Velocity Section 2 Acceleration Section 3 Falling Objects

Section 1 Displacement and Velocity
Chapter 2 Objectives Describe motion in terms of frame of reference, displacement, time, and velocity. Calculate the displacement of an object traveling at a known velocity for a specific time interval. Construct and interpret graphs of position versus time.

One Dimensional Motion
Section 1 Displacement and Velocity Chapter 2 One Dimensional Motion To simplify the concept of motion, we will first consider motion that takes place in one direction. One example is the motion of a commuter train on a straight track. To measure motion, you must choose a frame of reference. A frame of reference is a system for specifying the precise location of objects in space and time.

displacement = final position – initial position
Section 1 Displacement and Velocity Chapter 2 Displacement Displacement is a change in position. Displacement is not always equal to the distance traveled. The SI unit of displacement is the meter, m. Dx = xf – xi displacement = final position – initial position

Positive and Negative Displacements
Section 1 Displacement and Velocity Chapter 2 Positive and Negative Displacements

Chapter 2 Average Velocity
Section 1 Displacement and Velocity Chapter 2 Average Velocity Average velocity is the total displacement divided by the time interval during which the displacement occurred. In SI, the unit of velocity is meters per second, abbreviated as m/s.

Chapter 2 Velocity and Speed
Section 1 Displacement and Velocity Chapter 2 Velocity and Speed Velocity describes motion with both a direction and a numerical value (a magnitude). Speed has no direction, only magnitude. Average speed is equal to the total distance traveled divided by the time interval.

Interpreting Velocity Graphically
Section 1 Displacement and Velocity Chapter 2 Interpreting Velocity Graphically For any position-time graph, we can determine the average velocity by drawing a straight line between any two points on the graph. If the velocity is constant, the graph of position versus time is a straight line. The slope indicates the velocity. Object 1: positive slope = positive velocity Object 2: zero slope= zero velocity Object 3: negative slope = negative velocity

Interpreting Velocity Graphically, continued
Section 1 Displacement and Velocity Chapter 2 Interpreting Velocity Graphically, continued The instantaneous velocity is the velocity of an object at some instant or at a specific point in the object’s path. The instantaneous velocity at a given time can be determined by measuring the slope of the line that is tangent to that point on the position-versus-time graph.

Chapter 2 Objectives Describe motion in terms of changing velocity.
Section 2 Acceleration Chapter 2 Objectives Describe motion in terms of changing velocity. Compare graphical representations of accelerated and nonaccelerated motions. Apply kinematic equations to calculate distance, time, or velocity under conditions of constant acceleration.

Chapter 2 Changes in Velocity
Section 2 Acceleration Chapter 2 Changes in Velocity Acceleration is the rate at which velocity changes over time. An object accelerates if its speed, direction, or both change. Acceleration has direction and magnitude. Thus, acceleration is a vector quantity.

Changes in Velocity, continued
Section 2 Acceleration Chapter 2 Changes in Velocity, continued Consider a train moving to the right, so that the displacement and the velocity are positive. The slope of the velocity-time graph is the average acceleration. When the velocity in the positive direction is increasing, the acceleration is positive, as at A. When the velocity is constant, there is no acceleration, as at B. When the velocity in the positive direction is decreasing, the acceleration is negative, as at C.

Velocity and Acceleration
Section 2 Acceleration Chapter 2 Velocity and Acceleration

Motion with Constant Acceleration
Section 2 Acceleration Chapter 2 Motion with Constant Acceleration When velocity changes by the same amount during each time interval, acceleration is constant. The relationships between displacement, time, velocity, and constant acceleration are expressed by the equations shown on the next slide. These equations apply to any object moving with constant acceleration. These equations use the following symbols: Dx = displacement vi = initial velocity vf = final velocity Dt = time interval

Equations for Constantly Accelerated Straight-Line Motion
Section 2 Acceleration Chapter 2 Equations for Constantly Accelerated Straight-Line Motion

Chapter 2 Sample Problem Final Velocity After Any Displacement
Section 2 Acceleration Chapter 2 Sample Problem Final Velocity After Any Displacement A person pushing a stroller starts from rest, uniformly accelerating at a rate of m/s2. What is the velocity of the stroller after it has traveled 4.75 m?

Sample Problem, continued
Section 2 Acceleration Chapter 2 Sample Problem, continued 1. Define Given: vi = 0 m/s a = m/s2 Dx = 4.75 m Unknown: vf = ? Diagram: Choose a coordinate system. The most convenient one has an origin at the initial location of the stroller, as shown above. The positive direction is to the right.

Section 2 Acceleration Chapter 2 Sample Problem, continued 2. Plan Choose an equation or situation: Because the initial velocity, acceleration, and displacement are known, the final velocity can be found using the following equation: Rearrange the equation to isolate the unknown: Take the square root of both sides to isolate vf .

Section 2 Acceleration Chapter 2 Sample Problem, continued 3. Calculate Substitute the values into the equation and solve: Tip: Think about the physical situation to determine whether to keep the positive or negative answer from the square root. In this case, the stroller starts from rest and ends with a speed of 2.18 m/s. An object that is speeding up and has a positive acceleration must have a positive velocity. So, the final velocity must be positive. 4. Evaluate The stroller’s velocity after accelerating for 4.75 m is 2.18 m/s to the right.

Section 3 Falling Objects
Chapter 2 Objectives Relate the motion of a freely falling body to motion with constant acceleration. Calculate displacement, velocity, and time at various points in the motion of a freely falling object. Compare the motions of different objects in free fall.

Section 3 Falling Objects
Chapter 2 Free Fall Free fall is the motion of a body when only the force due to gravity is acting on the body. The acceleration on an object in free fall is called the acceleration due to gravity, or free-fall acceleration. Free-fall acceleration is denoted with the symbols ag (generally) or g (on Earth’s surface).

