Motion In 3D PDF

Unit 4: Motion in Two Dimensions Lesson 4.2 Motion Descriptors in Three Dimensions Contents Introduction 1 Learning Objectives 2 Warm Up...

Unit 4: Motion in Two Dimensions Lesson 4.2 Motion Descriptors in Three Dimensions Contents Introduction 1 Learning Objectives 2 Warm Up 2 Learn about It! 4 Position Vectors 4 Velocity Vectors 6 Independence of Motion 15 Acceleration Vector 15 Parallel and Perpendicular Components of Acceleration 23 Key Points 25 Key Formulas 26 Check Your Understanding 29 Challenge Yourself 32 Bibliography 33 Key to Try It! 34 Unit 4: Motion in Two Dimensions Lesson 4.2 Motion Descriptors in Three Dimensions Introduction In the past lesson, you learned about motion in one or two dimensions. It is easier to drive a car on a freeway since it is limited to one or two dimensions. However, stability and control for airplanes are much more complicated since they can freely move in three dimensions. Airplanes can rotate in a front-to-back axis, side-to-side axis, or around a vertical axis. You can also think of it as three imaginary lines passing through the airplane, which intersect at right angles. Motions in these dimensions should be carefully controlled 4.2. Motion Descriptors in Three Dimensions 1 Unit 4: Motion in Two Dimensions and maintained by the pilot. In this lesson, you will learn about how the different motion descriptors are defined in three dimensions. Learning Objectives DepEd Competency In this lesson, you should be able to do the Extend the definition of position, velocity, and acceleration to 2D and following: 3D using vector representation Define position, velocity, and (STEM_GP12KIN-Ic21). acceleration in three dimensions. Use vectors to represent the position and velocity of an object in three dimensions. Solve problems involving position, velocity, and acceleration vectors in two to three dimensions. Warm Up Ladybug Motion in 2D 10 minutes In this activity, you will review the motion descriptors in one and two dimensions. Materials laptop/tablet/phone paper pen or pencil 4.2. Motion Descriptors in Three Dimensions 2 Unit 4: Motion in Two Dimensions Procedure 1. Download the PhET simulation using the link below: Ladybug Motion 2D University of Colorado Boulder, “Ladybug Motion 2D,” PhET Interactive Simulations, https://phet.colorado.edu/en/simulation/ladybug-motion-2d, last accessed on March 17 2020. 2. Begin by enabling the following options: “Show both” for Vectors, “Manual” for Choose Motion, “Line” for Trace, and “Position” for Remote Control, as shown in Fig. 4.2.1. Fig. 4.2.1. Ladybug in motion 2D simulation 3. Move the ladybug manually and observe how the vectors change. 4. Try also the “Velocity” and “Acceleration” for Remote Control. Observe how the vectors change as well. 5. Try the linear, circular, and the elliptical motion options. Observe how the velocity and the acceleration changes for each motion. 6. Answer the guide questions below with your seatmate and write your answers on a 4.2. Motion Descriptors in Three Dimensions 3 Unit 4: Motion in Two Dimensions piece of paper. Make sure to discuss your observations. Guide Questions 1. What happens to the velocity and acceleration vectors as you move the ladybug manually? Describe in detail. 2. How did the velocity and acceleration change when you allow the ladybug to move in a linear motion? in a circular motion? in an elliptical motion? 3. How will you describe the direction of the velocity and acceleration vectors with respect to the ladybug’s position in a circular motion? Learn about It! In the past lessons you learned about motion descriptors in one and two dimensions. We can extend these concepts into three dimensions in the following sections. How do you define position, velocity, and acceleration in three dimensions? Position Vectors In describing a three dimensional motion, the three coordinates x, y, and z are utilized to locate a particle at a specific point. In this case, the position vector of a particle point P would be P (x, y, z), as shown in Fig. 4.2.2. When the particle is in motion, all these variables x, y, and z will be functions of time (t). Remember that in the past lesson, the unit vectors , , and correspond to the coordinate axes x, y, z, respectively. Using these concepts, the position vector is expressed as: Equation 