PHYSICS Principles and Problems Chapter 6 Motion in

PHYSICS Principles and Problems Chapter 6: Motion in Two Dimensions

CHAPTER 6 Motion in Two Dimensions BIG IDEA You can use vectors and Newton’s laws to describe projectile motion and circular motion.

CHAPTER 6 Table Of Contents Section 6. 1 Projectile Motion Section 6. 2 Circular Motion Section 6. 3 Relative Velocity Click a hyperlink to view the corresponding slides. Exit

Projectile Motion 6. SECTION 1 MAIN IDEA A projectile’s horizontal motion is independent of its vertical motion. Essential Questions • How are the vertical and horizontal motions of a projectile related? • What are the relationships between a projectile’s height, time in the air, initial velocity and horizontal distance traveled?

Projectile Motion 6. SECTION 1 Review Vocabulary • Motion diagram a series of images showing the positions of a moving object taken at regular time intervals. New Vocabulary • Projectile • Trajectory

6. SECTION 1 Projectile Motion Path of a Projectile • If you observed the movement of a golf ball being hit from a tee, a frog hopping, or a free throw being shot with a basketball, you would notice that all of these objects move through the air along similar paths, as do baseballs, and arrows. • Each path rises and then falls, always curving downward along a parabolic path.

6. SECTION 1 Projectile Motion Path of a Projectile (cont. ) • An object shot through the air is called a projectile. • A projectile can be a football or a drop of water. • You can draw a free-body diagram of a launched projectile and identify all the forces that are acting on it.

6. SECTION 1 Projectile Motion Path of a Projectile (cont. ) • No matter what the object is, after a projectile has been given an initial thrust, if you ignore air resistance, it moves through the air only under the force of gravity. • The force of gravity is what causes the object to curve downward in a parabolic flight path. Its path through space is called its trajectory.

6. SECTION 1 Projectile Motion Path of a Projectile (cont. ) • You can determine an object’s trajectory if you know its initial velocity. • Two types of projectile motion are horizontal and angled.

6. SECTION 1 Projectile Motion Independence of Motion in Two Dimensions Click image to view movie.

6. SECTION 1 Projectile Motion Angled Launches • When a projectile is launched at an angle, the initial velocity has a vertical component as well as a horizontal component. • If the object is launched upward, like a ball tossed straight up in the air, it rises with slowing speed, reaches the top of its path, and descends with increasing speed.

6. SECTION 1 Projectile Motion Angled Launches (cont. ) • The adjoining figure shows the separate vertical- and horizontal -motion diagrams for the trajectory of the ball.

6. SECTION 1 Projectile Motion Angled Launches (cont. ) • At each point in the vertical direction, the velocity of the object as it is moving upward has the same magnitude as when it is moving downward. • The only difference is that the directions of the two velocities are opposite.

6. SECTION 1 Projectile Motion Angled Launches (cont. ) • The adjoining figure defines two quantities associated with a trajectory.

6. SECTION 1 Projectile Motion Angled Launches (cont. ) • One is the maximum height, which is the height of the projectile when the vertical velocity is zero and the projectile has only its horizontal-velocity component.

6. SECTION 1 Projectile Motion Angled Launches (cont. ) • The other quantity depicted is the range, R, which is the horizontal distance that the projectile travels when the initial and final heights are the same. • Not shown is the flight time, which is how much time the projectile is in the air.

6. SECTION 1 Projectile Motion Angled Launches (cont. ) • For football punts, flight time is often called hang time.

6. SECTION 1 Projectile Motion The Flight of a Ball A ball is launched at 4. 5 m/s at 66° above the horizontal. What are the maximum height and flight time of the ball?

6. SECTION 1 Projectile Motion The Flight of a Ball (cont. ) Step 1: Analyze and Sketch the Problem • Establish a coordinate system with the initial position of the ball at the origin.

6. SECTION 1 Projectile Motion The Flight of a Ball (cont. ) • Show the positions of the ball at the beginning, at the maximum height, and at the end of the flight.