Free-Fall Acceleration
Section 3 Falling Objects Chapter 2 Free-Fall Acceleration Free-fall acceleration is the same for all objects, regardless of mass. This book will use the value g = 9.81 m/s2. Free-fall acceleration on Earth’s surface is –9.81 m/s2 at all points in the object’s motion. Consider a ball thrown up into the air. Moving upward: velocity is decreasing, acceleration is –9.81 m/s2 Top of path: velocity is zero, acceleration is –9.81 m/s2 Moving downward: velocity is increasing, acceleration is –9.81 m/s2

Chapter 2 Sample Problem Falling Object
Section 3 Falling Objects Chapter 2 Sample Problem Falling Object Jason hits a volleyball so that it moves with an initial velocity of 6.0 m/s straight upward. If the volleyball starts from 2.0 m above the floor, how long will it be in the air before it strikes the floor?

Section 3 Falling Objects Chapter 2 Sample Problem, continued 1. Define Given: Unknown: vi = +6.0 m/s Dt = ? a = –g = –9.81 m/s2 Dy = –2.0 m Diagram: Place the origin at the Starting point of the ball (yi = 0 at ti = 0).

Section 3 Falling Objects Chapter 2 Sample Problem, continued 2. Plan Choose an equation or situation: Both ∆t and vf are unknown. Therefore, first solve for vf using the equation that does not require time. Then, the equation for vf that does involve time can be used to solve for ∆t. Rearrange the equation to isolate the unknown: Take the square root of the first equation to isolate vf. The second equation must be rearranged to solve for ∆t.

Section 3 Falling Objects Chapter 2 Sample Problem, continued 3. Calculate Substitute the values into the equation and solve: First find the velocity of the ball at the moment that it hits the floor. Tip: When you take the square root to find vf , select the negative answer because the ball will be moving toward the floor, in the negative direction.

Section 3 Falling Objects Chapter 2 Sample Problem, continued Next, use this value of vf in the second equation to solve for ∆t. 4. Evaluate The solution, 1.50 s, is a reasonable amount of time for the ball to be in the air.

Chapter 2 Multiple Choice Use the graphs to answer questions 1–3.
Standardized Test Prep Multiple Choice Use the graphs to answer questions 1–3. 1. Which graph represents an object moving with a constant positive velocity? A. I C. III B. II D. IV

Multiple Choice, continued
Chapter 2 Standardized Test Prep Multiple Choice, continued Use the graphs to answer questions 1–3. 2. Which graph represents an object at rest? F. I H. III G. II J. IV

Chapter 2 Standardized Test Prep Multiple Choice, continued Use the graphs to answer questions 1–3. 3. Which graph represents an object moving with a constant positive acceleration? A. I C. III B. II D. IV

Chapter 2 Standardized Test Prep Multiple Choice, continued 4. A bus travels from El Paso, Texas, to Chihuahua, Mexico, in 5.2 h with an average velocity of 73 km/h to the south.What is the bus’s displacement? F. 73 km to the south G. 370 km to the south H. 380 km to the south J. 14 km/h to the south

Chapter 2 Standardized Test Prep Multiple Choice, continued Use the position-time graph of a squirrel running along a clothesline to answer questions 5–6. 5. What is the squirrel’s displacement at time t = 3.0 s? A. –6.0 m B. –2.0 m C m D m

Chapter 2 Standardized Test Prep Multiple Choice, continued Use the position-time graph of a squirrel running along a clothesline to answer questions 5–6. 6. What is the squirrel’s average velocity during the time interval between 0.0 s and 3.0 s? F. –2.0 m/s G. –0.67 m/s H. 0.0 m/s J m/s

Chapter 2 Standardized Test Prep Multiple Choice, continued 7. Which of the following statements is true of acceleration? A. Acceleration always has the same sign as displacement. B. Acceleration always has the same sign as velocity. C. The sign of acceleration depends on both the direction of motion and how the velocity is changing. D. Acceleration always has a positive sign.

Chapter 2 Standardized Test Prep Multiple Choice, continued 8. A ball initially at rest rolls down a hill and has an acceleration of 3.3 m/s2. If it accelerates for 7.5 s, how far will it move during this time? F. 12 m G. 93 m H. 120 m J. 190 m

Chapter 2 Standardized Test Prep Multiple Choice, continued 9. Which of the following statements is true for a ball thrown vertically upward? A. The ball has a negative acceleration on the way up and a positive acceleration on the way down. B. The ball has a positive acceleration on the way up and a negative acceleration on the way down. C. The ball has zero acceleration on the way up and a positive acceleration on the way down. D. The ball has a constant acceleration throughout its flight.

Chapter 2 Short Response
Standardized Test Prep Short Response 10. In one or two sentences, explain the difference between displacement and distance traveled.

Standardized Test Prep Short Response 10. In one or two sentences, explain the difference between displacement and distance traveled. Answer: Displacement measures only the net change in position from starting point to end point. The distance traveled is the total length of the path followed from starting point to end point and may be greater than or equal to the displacement.

Short Response, continued
Chapter 2 Standardized Test Prep Short Response, continued 11. The graph shows the position of a runner at different times during a run. Use the graph to determine the runner’s displacement and average velocity: a. for the time interval from t = 0.0 min to t = 10.0 min b. for the time interval from t = 10.0 min to t = 20.0 min c. for the time interval from t = 20.0 min to t = 30.0 min d. for the entire run

Chapter 2 Standardized Test Prep Short Response, continued 11. The graph shows the position of a runner at different times during a run. Use the graph to determine the runner’s displacement and average velocity. Answers will vary but should be approximately as follows: a. for t = 0.0 min to t = 10.0 min Answer: m, +4.0 m/s b. for t = 10.0 min to t = 20.0 min Answer: m, +2.5 m/s c. for t = 20.0 min to t = 30.0 min Answer: +900 m, +2 m/s d. for the entire run Answer: m, +2.7 m/s

Chapter 2 Standardized Test Prep Short Response, continued 12. For an object moving with constant negative acceleration, draw the following: a. a graph of position vs. time b. a graph of velocity vs. time For both graphs, assume the object starts with a positive velocity and a positive displacement from the origin.