4.2.1 4.2. Motion Descriptors in Three Dimensions 4 Unit 4: Motion in Two Dimensions Fig. 4.2.2. The components of position vector in three dimensions In Equation 4.2.1, the coefficients x, y, and z indicate the location of a particle in the coordinate system relative to the origin. As the particle moves, the position vector always changes and extends from the particle to the origin. You learned in the past lessons that displacement is the change in position of a particle or an object. When a particle moves from position P1 with a position vector to P2 with a position vector , at a time interval Δt, the change in position or displacement is expressed as: or Equation 4.2.2 4.2. Motion Descriptors in Three Dimensions 5 Unit 4: Motion in Two Dimensions Velocity Vectors The average velocity at a given time interval is similar to how we deal with the average velocity in one dimension, as illustrated in Fig. 4.2.3. It can be expressed as the change in position vector divided by the time interval as shown by: Fig. 4.2.3. Average velocity is the change in position vector per time interval This expression can be expanded depending on the components of the given position vector. In the past lesson, you learned that instantaneous velocity is the velocity of an object or a particle at a specific instant. This occurs as time interval approaches zero at that instant. In the limit , the average velocity approaches the instantaneous velocity. Instantaneous velocity in two or three dimensions is also expressed similar to one dimensional motion. The main difference is that both the position and instantaneous velocity are vectors. It can be expressed as: 4.2. Motion Descriptors in Three Dimensions 6 Unit 4: Motion in Two Dimensions The magnitude of the instantaneous velocity is equivalent to the speed, while its direction is given by the direction in which the particle is moving at that specific instant. The direction of the instantaneous velocity of any particle is always tangent to the path of the particle at specific particle’s position. When illustrated using an arrow, its tail represents the direction of travel of a particle, while its length represents the magnitude, as shown in Fig. 4.2.4. Fig. 4.2.4. Velocity of a particle Remember The direction of instantaneous velocity is always tangent to the path of the particle’s position. Instantaneous velocity can be divided into its components. It is expressed in terms of unit vectors as: Equation 4.2.3 4.2. Motion Descriptors in Three Dimensions 7 Unit 4: Motion in Two Dimensions The same equation can also be obtained by getting the derivative of Equation 4.2.1. The unit vectors are not dependent on time, therefore, their derivatives are zero. It can also be expressed in terms of unit vectors. The scalar components are given by: The magnitude of the instantaneous velocity is simply the speed of the object. It can be calculated using the Pythagorean theorem using the components of the speed. How will you define position and velocity vectors in three dimensions? Let’s Practice! Example 1 A dog runs across an open field. The coordinates (in meters) of the dog’s position as function of time t (in seconds) are given by x = ‒0.40t2 + 6.5t + 30 y = 0.11t2 ‒ 8.5t + 25 (a) What is the dog’s position vector at t = 10 s in vector notation? (b) What is the magnitude of this position vector? 4.2. Motion Descriptors in Three Dimensions 8 Unit 4: Motion in Two Dimensions Solution Step 1: Identify what is required in the problem. You are asked to calculate the dog’s position vector in vector notation and the magnitude of the position vector. Step 2: Identify the given in the problem. The coordinates of the dog’s position as a function of time are given. Step 3: Write the working equation. The position vector of the dog as a function of time is given by: To determine the magnitude of the position vector, use the Pythagorean theorem. Step 4: Substitute the given values. At t = 10 s, the components are: To solve the magnitude of the position vector: Step 5: Find the answer. The position vector in vector notation is. The magnitude of the position vector is 74 m. 4.2. Motion Descriptors in Three Dimensions 9 Unit 4: Motion in Two Dimensions 1 Try It! What is the (a) position vector in vector notation and (b) magnitude of the position vector of a particle at