6. SECTION 1 Projectile Motion The Flight of a Ball (cont. ) • Draw a motion diagram showing v, a, and Fnet.

6. SECTION 1 Projectile Motion The Flight of a Ball (cont. ) Identify the known and unknown variables. Known: Unknown: yi = 0. 0 m ymax = ? θi = 66° t=? vi = 4. 5 m/s ay = − 9. 8 m/s 2 Vy, max = 0. 0 m/s

6. SECTION 1 Projectile Motion The Flight of a Ball (cont. ) Step 2: Solve for the Unknown

6. SECTION 1 Projectile Motion The Flight of a Ball (cont. ) Find the y-component of vi.

6. SECTION 1 Projectile Motion The Flight of a Ball (cont. ) Substitute vi = 4. 5 m/s, θi = 66°

6. SECTION 1 Projectile Motion The Flight of a Ball (cont. ) Use symmetry to find the y-component of vf vyf = - vyi = - 4. 1 m/s

6. SECTION 1 Projectile Motion The Flight of a Ball (cont. ) Solve for the maximum height. Vy, max 2 = vyi 2 + 2 a(ymax – yi) (0. 0 m/s 2) = vyi + 2 a(ymax – 0. 0 m/s) ymax = - vyi / 2 a

6. SECTION 1 Projectile Motion The Flight of a Ball (cont. ) Substitute vyi = 4. 1 m/s, a = -9. 8 m/s 2 ymax = (4. 1 m/s)2 = 0. 86 m 2(-9. 8 m/s 2)

6. SECTION 1 Projectile Motion The Flight of a Ball (cont. ) Solve for the time to return to the launching height. Vyf = vyi + at

Projectile Motion 6. SECTION 1 The Flight of a Ball (cont. ) Solve for t. t = vyf – vyi a

6. SECTION 1 Projectile Motion The Flight of a Ball (cont. ) Substitute vyf = - 4. 1 m/s, vyi = 4. 1 m/s, a = - 9. 80 m/s 2 -4. 1 m/s – 4. 1 m/s = 0. 84 s -9. 8 m/s 2

6. SECTION 1 Projectile Motion The Flight of a Ball (cont. ) Step 3: Evaluate the Answer

6. SECTION 1 Projectile Motion The Flight of a Ball (cont. ) Are the units correct? Dimensional analysis verifies that the units are correct. Do the signs make sense? All of the signs should be positive. Are the magnitudes realistic? 0. 84 s is fast, but an initial velocity of 4. 5 m/s makes this time reasonable.

6. SECTION 1 Projectile Motion The Flight of a Ball (cont. ) The steps covered were: Step 1: Analyze and Sketch the Problem Establish a coordinate system with the initial position of the ball at the origin. Show the positions of the ball at the beginning, at the maximum height, and at the end of the flight. Draw a motion diagram showing v, a, and Fnet.

6. SECTION 1 Projectile Motion The Flight of a Ball (cont. ) The steps covered were: Step 2: Solve for the Unknown Find the y-component of vi. Find an expression for time. Solve for the maximum height. Solve for the time to return to the launching height. Step 3: Evaluate the Answer

6. SECTION 1 Projectile Motion Forces from Air • So far, air resistance has been ignored in the analysis of projectile motion. • Forces from the air can significantly change the motion of an object. – Ex. Flying a kite, parachute for a skydiver.

6. SECTION 1 Projectile Motion Forces from Air (cont. ) • Picture water flowing from a hose: – If wind is blowing in the same direction as the water’s initial movement, the horizontal distance the water travels increases. – If wind is blowing in the opposite direction as the water’s initial movement, the horizontal distance the water travels decreases. • Even if the air is not moving, it can have a significant effect on some moving objects. – Ex. Paper falling to the ground.

6. SECTION 1 Section Check A boy standing on a balcony drops one ball and throws another with an initial horizontal velocity of 3 m/s. Which of the following statements about the horizontal and vertical motions of the balls is correct? (Neglect air resistance. ) A. The balls fall with a constant vertical velocity and a constant horizontal acceleration. B. The balls fall with a constant vertical velocity as well as a constant horizontal velocity. C. The balls fall with a constant vertical acceleration and a constant horizontal velocity. D. The balls fall with a constant vertical acceleration and an increasing horizontal velocity.