Chapter 2 Standardized Test Prep Short Response, continued 12. For an object moving with constant negative acceleration, draw the following: a. a graph of position vs. time b. a graph of velocity vs. time For both graphs, assume the object starts with a positive velocity and a positive displacement from the origin. Answers:

Chapter 2 Standardized Test Prep Short Response, continued 13. A snowmobile travels in a straight line. The snowmobile’s initial velocity is +3.0 m/s. a. If the snowmobile accelerates at a rate of m/s2 for 7.0 s, what is its final velocity? b. If the snowmobile accelerates at the rate of –0.60 m/s2 from its initial velocity of +3.0 m/s, how long will it take to reach a complete stop?

Chapter 2 Standardized Test Prep Short Response, continued 13. A snowmobile travels in a straight line. The snowmobile’s initial velocity is +3.0 m/s. a. If the snowmobile accelerates at a rate of m/s2 for 7.0 s, what is its final velocity? b. If the snowmobile accelerates at the rate of –0.60 m/s2 from its initial velocity of +3.0 m/s, how long will it take to reach a complete stop? Answers: a m/s b. 5.0 s

Chapter 2 Extended Response
Standardized Test Prep Extended Response 14. A car moving eastward along a straight road increases its speed uniformly from 16 m/s to 32 m/s in 10.0 s. a. What is the car’s average acceleration? b. What is the car’s average velocity? c. How far did the car move while accelerating? Show all of your work for these calculations.

Standardized Test Prep Extended Response 14. A car moving eastward along a straight road increases its speed uniformly from 16 m/s to 32 m/s in 10.0 s. a. What is the car’s average acceleration? b. What is the car’s average velocity? c. How far did the car move while accelerating? Answers: a. 1.6 m/s2 eastward b. 24 m/s c. 240 m

Extended Response, continued
Chapter 2 Standardized Test Prep Extended Response, continued 15. A ball is thrown vertically upward with a speed of 25.0 m/s from a height of 2.0 m. a. How long does it take the ball to reach its highest point? b. How long is the ball in the air? Show all of your work for these calculations.

Chapter 2 Standardized Test Prep Extended Response, continued 15. A ball is thrown vertically upward with a speed of 25.0 m/s from a height of 2.0 m. a. How long does it take the ball to reach its highest point? b. How long is the ball in the air? Show all of your work for these calculations. Answers: a s b s

Chapter 3 Two-Dimensional Motion and Vectors

Chapter 3 Table of Contents Section 1 Introduction to Vectors
Two-Dimensional Motion and Vectors Table of Contents Section 1 Introduction to Vectors Section 2 Vector Operations Section 3 Projectile Motion Section 4 Relative Motion

Chapter 3 Objectives Distinguish between a scalar and a vector.
Section 1 Introduction to Vectors Chapter 3 Objectives Distinguish between a scalar and a vector. Add and subtract vectors by using the graphical method. Multiply and divide vectors by scalars.

Chapter 3 Scalars and Vectors
Section 1 Introduction to Vectors Chapter 3 Scalars and Vectors A scalar is a physical quantity that has magnitude but no direction. Examples: speed, volume, the number of pages in your textbook A vector is a physical quantity that has both magnitude and direction. Examples: displacement, velocity, acceleration In this book, scalar quantities are in italics. Vectors are represented by boldface symbols.

Graphical Addition of Vectors
Section 1 Introduction to Vectors Chapter 3 Graphical Addition of Vectors A resultant vector represents the sum of two or more vectors. Vectors can be added graphically. A student walks from his house to his friend’s house (a), then from his friend’s house to the school (b). The student’s resultant displacement (c) can be found by using a ruler and a protractor.

Triangle Method of Addition
Section 1 Introduction to Vectors Chapter 3 Triangle Method of Addition Vectors can be moved parallel to themselves in a diagram. Thus, you can draw one vector with its tail starting at the tip of the other as long as the size and direction of each vector do not change. The resultant vector can then be drawn from the tail of the first vector to the tip of the last vector.

Chapter 3 Properties of Vectors Vectors can be added in any order.
Section 1 Introduction to Vectors Chapter 3 Properties of Vectors Vectors can be added in any order. To subtract a vector, add its opposite. Multiplying or dividing vectors by scalars results in vectors.

Section 2 Vector Operations
Chapter 3 Objectives Identify appropriate coordinate systems for solving problems with vectors. Apply the Pythagorean theorem and tangent function to calculate the magnitude and direction of a resultant vector. Resolve vectors into components using the sine and cosine functions. Add vectors that are not perpendicular.

Coordinate Systems in Two Dimensions
Section 2 Vector Operations Chapter 3 Coordinate Systems in Two Dimensions One method for diagraming the motion of an object employs vectors and the use of the x- and y-axes. Axes are often designated using fixed directions. In the figure shown here, the positive y-axis points north and the positive x-axis points east.

Determining Resultant Magnitude and Direction
Section 2 Vector Operations Chapter 3 Determining Resultant Magnitude and Direction In Section 1, the magnitude and direction of a resultant were found graphically. With this approach, the accuracy of the answer depends on how carefully the diagram is drawn and measured. A simpler method uses the Pythagorean theorem and the tangent function.

Determining Resultant Magnitude and Direction, continued
Section 2 Vector Operations Chapter 3 Determining Resultant Magnitude and Direction, continued The Pythagorean Theorem Use the Pythagorean theorem to find the magnitude of the resultant vector. The Pythagorean theorem states that for any right triangle, the square of the hypotenuse—the side opposite the right angle—equals the sum of the squares of the other two sides, or legs.

Determining Resultant Magnitude and Direction, continued
Section 2 Vector Operations Chapter 3 Determining Resultant Magnitude and Direction, continued The Tangent Function Use the tangent function to find the direction of the resultant vector. For any right triangle, the tangent of an angle is defined as the ratio of the opposite and adjacent legs with respect to a specified acute angle of a right triangle.