t = 2.5 s? The coordinates of the particle’s position (in meters) as function of time are as follows: x = 5t2 ‒ 3.2t + 15 y = ‒2t2 ‒ 4.2t + 5 z = 0.5t2 + 6t ‒ 10 Example 2 What is the instantaneous velocity of the dog in vector notation in Let’s Practice! Example 1 at time t = 10 s? Solution Step 1: Identify what is required in the problem. You are asked to calculate the instantaneous velocity of the dog in vector notation. Step 2: Identify the given in the problem. The coordinates of the dog’s position as a function of time are given. Step 3: Write the working equation. The determine the velocity in vector notation, calculate the first derivative of the given components using: Step 4: Substitute the given values. Determine the first derivative of the x and y-components. 4.2. Motion Descriptors in Three Dimensions 10 Unit 4: Motion in Two Dimensions Substitute t=10 s to the x- and y-components of velocity. Step 5: Find the answer. The instantaneous velocity of the dog in vector notation is. 2 Try It! Find the velocity of a particle in vector notation at t = 3 s if the coordinates of the particle’s position (in meters) as a function of time are given as: x = 0.5t2 + 2.5t + 5 y = ‒0.8t2 + 3t + 4.5 z = 0.6t2 ‒ 9t ‒ 10 Example 3 A car, represented as a particle, has the following components (in meters) as a function of time (in seconds). x = 2.1t2 ‒ 4.3t + 10 y = ‒0.9t2 + 8t ‒ 5 z = 3.5t2 ‒ 2.5t ‒ 15 4.2. Motion Descriptors in Three Dimensions 11 Unit 4: Motion in Two Dimensions (a) What is the position vector in vector notation form, and the magnitude of the position of the car at time t = 20 s? (b) What is the velocity of the car in vector notation form at t = 20 s? (c) What is the magnitude of the velocity vector at t = 20 s? Solution Step 1: Identify what is required in the problem. You are asked to calculate the position vector in vector notation form, magnitude of the position vector, velocity of the car in vector notation form, and the magnitude of the velocity vector. Step 2: Identify the given in the problem. The components of the car’s position are given. Step 3: Write the working equation. The position vector of the dog as a function of time is given by: To determine the magnitude of the position vector, use the Pythagorean theorem. To determine the velocity in vector notation, calculate the first derivative of the given components as: To determine the magnitude of the velocity , use. 4.2. Motion Descriptors in Three Dimensions 12 Unit 4: Motion in Two Dimensions Step 4: Substitute the given values. At t = 20 s, the components of the position vector are: For the magnitude of the position vector: The first derivative of the x, y, and z-components are: Substitute t = 20 s to the components of the velocity. For the magnitude of the velocity vector, use. 4.2. Motion Descriptors in Three Dimensions 13 Unit 4: Motion in Two Dimensions Step 5: Find the answer. The position vector of the car in vector notation is. The magnitude of the position vector is 1551.76 m. The velocity of the car in vector notation is. The magnitude of the velocity vector is 161.38 m/s. 3 Try It! The components of a the position of a moving particle (in meters) as a function of time (in seconds) are given by x = 10t2 + 0.5t + 10 y = ‒5t2 + 6.1t ‒ 10 z = ‒0.3t2 ‒ 9t + 20 (a) What is the position vector in vector notation form, and the magnitude of the position of the car at time t = 0.5 s? (b) What is the velocity of the car in vector notation form at t = 0.5 s? (c) What is the magnitude of the velocity vector at t = 0.5 s? 4.2. Motion Descriptors in Three Dimensions 14 Unit 4: Motion in Two Dimensions Independence of Motion Notice that in both position and velocity vectors, the components are separate functions of time. It means that a motion along the x-axis has no effect on the motion of the object along the y- and z-axes. Thus, the motion of the object can be divided into separate components along each corresponding coordinate axis. For example, a person walks from point A to point B. From point A, her path includes several meters in the north direction, then another few meters to the east direction to reach point B. Her motion along the north direction is independent of her motion along the east direction. These independence of motion will be deemed useful in discussing free-fall and