6. SECTION 1 Section Check Answer Reason: The vertical and horizontal motions of a projectile are independent. The only force acting on the two balls is the force of gravity. Because it acts in the vertical direction, the balls accelerates in the vertical direction. The horizontal velocity remains constant throughout the flight of the balls.

6. SECTION 1 Section Check Which of the following conditions is met when a projectile reaches its maximum height? A. The vertical component of the velocity is zero. B. The vertical component of the velocity is maximum. C. The horizontal component of the velocity is maximum. D. The acceleration in the vertical direction is zero.

6. SECTION 1 Section Check Answer Reason: The maximum height is the height at which the object stops its upward motion and starts falling down, i. e. when the vertical component of the velocity becomes zero.

6. SECTION 1 Section Check Suppose you toss a ball up and catch it while riding in a bus. Why does the ball fall in your hands rather than falling at the place where you tossed it?

6. SECTION 1 Section Check Answer Trajectory depends on the frame of reference. For an observer on the ground, when the bus is moving, your hand is also moving with the same velocity as the bus, i. e. the bus, your hand, and the ball will have the same horizontal velocity. Therefore, the ball will follow a trajectory and fall back into your hands.

Circular Motion 6. SECTION 2 MAIN IDEA An object in circular motion has an acceleration toward the circle’s center due to an unbalanced force toward the circle’s center. Essential Questions • Why is an object moving in a circle at a constant speed accelerating? • How does centripetal acceleration depend upon the object’s speed and the radius of the circle? • What causes centripetal acceleration?

Circular Motion 6. SECTION 2 Review Vocabulary • Average velocity the change in position divided by the time during which the change occurred; the slope of an object’s position-time graph. New Vocabulary • Uniform circular motion • Centripetal acceleration • Centripetal force

6. SECTION 2 Circular Motion Describing Circular Motion Click image to view movie.

6. SECTION 2 Circular Motion Centripetal Acceleration • The angle between position vectors r 1 and r 2 is the same as that between velocity vectors v 1 and v 2. • Thus, ∆r/r = ∆v/v. The equation does not change if both sides are divided by ∆t.

6. SECTION 2 Circular Motion Centripetal Acceleration (cont. ) • However, v = ∆r/∆t and a = ∆v/∆t • Substituting v = ∆r/∆t in the left-hand side and a = ∆v/∆t in the right-hand side gives the following equation:

6. SECTION 2 Circular Motion Centripetal Acceleration (cont. ) • Solve the equation for acceleration and give it the special symbol ac, for centripetal acceleration. • Centripetal acceleration always points to the center of the circle. Its magnitude is equal to the square of the speed, divided by the radius of motion.

6. SECTION 2 Circular Motion Centripetal Acceleration (cont. ) • One way of measuring the speed of an object moving in a circle is to measure its period, T, the time needed for the object to make one complete revolution. • During this time, the object travels a distance equal to the circumference of the circle, 2πr. The object’s speed, then, is represented by v = 2πr/T.

6. SECTION 2 Circular Motion Centripetal Acceleration (cont. ) • The acceleration of an object moving in a circle is always in the direction of the net force acting on it, there must be a net force toward the center of the circle. This force can be provided by any number of agents. • When an Olympic hammer thrower swings the hammer, the force is the tension in the chain attached to the massive ball.

6. SECTION 2 Circular Motion Centripetal Acceleration (cont. ) • When an object moves in a circle, the net force toward the center of the circle is called the centripetal force. • To analyze centripetal acceleration situations accurately, you must identify the agent of the force that causes the acceleration. Then you can apply Newton’s second law for the component in the direction of the acceleration in the following way.

6. SECTION 2 Circular Motion Centripetal Acceleration (cont. ) • Newton’s Second Law for Circular Motion • The net centripetal force on an object moving in a circle is equal to the object’s mass times the centripetal acceleration.

6. SECTION 2 Circular Motion Centripetal Acceleration (cont. ) • When solving problems, it is useful to choose a coordinate system with one axis in the direction of the acceleration. • For circular motion, the direction of the acceleration is always toward the center of the circle.

6. SECTION 2 Circular Motion Centripetal Acceleration (cont. ) • Rather than labeling this axis x or y, call it c, for centripetal acceleration. The other axis is in the direction of the velocity, tangent to the circle. It is labeled tang for tangential. • Centripetal force is just another name for the net force in the centripetal direction. It is the sum of all the real forces, those for which you can identify agents that act along the centripetal axis.