Chapter 3 Sample Problem Finding Resultant Magnitude and Direction
Section 2 Vector Operations Chapter 3 Sample Problem Finding Resultant Magnitude and Direction An archaeologist climbs the Great Pyramid in Giza, Egypt. The pyramid’s height is 136 m and its width is 2.30  102 m. What is the magnitude and the direction of the displacement of the archaeologist after she has climbed from the bottom of the pyramid to the top?

Section 2 Vector Operations Chapter 3 Sample Problem, continued 1. Define Given: Dy = 136 m Dx = 1/2(width) = 115 m Unknown: d = ? q = ? Diagram: Choose the archaeologist’s starting position as the origin of the coordinate system, as shown above.

Section 2 Vector Operations Chapter 3 Sample Problem, continued 2. Plan Choose an equation or situation: The Pythagorean theorem can be used to find the magnitude of the archaeologist’s displacement. The direction of the displacement can be found by using the inverse tangent function. Rearrange the equations to isolate the unknowns:

Section 2 Vector Operations Chapter 3 Sample Problem, continued 3. Calculate Evaluate Because d is the hypotenuse, the archaeologist’s displacement should be less than the sum of the height and half of the width. The angle is expected to be more than 45 because the height is greater than half of the width.

Resolving Vectors into Components
Section 2 Vector Operations Chapter 3 Resolving Vectors into Components You can often describe an object’s motion more conveniently by breaking a single vector into two components, or resolving the vector. The components of a vector are the projections of the vector along the axes of a coordinate system. Resolving a vector allows you to analyze the motion in each direction.

Resolving Vectors into Components, continued
Section 2 Vector Operations Chapter 3 Resolving Vectors into Components, continued Consider an airplane flying at 95 km/h. The hypotenuse (vplane) is the resultant vector that describes the airplane’s total velocity. The adjacent leg represents the x component (vx), which describes the airplane’s horizontal speed. The opposite leg represents the y component (vy), which describes the airplane’s vertical speed.

Resolving Vectors into Components, continued
Section 2 Vector Operations Chapter 3 Resolving Vectors into Components, continued The sine and cosine functions can be used to find the components of a vector. The sine and cosine functions are defined in terms of the lengths of the sides of right triangles.

Adding Vectors That Are Not Perpendicular
Section 2 Vector Operations Chapter 3 Adding Vectors That Are Not Perpendicular Suppose that a plane travels first 5 km at an angle of 35°, then climbs at 10° for 22 km, as shown below. How can you find the total displacement? Because the original displacement vectors do not form a right triangle, you can not directly apply the tangent function or the Pythagorean theorem. d2 d1

Adding Vectors That Are Not Perpendicular, continued
Section 2 Vector Operations Chapter 3 Adding Vectors That Are Not Perpendicular, continued You can find the magnitude and the direction of the resultant by resolving each of the plane’s displacement vectors into its x and y components. Then the components along each axis can be added together. As shown in the figure, these sums will be the two perpendicular components of the resultant, d. The resultant’s magnitude can then be found by using the Pythagorean theorem, and its direction can be found by using the inverse tangent function.

Chapter 3 Sample Problem Adding Vectors Algebraically
Section 2 Vector Operations Chapter 3 Sample Problem Adding Vectors Algebraically A hiker walks 27.0 km from her base camp at 35° south of east. The next day, she walks 41.0 km in a direction 65° north of east and discovers a forest ranger’s tower. Find the magnitude and direction of her resultant displacement

Section 2 Vector Operations Chapter 3 Sample Problem, continued 1 . Select a coordinate system. Then sketch and label each vector. Given: d1 = 27.0 km q1 = –35° d2 = 41.0 km q2 = 65° Tip: q1 is negative, because clockwise movement from the positive x-axis is negative by convention. Unknown: d = ? q = ?

Section 2 Vector Operations Chapter 3 Sample Problem, continued 2 . Find the x and y components of all vectors. Make a separate sketch of the displacements for each day. Use the cosine and sine functions to find the components.

Section 2 Vector Operations Chapter 3 Sample Problem, continued 3 . Find the x and y components of the total displacement. 4 . Use the Pythagorean theorem to find the magnitude of the resultant vector. This slide is part of SAMPLE PROBLEM custom show.

Section 2 Vector Operations Chapter 3 Sample Problem, continued 5 . Use a suitable trigonometric function to find the angle.

Chapter 3 Objectives Recognize examples of projectile motion.
Section 3 Projectile Motion Chapter 3 Objectives Recognize examples of projectile motion. Describe the path of a projectile as a parabola. Resolve vectors into their components and apply the kinematic equations to solve problems involving projectile motion.

Section 3 Projectile Motion
Chapter 3 Projectiles Objects that are thrown or launched into the air and are subject to gravity are called projectiles. Projectile motion is the curved path that an object follows when thrown, launched,or otherwise projected near the surface of Earth. If air resistance is disregarded, projectiles follow parabolic trajectories.

Projectiles, continued
Section 3 Projectile Motion Chapter 3 Projectiles, continued Projectile motion is free fall with an initial horizontal velocity. The yellow ball is given an initial horizontal velocity and the red ball is dropped. Both balls fall at the same rate. In this book, the horizontal velocity of a projectile will be considered constant. This would not be the case if we accounted for air resistance.

Kinematic Equations for Projectiles
Section 3 Projectile Motion Chapter 3 Kinematic Equations for Projectiles How can you know the displacement, velocity, and acceleration of a projectile at any point in time during its flight? One method is to resolve vectors into components, then apply the simpler one-dimensional forms of the equations for each component. Finally, you can recombine the components to determine the resultant.

Kinematic Equations for Projectiles, continued
Section 3 Projectile Motion Chapter 3 Kinematic Equations for Projectiles, continued To solve projectile problems, apply the kinematic equations in the horizontal and vertical directions. In the vertical direction, the acceleration ay will equal –g (–9.81 m/s2) because the only vertical component of acceleration is free-fall acceleration. In the horizontal direction, the acceleration is zero, so the velocity is constant.