projectile motions in the next lessons. Acceleration Vector In one-dimensional motion, we describe acceleration as the change in velocity of a particle divided by the unit time. It is characterized by the changes in the magnitude of the velocity and the changes in the direction of velocity. Suppose a particle has a velocity of at t1 and at t2. During the time interval t2 - t1 ( ), the change in velocity is given by ( ). Using these variables, the average acceleration can be expressed as: In the same way, instantaneous acceleration is given by: This occurs when the object at point P2 approaches point P1, wherein Δt approaches zero. 4.2. Motion Descriptors in Three Dimensions 15 Unit 4: Motion in Two Dimensions When dealing with acceleration in three dimensions, acceleration vector can be expressed in terms of unit vectors, as shown in Equation 4.2.4. It is also similar to the derivative of the components of velocity as a function of time. or Equation 4.2.4 The scalar components are given by: Remember that velocity is the first derivative of the position vector as a function of time. Hence, we can express acceleration as the second derivative of their corresponding components as shown below: How will you define acceleration in three dimensions? 4.2. Motion Descriptors in Three Dimensions 16 Unit 4: Motion in Two Dimensions Let’s Practice! Example 4 Using the given components of velocity in Let’s Practice! Example 2, calculate the acceleration in vector notation form at t = 5 s. Solution Step 1: Identify what is required in the problem. You are asked to calculate the acceleration of the dog in vector notation form. Step 2: Identify the given in the problem. The components of the dog’s velocity are given. Step 3: Write the working equation. The acceleration vector can be determined by calculating the first derivative of the components of the velocity: Step 4: Substitute the given values. To find the x-component of , use. To find the y-component of , use. Step 5: Find the answer. 4.2. Motion Descriptors in Three Dimensions 17 Unit 4: Motion in Two Dimensions The acceleration vector in vector notation form is. 4 Try It! What is the acceleration vector in the vector notation form of a particle if the components of its velocity vector are given as vx = 5t ‒ 10.2, vy = 2.5t + 6.2, and vz = ‒0.75t ‒ 3.1? Example 5 The components of an object’s position (in meters) as functions of time (in seconds) are given by the following: x = ‒5t2 + 3.1t ‒ 6 y = 0.4t2 ‒ 8.2t ‒ 3.5 z = 1.5t2 + 7t ‒ 2.8 (a) What is the acceleration of the object in vector notation form? (b) What is the magnitude of the object’s acceleration? Solution Step 1: Identify what is required in the problem. You are asked to calculate the acceleration in vector notation form and the magnitude of the object’s acceleration. Step 2: Identify the given in the problem. The components of the object’s position are given. Step 3: Write the working equation. Since the position vectors are given, the acceleration vector can be determined by calculating the second derivative of the components of the position vector. 4.2. Motion Descriptors in Three Dimensions 18 Unit 4: Motion in Two Dimensions The magnitude of the acceleration vector can be determined using the Pythagorean theorem. Step 4: Substitute the given values. Calculate the second derivative of the position vectors. To calculate the magnitude of the acceleration vector: Step 5: Find the answer. 4.2. Motion Descriptors in Three Dimensions 19 Unit 4: Motion in Two Dimensions The acceleration vector in terms of notation vector is. The magnitude of the acceleration is 10.47 m/s. 2 5 Try It! Find the acceleration vector in vector notation form as well as its magnitude at t = 2 s. The following components of the velocity vector (in m/s) as a function of time are given: vx= 6.5t2 ‒ 0.6t + 10 vy= 9.1t2 + 2t ‒ 5 vz= ‒0.8t2 + 0.9t ‒ 2 Example 6 A vehicle used to explore the surface of Mars has the following x-, y- and z- coordinates that vary with time x = 6.5 m ‒ (0.45 m/s2)t2 y = (2.5 m/s)t ‒ (0.30 m/s2)t2 z = (1.5 m/s)t + (0.25 m/s3)t3 (a) Find the components of the average acceleration for the interval t = 0.0 s to t = 3.0 s. (b) Find the instantaneous acceleration at t = 3.0 s. (c) Find the magnitude of the acceleration at t = 3.0 s. Solution Step 1: Identify what is required in the problem. You are asked to calculate the components of the average acceleration, the instantaneous acceleration at t = 3.0 s, and the magnitude of the acceleration. Step 2: Identify the given in the problem. The components of the position vector are given as well as the time interval. 