6. SECTION 2 Circular Motion Centrifugal “Force” • According to Newton’s first law, you will continue moving with the same velocity unless there is a net force acting on you. • The passenger in the car would continue to move straight ahead if it were not for the force of the car acting in the direction of the acceleration.

6. SECTION 2 Circular Motion Centrifugal “Force” (cont. ) • The so-called centrifugal, or outward force, is a fictitious, nonexistent force. • You feel as if you are being pushed only because you are accelerating relative to your surroundings. There is no real force because there is no agent exerting a force.

6. SECTION 2 Section Check Explain why an object moving in a circle at a constant speed is accelerating.

6. SECTION 2 Section Check Answer Acceleration is the rate of change of velocity, the object is accelerating due to its constant change in the direction of its motion.

6. SECTION 2 Section Check What is the relationship between the magnitude of centripetal acceleration (ac) and an object’s speed (v)? A. B. C. D.

6. SECTION 2 Section Check Answer Reason: From the equation for centripetal acceleration: Centripetal acceleration always points to the center of the circle. Its magnitude is equal to the square of the speed divided by the radius of the motion.

6. SECTION 2 Section Check What is the direction of the velocity vector of an accelerating object? A. toward the center of the circle B. away from the center of the circle C. along the circular path D. tangent to the circular path

Section Check 6. SECTION 2 Answer Reason: While constantly changing, the velocity vector for an object in uniform circular motion is always tangent to the circle. Vectors are never curved and therefore cannot be along a circular path.

Relative Velocity 6. SECTION 3 MAIN IDEA An object’s velocity depends on the reference frame chosen. Essential Questions • What is relative velocity? • How do you find the velocities of an object in different reference frames?

Relative Velocity 6. SECTION 3 Review Vocabulary • resultant a vector that results from the sum of two other vectors. New Vocabulary • Reference frame

6. SECTION 3 Relative Velocity Relative Motion in One Dimension • Suppose you are in a school bus that is traveling at a velocity of 8 m/s in a positive direction. You walk with a velocity of 1 m/s toward the front of the bus. • If a friend of yours is standing on the side of the road watching the bus go by, how fast would your friend say that you are moving?

6. SECTION 3 Relative Velocity Relative Motion in One Dimension (cont. ) • If the bus is traveling at 8 m/s, this means that the velocity of the bus is 8 m/s, as measured by your friend in a coordinate system fixed to the road. • When you are standing still, your velocity relative to the road is also 8 m/s, but your velocity relative to the bus is zero.

6. SECTION 3 Relative Velocity Relative Motion in One Dimension (cont. ) • In the previous example, your motion is viewed from different coordinate systems. • A coordinate system from which motion is viewed is a reference frame.

6. SECTION 3 Relative Velocity Relative Motion in One Dimension (cont. ) • Walking at 1 m/s towards the front of the bus means your velocity is measured in the reference from of the bus. • Your velocity in the road’s reference frame is different • You can rephrase the problem as follows: given the velocity of the bus relative to the road and your velocity relative to the bus, what is your velocity relative to the road?

6. SECTION 3 Relative Velocity Relative Motion in One Dimension (cont. ) • When a coordinate system is moving, two velocities are added if both motions are in the same direction, and one is subtracted from the other if the motions are in opposite directions. • In the given figure, you will find that your velocity relative to the street is 9 m/s, the sum of 8 m/s and 1 m/s.

6. SECTION 3 Relative Velocity Relative Motion in One Dimension (cont. ) • You can see that when the velocities are along the same line, simple addition or subtraction can be used to determine the relative velocity.

6. SECTION 3 Relative Velocity Relative Motion in One Dimension (cont. ) • Mathematically, relative velocity is represented as vy/b + vb/r = vy/r. • The more general form of this equation is: Relative Velocity va/b + vb/c + va/c • The relative velocity of object a to object c is the vector sum of object a’s velocity relative to object b and object b’s velocity relative to object c.

6. SECTION 3 Relative Velocity Relative Motion in Two Dimensions • The method for adding relative velocities also applies to motion in two dimensions. • As with one-dimensional motion, you first draw a vector diagram to describe the motion and then you solve the problem mathematically.