Kinematic Equations for Projectiles, continued
Section 3 Projectile Motion Chapter 3 Kinematic Equations for Projectiles, continued Projectiles Launched Horizontally The initial vertical velocity is 0. The initial horizontal velocity is the initial velocity. Projectiles Launched At An Angle Resolve the initial velocity into x and y components. The initial vertical velocity is the y component. The initial horizontal velocity is the x component.

Chapter 3 Sample Problem Projectiles Launched At An Angle
Section 3 Projectile Motion Chapter 3 Sample Problem Projectiles Launched At An Angle A zookeeper finds an escaped monkey hanging from a light pole. Aiming her tranquilizer gun at the monkey, she kneels 10.0 m from the light pole,which is 5.00 m high. The tip of her gun is 1.00 m above the ground. At the same moment that the monkey drops a banana, the zookeeper shoots. If the dart travels at 50.0 m/s,will the dart hit the monkey, the banana, or neither one?

Section 3 Projectile Motion Chapter 3 Sample Problem, continued 1 . Select a coordinate system. The positive y-axis points up, and the positive x-axis points along the ground toward the pole. Because the dart leaves the gun at a height of 1.00 m, the vertical distance is 4.00 m.

Section 3 Projectile Motion Chapter 3 Sample Problem, continued 2 . Use the inverse tangent function to find the angle that the initial velocity makes with the x-axis.

Section 3 Projectile Motion Chapter 3 Sample Problem, continued 3 . Choose a kinematic equation to solve for time. Rearrange the equation for motion along the x-axis to isolate the unknown Dt, which is the time the dart takes to travel the horizontal distance.

Section 3 Projectile Motion Chapter 3 Sample Problem, continued 4 . Find out how far each object will fall during this time. Use the free-fall kinematic equation in both cases. For the banana, vi = 0. Thus: Dyb = ½ay(Dt)2 = ½(–9.81 m/s2)(0.215 s)2 = –0.227 m The dart has an initial vertical component of velocity equal to vi sin q, so: Dyd = (vi sin q)(Dt) + ½ay(Dt)2 Dyd = (50.0 m/s)(sin 21.8)(0.215 s) +½(–9.81 m/s2)(0.215 s)2 Dyd = 3.99 m – m = 3.76 m

Section 3 Projectile Motion Chapter 3 Sample Problem, continued 5 . Analyze the results. Find the final height of both the banana and the dart. ybanana, f = yb,i+ Dyb = 5.00 m + (–0.227 m) ybanana, f = 4.77 m above the ground ydart, f = yd,i+ Dyd = 1.00 m m ydart, f = 4.76 m above the ground The dart hits the banana. The slight difference is due to rounding.

Section 4 Relative Motion
Chapter 3 Objectives Describe situations in terms of frame of reference. Solve problems involving relative velocity.

Chapter 3 Frames of Reference
Section 4 Relative Motion Chapter 3 Frames of Reference If you are moving at 80 km/h north and a car passes you going 90 km/h, to you the faster car seems to be moving north at 10 km/h. Someone standing on the side of the road would measure the velocity of the faster car as 90 km/h toward the north. This simple example demonstrates that velocity measurements depend on the frame of reference of the observer.

Frames of Reference, continued
Section 4 Relative Motion Chapter 3 Frames of Reference, continued Consider a stunt dummy dropped from a plane. (a) When viewed from the plane, the stunt dummy falls straight down. (b) When viewed from a stationary position on the ground, the stunt dummy follows a parabolic projectile path.

Chapter 3 Relative Velocity Section 4 Relative Motion
When solving relative velocity problems, write down the information in the form of velocities with subscripts. Using our earlier example, we have: vse = +80 km/h north (se = slower car with respect to Earth) vfe = +90 km/h north (fe = fast car with respect to Earth) unknown = vfs (fs = fast car with respect to slower car) Write an equation for vfs in terms of the other velocities. The subscripts start with f and end with s. The other subscripts start with the letter that ended the preceding velocity: vfs = vfe + ves

Relative Velocity, continued
Section 4 Relative Motion Chapter 3 Relative Velocity, continued An observer in the slow car perceives Earth as moving south at a velocity of 80 km/h while a stationary observer on the ground (Earth) views the car as moving north at a velocity of 80 km/h. In equation form: ves = –vse Thus, this problem can be solved as follows: vfs = vfe + ves = vfe – vse vfs = (+90 km/h n) – (+80 km/h n) = +10 km/h n A general form of the relative velocity equation is: vac = vab + vbc

Chapter 3 Multiple Choice
Standardized Test Prep Multiple Choice 1. Vector A has a magnitude of 30 units. Vector B is perpendicular to vector A and has a magnitude of 40 units. What would the magnitude of the resultant vector A + B be? A. 10 units B. 50 units C. 70 units D. zero

Chapter 3 Standardized Test Prep Multiple Choice, continued 2. What term represents the magnitude of a velocity vector? F. acceleration G. momentum H. speed J. velocity

Chapter 3 Standardized Test Prep Multiple Choice, continued Use the diagram to answer questions 3–4. 3. What is the direction of the resultant vector A + B? A. 15º above the x-axis B. 75º above the x-axis C. 15º below the x-axis D. 75º below the x-axis

Chapter 3 Standardized Test Prep Multiple Choice, continued Use the diagram to answer questions 3–4. 4. What is the direction of the resultant vector A – B? F. 15º above the x-axis G. 75º above the x-axis H. 15º below the x-axis J. 75º below the x-axis

Chapter 3 Standardized Test Prep Multiple Choice, continued Use the passage below to answer questions 5–6. A motorboat heads due east at 5.0 m/s across a river that flows toward the south at a speed of 5.0 m/s. 5. What is the resultant velocity relative to an observer on the shore ? A. 3.2 m/s to the southeast B. 5.0 m/s to the southeast C. 7.1 m/s to the southeast D m/s to the southeast

Chapter 3 Standardized Test Prep Multiple Choice, continued Use the passage below to answer questions 5–6. A motorboat heads due east at 5.0 m/s across a river that flows toward the south at a speed of 5.0 m/s. 6. If the river is 125 m wide, how long does the boat take to cross the river? F. 39 s G. 25 s H. 17 s J. 12 s