4.2. Motion Descriptors in Three Dimensions 20 Unit 4: Motion in Two Dimensions Step 3: Write the working equation. To get the components of the average acceleration, use. To calculate the instantaneous acceleration, use. To determine the magnitude of the acceleration, use. Step 4: Substitute the given values. To identify the components of the average acceleration, the components of the instantaneous velocity at any time t should be determined by calculating the first derivative of the position vectors. Find the components of the velocity at t = 0 s and t = 3.0 s. At t = 0 s: 4.2. Motion Descriptors in Three Dimensions 21 Unit 4: Motion in Two Dimensions At t = 3.0 s: Solve for the average acceleration for each component. To determine the instantaneous acceleration, the derivative of the velocity of each component should be calculated. Therefore,. Substitute t = 3.0 s to the acceleration vector. Calculate the magnitude of the acceleration using. 4.2. Motion Descriptors in Three Dimensions 22 Unit 4: Motion in Two Dimensions Step 5: Find the answer. The components of the average acceleration from t = 0 s to t = 3.0 s: The instantaneous acceleration at t =3.0 s: The magnitude of the acceleration at t = 3.0 s: The components of the average acceleration are: aav-x = 0.9 m/s2, aav-y = ‒0.6 m/s2, aav-z = 2.25 m/s2. The instantaneous acceleration at t = 3.0 s is. The magnitude of the acceleration at t = 3.0 s is 21.42 m/s. 2 6 Try It! A car has the following components of its position as a function of time: x = (5.0 m/s)t ‒ (2.1 m/s2)t2 y = (6.5 m/s)t ‒ (0.10 m/s3)t3 z = (3.0 m/s2)t2 (a) Find the components of the average acceleration at t = 0 s to t = 0.5 s. (b) Find the instantaneous acceleration at t = 0.5 s in vector notation. (c) Find the magnitude of the instantaneous acceleration from (b). Parallel and Perpendicular Components of Acceleration Another way to describe acceleration is to think of with its components parallel to the particle’s path and , and another component perpendicular to the path and to , as shown in Fig. 4.2.5. The parallel component of acceleration signifies the changes in the 4.2. Motion Descriptors in Three Dimensions 23 Unit 4: Motion in Two Dimensions speed of the particle, while the perpendicular component represents the changes in the direction of the particle’s motion. Fig. 4.2.5 Components of acceleration To explain this, consider Fig. 4.2.6a, where the acceleration vector is similar to the direction of. In this case, it has only a parallel component, while its perpendicular component is zero. The figure shows that vector is smaller than vector , indicating that the object increases its speed while traveling in the same direction. The increase in speed is given by. Fig. 4.2.6. Acceleration when directed (a) parallel and (b) perpendicular to an object’s velocity 4.2. Motion Descriptors in Three Dimensions 24 Unit 4: Motion in Two Dimensions In Fig. 4.2.6b, acceleration is perpendicular to the velocity, Hence, it only has a perpendicular component, while its parallel component is zero. At a small interval, the change in velocity is almost perpendicular to , making the directions of both and different. As the time interval gets smaller, the angle also approaches zero, making the perpendicular to both vectors, while and have the same magnitude. In this case, the magnitude of the velocity (speed) stays the same but its direction changes. This can be observed in a curved path where the speed is constant while the direction of the object changes at each instant. In most cases, acceleration can have both parallel and perpendicular components. It means that the speed of the particle varies as well as the direction of the particle’s motion. What do the parallel and perpendicular components of acceleration signify? Tips It is always easier to solve motion problems while considering separate components for each variable. Position, velocity, and acceleration vectors can always be separated based on their components along the x-, y-, and z-axes. Key Points ___________________________________________________________________________________________ The position vector illustrates the position of a particle in two or three dimensions. When the particle is moving, the position vector is considered as a function of time. In three dimensions, the position vector is simply the vector sum of one dimensional position in the x-, y-, and z-axes. The displacement vector is the change in position of a particle or an object at a specific time interval in two to three dimensions. Instantaneous velocity is the rate of change in the position at an instant 4.2. Motion Descriptors in Three Dimensions 25 Unit 4: Motion in Two Dimensions (infinitesimal time). Both the position and the velocity are vectors. In three dimensions, is simply the sum of each component of a particle’s instantaneous velocity vector along the x-, y-, and z- axes. Instantaneous acceleration is the change in velocity over a very small amount of time (infinitesimal). It is expressed as the derivative of the velocity function with respect to time. Acceleration in three dimensions can be written as the vector sum of one-dimensional acceleration in the x-, y-, and z-axes. Acceleration is also described based on its parallel and perpendicular components. ___________________________________________________________________________________________ Key Formulas ___________________________________________________________________________________________ Concept Formula Description Position Vector Use this formula to solve for the position vector in three dimensions if the where components are given. is the position vector; x is the position along the x-axis; y is the position along the y-axis, and z is the position along the z-axis. Use this formula to solve for where displacement if the components of the position is the displacement; vectors are given. 4.2. Motion Descriptors in Three Dimensions 26 Unit 4: Motion in Two Dimensions x1 is the position of r1 along the x-axis; x2 is the position of r2 along the x-axis; y1 is the position of r1 along the y-axis; y2 is the position of r2 along the x-axis; z1 is the position of r1 along the y-axis, and z2 is the position of r2 along the x-axis. Velocity Vector Use this formula to solve for instantaneous velocity if or the components of the velocity or the components of the position vector are where given. is the instantaneous velocity; is the position vector; vx is the x-component of velocity; vy is the y-component of velocity; vz is the z-component of velocity; x is the x-component of the object’s position; y is the y-component of the object’s position, and z is the z-component of the 4.2. Motion Descriptors in Three Dimensions 27 Unit 4: Motion in Two Dimensions object’s position. Acceleration vector Use this formula to solve for the acceleration vector if or the components of the acceleration or the components of velocity are given. where is the acceleration vector; ax is the x-component of acceleration; ay is the y-component of acceleration; az is the z-component of acceleration; is the instantaneous velocity; vx is the x-component of velocity; vy is the y-component of velocity, and vz is the z-component of velocity. ___________________________________________________________________________________________ 4.2. Motion Descriptors in Three Dimensions 28 Unit 4: Motion in Two Dimensions Check Your Understanding A. Write true if the statement is correct. If the statement is incorrect, change the underline word with the correct one. 1. When the particle is moving, the position vector is considered as a function of time. 2. Instantaneous velocity is the change in position vector divided by the time interval. 3. Displacement is the change in position of a particle. 4. Average acceleration is the change in velocity divided by a very small amount of time. 5. Position, velocity, and acceleration vectors in three dimensions are characterized by the sum of their one-dimensional counterparts along the x-, y-, and z-axes. 6. In instantaneous velocity, both the position and the velocity are vectors. 7. The perpendicular component of acceleration signifies the changes in the speed of a particle. 8. The parallel component of acceleration represents the changes in direction of the particle’s motion. 