6. SECTION 3 Relative Velocity Relative Motion in Two Dimensions (cont. ) • For example, airline pilots must take into account the plane’s speed relative to the air, and their direction of flight relative to the air. They also must consider the velocity of the wind at the altitude they are flying relative to the ground.

6. SECTION 3 Relative Velocity Relative Motion in Two Dimensions (cont. ) • You can use the equations in the figure to solve problems for relative motion in two dimensions. • Velocity of a reference frame moving relative to the ground is. • Velocity of an object in the moving frame is.

6. SECTION 3 Relative Velocity of a Marble Ana and Sandra are riding on a ferry boat that is traveling east at a speed of 4. 0 m/s. Sandra rolls a marble with a velocity of 0. 75 m/s north, straight across the deck of the boat to Ana. What is the velocity of the marble relative to the water?

6. SECTION 3 Relative Velocity of a Marble (cont. ) Step 1: Analyze and Sketch the Problem • Establish a coordinate system.

6. SECTION 3 Relative Velocity of a Marble (cont. ) • Draw vectors to represent the velocities of the boat relative to the water and the marble relative to the boat.

6. SECTION 3 Relative Velocity of a Marble (cont. ) Identify known and unknown variables. Known: Unknown: vb/w = 4. 0 m/s vm/w = ? vm/b = 0. 75 m/s

6. SECTION 3 Relative Velocity of a Marble (cont. ) Step 2: Solve for the Unknown

6. SECTION 3 Relative Velocity of a Marble (cont. ) Because the two velocities are at right angles, use the Pythagorean theorem.

6. SECTION 3 Relative Velocity of a Marble (cont. ) Substitute vb/w = 4. 0 m/s, vm/b = 0. 75 m/s

6. SECTION 3 Relative Velocity of a Marble (cont. ) Find the angle of the marble’s motion.

6. SECTION 3 Relative Velocity of a Marble (cont. ) Substitute vb/w = 4. 0 m/s, vm/b = 0. 75 m/s = 11° north of east The marble is traveling 4. 1 m/s at 11° north of east.

6. SECTION 3 Relative Velocity of a Marble (cont. ) Step 3: Evaluate the Answer

6. SECTION 3 Relative Velocity of a Marble (cont. ) Are the units correct? Dimensional analysis verifies that the velocity is in m/s. Do the signs make sense? The signs should all be positive. Are the magnitudes realistic? The resulting velocity is of the same order of magnitude as the velocities given in the problem.

6. SECTION 3 Relative Velocity of a Marble (cont. ) The steps covered were: Step 1: Analyze and Sketch the Problem Establish a coordinate system. Draw vectors to represent the velocities of the boat relative to the water and the marble relative to the boat.

6. SECTION 3 Relative Velocity of a Marble (cont. ) The steps covered were: Step 2: Solve for the Unknown Use the Pythagorean theorem. Step 3: Evaluate the Answer

6. SECTION 3 Section Check Steven is walking on the top level of a double-decker bus with a velocity of 2 m/s toward the rear end of the bus. The bus is moving with a velocity of 10 m/s. What is the velocity of Steven with respect to Anudja, who is sitting on the top level of the bus and to Mark, who is standing on the street? A. The velocity of Steven with respect to Anudja is 2 m/s and is 12 m/s with respect to Mark. B. The velocity of Steven with respect to Anudja is 2 m/s and is 8 m/s with respect to Mark. C. The velocity of Steven with respect to Anudja is 10 m/s and is 12 m/s with respect to Mark. D. The velocity of Steven with respect to Anudja is 10 m/s and is 8 m/s with respect to Mark.

6. SECTION 3 Section Check Answer Reason: The velocity of Steven with respect to Anudja is 2 m/s since Steven is moving with a velocity of 2 m/s with respect to the bus, and Anudja is at rest with respect to the bus. The velocity of Steven with respect to Mark can be understood with the help of the following vector representation.