Chapter 3 Standardized Test Prep Multiple Choice, continued 7. The pilot of a plane measures an air velocity of 165 km/h south relative to the plane. An observer on the ground sees the plane pass overhead at a velocity of 145 km/h toward the north.What is the velocity of the wind that is affecting the plane relative to the observer? A. 20 km/h to the north B. 20 km/h to the south C. 165 km/h to the north D. 310 km/h to the south

Chapter 3 Standardized Test Prep Multiple Choice, continued 8. A golfer takes two putts to sink his ball in the hole once he is on the green. The first putt displaces the ball 6.00 m east, and the second putt displaces the ball 5.40 m south. What displacement would put the ball in the hole in one putt? F m southeast G m at 48.0º south of east H m at 42.0º south of east J m at 42.0º south of east

Chapter 3 Standardized Test Prep Multiple Choice, continued Use the passage to answer questions 9–12. A girl riding a bicycle at 2.0 m/s throws a tennis ball horizontally forward at a speed of 1.0 m/s from a height of 1.5 m. At the same moment, a boy standing on the sidewalk drops a tennis ball straight down from a height of 1.5 m. 9. What is the initial speed of the girl’s ball relative to the boy? A. 1.0 m/s C. 2.0 m/s B. 1.5 m/s D. 3.0 m/s

Chapter 3 Standardized Test Prep Multiple Choice, continued Use the passage to answer questions 9–12. A girl riding a bicycle at 2.0 m/s throws a tennis ball horizontally forward at a speed of 1.0 m/s from a height of 1.5 m. At the same moment, a boy standing on the sidewalk drops a tennis ball straight down from a height of 1.5 m. 10. If air resistance is disregarded, which ball will hit the ground first? F. the boy’s ball H. neither G. the girl’s ball J. cannot be determined

Chapter 3 Standardized Test Prep Multiple Choice, continued Use the passage to answer questions 9–12. A girl riding a bicycle at 2.0 m/s throws a tennis ball horizontally forward at a speed of 1.0 m/s from a height of 1.5 m. At the same moment, a boy standing on the sidewalk drops a tennis ball straight down from a height of 1.5 m. 11. If air resistance is disregarded, which ball will have a greater speed (relative to the ground) when it hits the ground? A. the boy’s ball C. neither B. the girl’s ball D. cannot be determined

Chapter 3 Standardized Test Prep Multiple Choice, continued Use the passage to answer questions 9–12. A girl riding a bicycle at 2.0 m/s throws a tennis ball horizontally forward at a speed of 1.0 m/s from a height of 1.5 m. At the same moment, a boy standing on the sidewalk drops a tennis ball straight down from a height of 1.5 m. 12. What is the speed of the girl’s ball when it hits the ground? F. 1.0 m/s H. 6.2 m/s G. 3.0 m/s J. 8.4 m/s

Standardized Test Prep Short Response 13. If one of the components of one vector along the direction of another vector is zero, what can you conclude about these two vectors?

Standardized Test Prep Short Response 13. If one of the components of one vector along the direction of another vector is zero, what can you conclude about these two vectors? Answer: They are perpendicular.

Chapter 3 Standardized Test Prep Short Response, continued 14. A roller coaster travels 41.1 m at an angle of 40.0° above the horizontal. How far does it move horizontally and vertically?

Chapter 3 Standardized Test Prep Short Response, continued 14. A roller coaster travels 41.1 m at an angle of 40.0° above the horizontal. How far does it move horizontally and vertically? Answer: 31.5 m horizontally, 26.4 m vertically

Chapter 3 Standardized Test Prep Short Response, continued 15. A ball is thrown straight upward and returns to the thrower’s hand after 3.00 s in the air. A second ball is thrown at an angle of 30.0° with the horizontal. At what speed must the second ball be thrown to reach the same height as the one thrown vertically?

Chapter 3 Standardized Test Prep Short Response, continued 15. A ball is thrown straight upward and returns to the thrower’s hand after 3.00 s in the air. A second ball is thrown at an angle of 30.0° with the horizontal. At what speed must the second ball be thrown to reach the same height as the one thrown vertically? Answer: 29.4 m/s

Standardized Test Prep Extended Response 16. A human cannonball is shot out of a cannon at 45.0° to the horizontal with an initial speed of 25.0 m/s. A net is positioned at a horizontal distance of 50.0 m from the cannon. At what height above the cannon should the net be placed in order to catch the human cannonball? Show your work.

Standardized Test Prep Extended Response 16. A human cannonball is shot out of a cannon at 45.0° to the horizontal with an initial speed of 25.0 m/s. A net is positioned at a horizontal distance of 50.0 m from the cannon. At what height above the cannon should the net be placed in order to catch the human cannonball? Show your work. Answer: 10.8 m

Chapter 3 Standardized Test Prep Extended Response, continued Read the following passage to answer question 17. Three airline executives are discussing ideas for developing flights that are more energy efficient. Executive A: Because the Earth rotates from west to east, we could operate “static flights”—a helicopter or airship could begin by rising straight up from New York City and then descend straight down four hours later when San Francisco arrives below. Executive B: This approach could work for one-way flights, but the return trip would take 20 hours. continued on the next slide

Chapter 3 Standardized Test Prep Extended Response, continued Executive C: That approach will never work. Think about it.When you throw a ball straight up in the air, it comes straight back down to the same point. Executive A: The ball returns to the same point because Earth’s motion is not significant during such a short time. 17. State which of the executives is correct, and explain why.

Chapter 3 Standardized Test Prep Extended Response, continued 17. State which of the executives is correct, and explain why. Answer: Executive C is correct. Explanations should include the concept of relative velocity—when a helicopter lifts off straight up from the ground, it is already moving horizontally with Earth’s horizontal velocity. (We assume that Earth’s motion is constant for the purposes of this scenario and does not depend on time.)