9. The direction of the instantaneous velocity is always tangent to the particle’s path. 10. Motion along each component is dependent on each other. B. Solve the following problems. 1. Find the position vector of a particle at t = 1.2 s. Write it in vector notation form and provide the magnitude of the position vector. The coordinates of the particle’s position (in meters) as functions of time 9in seconds) are: x = 0.2t2 + 0.5t + 2 y = ‒0.4t2 + 2.2t z = ‒0.9t2 ‒ 1.5t ‒ 2 2. What is the instantaneous velocity in vector notation form at t = 2.5 s if the same coordinates of the particle’s position from item no. 1 are given? What is the magnitude of the instantaneous velocity? 4.2. Motion Descriptors in Three Dimensions 29 Unit 4: Motion in Two Dimensions 3. The coordinates of the position of a moving vehicle (in meters) as a function of time (in seconds) are given by: x = ‒4.5t2 + 6.2t + 12 y = 3.5t2 + 2.5t ‒ 15 z = ‒0.6t2 ‒ 3.0t + 8 a. What is the position vector in vector notation form, and the magnitude of the position of the car at time t = 2 s? b. What is the velocity of the car in vector notation form at t = 2 s? c. What is the magnitude of the velocity vector at t = 2 s? 4. A car has the following components of its position as a function of time: x = (3.0 m/s)t + (1.5 m/s2)t2 y = (2.5 m/s)t ‒ (1.2 m/s3)t3 z = ‒(4.0 m/s2)t2 Find the components of the average acceleration of the car from t = 0 s to t = 2 s. 5. Using the same given components of position in #4, find (a) the instantaneous acceleration in vector notation form at t = 6 s, and (b) find the magnitude of the instantaneous acceleration at t = 6 s. C. Provide what is being asked in each item. 1. Illustrate the position vector in the figure on the right. Identify its components in the x-, y-, and z-axes. 4.2. Motion Descriptors in Three Dimensions 30 Unit 4: Motion in Two Dimensions For items 2, 3 and 4, refer to the figure below. 2. Draw the instantaneous velocity vector of the particle shown. 3. Draw the acceleration vector of the particle shown. Assume that the speed of the particle is constant. 4. How will you draw the displacement as the particle in the figure changed from one position to another? 5. A car is moving along a curved path (from P1 to P2) shown below. Its path and velocity at two instants were also plotted. What can you conclude about the car’s acceleration? 4.2. Motion Descriptors in Three Dimensions 31 Unit 4: Motion in Two Dimensions Challenge Yourself Answer the following questions. 1. A rabbit follows a path shown in the figure below. Rank the points a, b, and c (from highest to lowest) according to the magnitude of the average velocity as it reaches each point from the initial point i. Assume that the rabbit takes the same amount of time to move from one point to another. 2. Illustrate the acceleration vector as the object’s speed increases and decreases as it follows a curved path. 3. Provide another position vector that will have an equal acceleration as. 4. Is the term “acceleration” only applicable when the speed of an object changes? Why or why not? 5. A roller coaster reaches the highest point of its speed hill, as shown in the figure below. Which arrow correctly represents the acceleration of the cart at this point? Explain your answer. 4.2. Motion Descriptors in Three Dimensions 32 Unit 4: Motion in Two Dimensions Bibliography Faughn, Jerry S. and Raymond A. Serway. Serway’s College Physics (7th ed). Singapore: Brooks/Cole, 2006. Giancoli, Douglas C. Physics Principles with Applications (7th ed). USA: Pearson Education, 2014. Halliday, David, Robert Resnick and Kenneth Krane. Fundamentals of Physics (5th ed). USA: Wiley, 2002. Knight, Randall D. Physics for Scientists and Engineers: A Strategic Approach (4th ed). USA: Pearson Education, 2017. Serway, Raymond A. and John W. Jewett, Jr. Physics for Scientists and Engineers with Modern Physics (9th ed). USA: Brooks/Cole, 2014. Walker, James S. Physics (5th ed). USA: Pearson Education, 2017. Young, Hugh D., Roger A. Freedman, and A. Lewis Ford. Sears and Zemansky’s University Physics with Modern Physics (13th ed). USA: Pearson Education, 2012. 4.2. Motion Descriptors in Three Dimensions 33 Unit 4: Motion in Two Dimensions Key to Try It! 1. ; r = 43 m 2. 3. ; r = 21.63 m; ; v = 14.07 m/s 4. 5. ; a = 46.10 m/s2 6. , , ; ; a = 7.33 m/s2 4.2. Motion Descriptors in Three Dimensions 34