6. SECTION 3 Section Check Which of the following formulas correctly relates the relative velocities of objects a, b, and c to each other? A. va/b + va/c = vb/c B. va/b vb/c = va/c C. va/b + vb/c = va/c D. va/b va/c = vb/c

6. SECTION 3 Section Check Answer Reason: The relative velocity equation is va/b + vb/c = va/c. The relative velocity of object a to object c is the vector sum of object a’s velocity relative to object b and object b’s velocity relative to object c.

6. SECTION 3 Section Check An airplane flies due south at 100 km/hr relative to the air. Wind is blowing at 20 km/hr to the west relative to the ground. What is the plane’s speed with respect to the ground? A. (100 + 20) km/hr B. (100 − 20) km/hr C. D.

Section Check 6. SECTION 3 Answer Reason: Since the two velocities are at right angles, we can apply the Pythagorean theorem. By using relative velocity law, we can write: vp/a 2 + va/g 2 = vp/g 2

Motion in Two Dimensions CHAPTER 6 Resources Physics Online Study Guide Chapter Assessment Questions Standardized Test Practice

Projectile Motion 6. SECTION 1 Study Guide • The vertical and horizontal motions of a projectile are independent. When there is no air resistance, the horizontal motion component does not experience an acceleration and has constant velocity; the vertical motion component of a projectile experiences a constant acceleration under these same conditions.

Projectile Motion 6. SECTION 1 Study Guide • The curved flight path a projectile follows is called a trajectory and is a parabola. The height, time of flight, initial velocity and horizontal distance of this path are related by the equations of motion. The horizontal distance a projectile travels before returning to its initial height depends on the acceleration due to gravity an on both components on the initial velocity.

Circular Motion 6. SECTION 2 Study Guide • An object moving in a circle at a constant speed has an acceleration toward the center of the circle because the direction of its velocity is constantly changing. • Acceleration toward the center of the circle is called centripetal acceleration. It depends directly on the square of the object’s speed and inversely on the radius of the circle.

Circular Motion 6. SECTION 2 Study Guide • A net force must be exerted by external agents toward the circle’s center to cause centripetal acceleration.

Relative Velocity 6. SECTION 3 Study Guide • A coordinate system from which you view motion is called a reference frame. Relative velocity is the velocity of an object observed in a different, moving reference frame. • You can use vector addition to solve motion problems of an object in a moving reference frame.

CHAPTER 6 Motion in Two Dimensions Chapter Assessment What is the range of a projectile? A. the total trajectory that the projectile travels B. the vertical distance that the projectile travels C. the horizontal distance that the projectile travels D. twice the maximum height of the projectile

CHAPTER 6 Motion in Two Dimensions Chapter Assessment Reason: When a projectile is launched at an angle, the straight (horizontal) distance the projectile travels is known as the range of the projectile.

CHAPTER 6 Motion in Two Dimensions Chapter Assessment Define the flight time of a trajectory. A. time taken by the projectile to reach the maximum height B. the maximum height reached by the projectile divided by the magnitude of the vertical velocity C. the total time the projectile was in the air D. half the total time the projectile was in the air

CHAPTER 6 Motion in Two Dimensions Chapter Assessment Reason: The flight time of a trajectory is defined as the total time the projectile was in the air.

CHAPTER 6 Motion in Two Dimensions Chapter Assessment What is centripetal force? Answer: When an object moves in a circle, the net force toward the center of the circle is called centripetal force.

CHAPTER 6 Motion in Two Dimensions Chapter Assessment Donna is traveling in a train due north at 30 m/s. What is the magnitude of the velocity of Donna with respect to another train which is running due south at 30 m/s? A. 30 m/s + 30 m/s B. 30 m/s − 30 m/s C. 302 m/s + 302 m/s D. 302 m/s − 302 m/s

CHAPTER 6 Motion in Two Dimensions Chapter Assessment Reason: The magnitude of the velocity of Donna relative to the ground is 30 m/s and the magnitude of velocity of the other train relative to the ground is also 30 m/s. Now, with this speed, if the trains move in the same direction, then the relative speed of one train to another will be zero.

CHAPTER 6 Motion in Two Dimensions Chapter Assessment Reason: In this case, since the two trains are moving in opposite directions, the relative speed (magnitude of the velocity) of Donna relative to another train is 30 m/s + 30 m/s.