Graphical Addition of Vectors
Section 1 Introduction to Vectors Chapter 3 Graphical Addition of Vectors

Adding Vectors That Are Not Perpendicular
Section 2 Vector Operations Chapter 3 Adding Vectors That Are Not Perpendicular

Section 3 Projectile Motion
Chapter 3 Projectiles

Section 4 Relative Motion
Chapter 3 Frames of Reference

Chapter 7 Circular Motion and Gravitation

Chapter 7 Table of Contents Section 1 Circular Motion
Circular Motion and Gravitation Table of Contents Section 1 Circular Motion Section 2 Newton’s Law of Universal Gravitation Section 3 Motion in Space Section 4 Torque and Simple Machines

Section 1 Circular Motion
Chapter 7 Objectives Solve problems involving centripetal acceleration. Solve problems involving centripetal force. Explain how the apparent existence of an outward force in circular motion can be explained as inertia resisting the centripetal force.

Chapter 7 Tangential Speed
Section 1 Circular Motion Chapter 7 Tangential Speed The tangential speed (vt) of an object in circular motion is the object’s speed along an imaginary line drawn tangent to the circular path. Tangential speed depends on the distance from the object to the center of the circular path. When the tangential speed is constant, the motion is described as uniform circular motion.

Centripetal Acceleration
Section 1 Circular Motion Chapter 7 Centripetal Acceleration

Section 1 Circular Motion Chapter 7 Centripetal Acceleration The acceleration of an object moving in a circular path and at constant speed is due to a change in direction. An acceleration of this nature is called a centripetal acceleration.

Centripetal Acceleration, continued
Section 1 Circular Motion Chapter 7 Centripetal Acceleration, continued (a) As the particle moves from A to B, the direction of the particle’s velocity vector changes. (b) For short time intervals, ∆v is directed toward the center of the circle. Centripetal acceleration is always directed toward the center of a circle.

Centripetal Acceleration, continued
Section 1 Circular Motion Chapter 7 Centripetal Acceleration, continued You have seen that centripetal acceleration results from a change in direction. In circular motion, an acceleration due to a change in speed is called tangential acceleration. To understand the difference between centripetal and tangential acceleration, consider a car traveling in a circular track. Because the car is moving in a circle, the car has a centripetal component of acceleration. If the car’s speed changes, the car also has a tangential component of acceleration.

Chapter 7 Centripetal Force Section 1 Circular Motion
Consider a ball of mass m that is being whirled in a horizontal circular path of radius r with constant speed. The force exerted by the string has horizontal and vertical components. The vertical component is equal and opposite to the gravitational force. Thus, the horizontal component is the net force. This net force, which is is directed toward the center of the circle, is a centripetal force.

Centripetal Force, continued
Section 1 Circular Motion Chapter 7 Centripetal Force, continued Newton’s second law can be combined with the equation for centripetal acceleration to derive an equation for centripetal force:

Section 1 Circular Motion Chapter 7 Centripetal Force, continued Centripetal force is simply the name given to the net force on an object in uniform circular motion. Any type of force or combination of forces can provide this net force. For example, friction between a race car’s tires and a circular track is a centripetal force that keeps the car in a circular path. As another example, gravitational force is a centripetal force that keeps the moon in its orbit.

Section 1 Circular Motion Chapter 7 Centripetal Force, continued If the centripetal force vanishes, the object stops moving in a circular path. A ball that is on the end of a string is whirled in a vertical circular path. If the string breaks at the position shown in (a), the ball will move vertically upward in free fall. If the string breaks at the top of the ball’s path, as in (b), the ball will move along a parabolic path.

Describing a Rotating System
Section 1 Circular Motion Chapter 7 Describing a Rotating System To better understand the motion of a rotating system, consider a car traveling at high speed and approaching an exit ramp that curves to the left. As the driver makes the sharp left turn, the passenger slides to the right and hits the door. What causes the passenger to move toward the door?

Describing a Rotating System, continued
Section 1 Circular Motion Chapter 7 Describing a Rotating System, continued As the car enters the ramp and travels along a curved path, the passenger, because of inertia, tends to move along the original straight path. If a sufficiently large centripetal force acts on the passenger, the person will move along the same curved path that the car does. The origin of the centripetal force is the force of friction between the passenger and the car seat. If this frictional force is not sufficient, the passenger slides across the seat as the car turns underneath.

Section 2 Newton’s Law of Universal Gravitation
Chapter 7 Objectives Explain how Newton’s law of universal gravitation accounts for various phenomena, including satellite and planetary orbits, falling objects, and the tides. Apply Newton’s law of universal gravitation to solve problems.

Chapter 7 Gravitational Force Orbiting objects are in free fall.
Section 2 Newton’s Law of Universal Gravitation Chapter 7 Gravitational Force Orbiting objects are in free fall. To see how this idea is true, we can use a thought experiment that Newton developed. Consider a cannon sitting on a high mountaintop. Each successive cannonball has a greater initial speed, so the horizontal distance that the ball travels increases. If the initial speed is great enough, the curvature of Earth will cause the cannonball to continue falling without ever landing.

Gravitational Force, continued
Section 2 Newton’s Law of Universal Gravitation Chapter 7 Gravitational Force, continued The centripetal force that holds the planets in orbit is the same force that pulls an apple toward the ground—gravitational force. Gravitational force is the mutual force of attraction between particles of matter. Gravitational force depends on the masses and on the distance between them.

Section 2 Newton’s Law of Universal Gravitation Chapter 7 Gravitational Force, continued Newton developed the following equation to describe quantitatively the magnitude of the gravitational force if distance r separates masses m1 and m2: The constant G, called the constant of universal gravitation, equals  10–11 N•m2/kg.

Newton’s Law of Universal Gravitation
Section 2 Newton’s Law of Universal Gravitation Chapter 7 Newton’s Law of Universal Gravitation

Section 2 Newton’s Law of Universal Gravitation Chapter 7 Gravitational Force, continued The gravitational forces that two masses exert on each other are always equal in magnitude and opposite in direction. This is an example of Newton’s third law of motion. One example is the Earth-moon system, shown on the next slide. As a result of these forces, the moon and Earth each orbit the center of mass of the Earth-moon system. Because Earth has a much greater mass than the moon, this center of mass lies within Earth.