CHAPTER 6 Motion in Two Dimensions Chapter Assessment What is the relationship between the magnitude of centripetal acceleration and the radius of a circle? A. B. C. D.

CHAPTER 6 Motion in Two Dimensions Chapter Assessment Reason: Centripetal acceleration always points to the center of the circle. Its magnitude is equal to the square of the speed, divided by the radius of motion. That is, Therefore,

CHAPTER 6 Motion in Two Dimensions Chapter Assessment Which of the following formulas can be used to calculate the period (T) of a rotating object if the centripetal acceleration (ac) and radius (r) are given? A. C. B. D.

CHAPTER 6 Motion in Two Dimensions Chapter Assessment Reason: We know that the centripetal acceleration always points toward the center of the circle. Its magnitude is equal to the square of the speed divided by the radius of motion. That is, ac = v 2/r. To measure the speed of an object moving in a circle, we measure the period, T, the time needed for the object to make one complete revolution. That is, v = 2πr/T, where 2πr is the circumference of the circle.

CHAPTER 6 Motion in Two Dimensions Chapter Assessment Reason: Substituting for v in the equation v = 2πr/T, we get,

CHAPTER 6 Motion in Two Dimensions Standardized Test Practice A 1. 60 -m tall girl throws a football at an angle of 41. 0° from the horizontal and at an initial velocity of 9. 40 m/s. How far away from the girl will it land? A. 4. 55 m B. 5. 90 m C. 8. 90 m D. 10. 5 m

CHAPTER 6 Motion in Two Dimensions Standardized Test Practice A dragonfly is sitting on a merry-go-round, 2. 8 m from the center. If the tangential velocity of the ride is 0. 89 m/s, what is the centripetal acceleration of the dragonfly? A. 0. 11 m/s 2 B. 0. 28 m/s 2 C. 0. 32 m/s 2 D. 2. 2 m/s 2

CHAPTER 6 Motion in Two Dimensions Standardized Test Practice The centripetal force on a 0. 82 -kg object on the end of a 2. 0 -m massless string being swung in a horizontal circle is 4. 0 N. What is the tangential velocity of the object? A. 2. 8 m/s 2 B. 3. 1 m/s 2 C. 4. 9 m/s 2 D. 9. 8 m/s 2

CHAPTER 6 Motion in Two Dimensions Standardized Test Practice A 1000 -kg car enters an 80 -m radius curve at 20 m/s. What centripetal force must be supplied by friction so the car does not skid? A. 5. 0 N B. 2. 5× 102 N C. 5. 0× 103 N D. 1. 0× 103 N

CHAPTER 6 Motion in Two Dimensions Standardized Test Practice A jogger on a riverside path sees a rowing team coming toward him. If the jogger is moving at 10 km/h, and the boat is moving at 20 km/h, how quickly does the jogger approach the boat? A. 3 m/s B. 8 m/s C. 40 m/s D. 100 m/s

CHAPTER 6 Motion in Two Dimensions Standardized Test Practice Test-Taking Tip Practice Under Testlike Conditions Answer all of the questions in the time provided without referring to your book. Did you complete the test? Could you have made better use of your time? What topics do you need to review?

CHAPTER 6 Motion in Two Dimensions Chapter Resources Relative Velocity Another example of combined relative velocities is the navigation of migrating neotropical songbirds. In addition to knowing in which direction to fly, a bird must account for its speed relative to the air and its direction relative to the ground.

CHAPTER 6 Motion in Two Dimensions Chapter Resources Relative Velocity If a bird tries to fly over the Gulf of Mexico into a headwind that is too strong, it will run out of energy before it reaches the other shore and will perish. Similarly, the bird must account for crosswinds or it will not reach its destination.

CHAPTER 6 Motion in Two Dimensions Chapter Resources The Flight of a Ball A ball is launched at 4. 5 m/s at 66° above the horizontal. What are the maximum height and flight time of the ball?

CHAPTER 6 Motion in Two Dimensions Chapter Resources Relative Velocity of a Marble Ana and Sandra are riding on a ferry boat that is traveling east at a speed of 4. 0 m/s. Sandra rolls a marble with a velocity of 0. 75 m/s north, straight across the deck of the boat to Ana. What is the velocity of the marble relative to the water?