Newton’s Law of Universal Gravitation
Section 2 Newton’s Law of Universal Gravitation Chapter 7 Newton’s Law of Universal Gravitation

Applying the Law of Gravitation
Section 2 Newton’s Law of Universal Gravitation Chapter 7 Applying the Law of Gravitation Newton’s law of gravitation accounts for ocean tides. High and low tides are partly due to the gravitational force exerted on Earth by its moon. The tides result from the difference between the gravitational force at Earth’s surface and at Earth’s center.

Applying the Law of Gravitation, continued
Section 2 Newton’s Law of Universal Gravitation Chapter 7 Applying the Law of Gravitation, continued Cavendish applied Newton’s law of universal gravitation to find the value of G and Earth’s mass. When two masses, the distance between them, and the gravitational force are known, Newton’s law of universal gravitation can be used to find G. Once the value of G is known, the law can be used again to find Earth’s mass.

Section 2 Newton’s Law of Universal Gravitation Chapter 7 Applying the Law of Gravitation, continued Gravity is a field force. Gravitational field strength, g, equals Fg/m. The gravitational field, g, is a vector with magnitude g that points in the direction of Fg. Gravitational field strength equals free-fall acceleration. The gravitational field vectors represent Earth’s gravitational field at each point.

Section 2 Newton’s Law of Universal Gravitation Chapter 7 Applying the Law of Gravitation, continued weight = mass  gravitational field strength Because it depends on gravitational field strength, weight changes with location: On the surface of any planet, the value of g, as well as your weight, will depend on the planet’s mass and radius.

Section 3 Motion in Space Chapter 7 Sample Problem, continued 1. Define Given: r1 = 361 km = 3.61  105 m Unknown: T = ? vt = ? 2. Plan Choose an equation or situation: Use the equations for the period and speed of an object in a circular orbit.

Section 3 Motion in Space Chapter 7 Sample Problem, continued Use Table 1 in the textbook to find the values for the radius (r2) and mass (m) of Venus. r2 = 6.05  106 m m = 4.87  1024 kg Find r by adding the distance between the spacecraft and Venus’s surface (r1) to Venus’s radius (r2). r = r1 + r2 r = 3.61  105 m  106 m = 6.41  106 m

Section 3 Motion in Space Chapter 7 Sample Problem, continued 3. Calculate 4. Evaluate Magellan takes (5.66  103 s)(1 min/60 s)  94 min to complete one orbit.

Weight and Weightlessness
Section 3 Motion in Space Chapter 7 Weight and Weightlessness To learn about apparent weightlessness, imagine that you are in an elevator: When the elevator is at rest, the magnitude of the normal force acting on you equals your weight. If the elevator were to accelerate downward at 9.81 m/s2, you and the elevator would both be in free fall. You have the same weight, but there is no normal force acting on you. This situation is called apparent weightlessness. Astronauts in orbit experience apparent weightlessness.

Weight and Weightlessness
Section 3 Motion in Space Chapter 7 Weight and Weightlessness

Chapter 7 Multiple Choice
Standardized Test Prep Multiple Choice 1. An object moves in a circle at a constant speed. Which of the following is not true of the object? A. Its acceleration is constant. B. Its tangential speed is constant. C. Its velocity is constant. D. A centripetal force acts on the object.

Chapter 7 Standardized Test Prep Multiple Choice, continued Use the passage below to answer questions 2–3. A car traveling at 15 m/s on a flat surface turns in a circle with a radius of 25 m. 2. What is the centripetal acceleration of the car? F. 2.4  10-2 m/s2 G m/s2 H. 9.0 m/s2 J. zero

Chapter 7 Standardized Test Prep Multiple Choice, continued Use the passage below to answer questions 2–3. A car traveling at 15 m/s on a flat surface turns in a circle with a radius of 25 m. 3. What is the most direct cause of the car’s centripetal acceleration? A. the torque on the steering wheel B. the torque on the tires of the car C. the force of friction between the tires and the road D. the normal force between the tires and the road

Chapter 7 Standardized Test Prep Multiple Choice, continued 4. Earth (m = 5.97  1024 kg) orbits the sun (m =  1030 kg) at a mean distance of 1.50  1011 m. What is the gravitational force of the sun on Earth? (G =  N•m2/kg2) F  1032 N G  1022 N H  10–2 N J  10–8 N

Chapter 7 Standardized Test Prep Multiple Choice, continued 5. Which of the following is a correct interpretation of the expression ? A. Gravitational field strength changes with an object’s distance from Earth. B. Free-fall acceleration changes with an object’s distance from Earth. C. Free-fall acceleration is independent of the falling object’s mass. D. All of the above are correct interpretations.

Chapter 7 Standardized Test Prep Multiple Choice, continued 12. Which of the following statements is correct? F. Mass and weight both vary with location. G. Mass varies with location, but weight does not. H. Weight varies with location, but mass does not. J. Neither mass nor weight varies with location.

Standardized Test Prep Short Response 14. Explain how it is possible for all the water to remain in a pail that is whirled in a vertical path, as shown below.

Standardized Test Prep Short Response 14. Explain how it is possible for all the water to remain in a pail that is whirled in a vertical path, as shown below. Answer: The water remains in the pail even when the pail is upside down because the water tends to move in a straight path due to inertia.

Chapter 7 Standardized Test Prep Short Response, continued 15. Explain why approximately two high tides take place every day at a given location on Earth.

Chapter 7 Standardized Test Prep Short Response, continued 15. Explain why approximately two high tides take place every day at a given location on Earth. Answer: The moon’s tidal forces create two bulges on Earth. As Earth rotates on its axis once per day, any given point on Earth passes through both bulges.

Section 1 Circular Motion Chapter 7 Centripetal Acceleration

Section 1 Circular Motion
Chapter 7 Centripetal Force

Section 3 Motion in Space
Chapter 7 Kepler’s Laws

Chapter 2,3,7 MASTER CLASS NOTES FOR MOTION Chapter 2,3 & 7

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