Worksheet 1. What You Need to Know About Motion Along the x-axis (Part 1)

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1 Worksheet 1. What You Need to Know About Motion Along the x-axis (Part 1) In discussing motion, there are three closely related concepts that you need to keep straight. These are: If x(t) represents the position of a particle along the x-axis at any time t, then the following statements are true. 1. Initially means when = At the origin means = At rest means = If the velocity of the particle is positive, then the particle is moving to the. 5. If the velocity of the particle is, then the particle is moving to the left. 6. To find average velocity over a time interval, divide the change in by the change in time. 7. Instantaneous velocity is the velocity at a single moment (instant!) in time. 8. If the acceleration of the particle is positive, then the is increasing. 9. If the acceleration of the particle is, then the velocity is decreasing. 10. In order for a particle to change direction, the must change signs. 11. One way to determine total distance traveled over a time interval is to find the sum of the absolute values of the differences in position between all resting points. Here s an example: If the position of a particle is given by: x(t) 1 3 t 3 t 2 3t 4, find the total distance traveled on the interval 0 t 6. Dixie Ross, Pflugerville High School, Pflugerville, Texas 5

2 Worksheet 1. Solutions and Notes for Students The three concepts are as follows. Position: x(t) determines where the particle is located on the x-axis at a given time t Velocity: v(t) x (t) determines how fast the position is changing at a time t as well as the direction of movement Acceleration: a(t) v (t) x (t) determines how fast the velocity is changing at time t; the sign indicates if the velocity is increasing or decreasing The true statements are as follows. 1. Initially means when time, t = At the origin means position, x(t) = At rest means velocity, v(t) = If the velocity of the particle is positive, then the particle is moving to the right. 5. If the velocity of the particle is negative, then the particle is moving to the left. 6. To find average velocity over a time interval, divide the change in position by the change in time. 7. Instantaneous velocity is the velocity at a single moment (instant!) in time. 8. If the acceleration of the particle is positive, then the velocity is increasing. 9. If the acceleration of the particle is negative, then the velocity is decreasing. 10. In order for a particle to change direction, the velocity must change signs. 11. First, find the times at which x (t) v(t) 0. That would be t 1 (which is out of our interval) and t 3. Next, evaluate the position at the end points and at each of the resting points. The particle moved to the left 9 units and then to the right by 27 units for a total distance traveled of 36 units. Point out to students how this is similar to a closed interval test, where you have to determine function values at the end points as well as at any critical points found. t x(t) Dixie Ross, Pflugerville High School, Pflugerville, Texas 6

3 Worksheet 2. Sample Practice Problems for the Topic of Motion (Part 1) Example 1 (numerical). The data in the table below give selected values for the velocity, in meters/minute, of a particle moving along the x-axis. The velocity v is a differentiable function of time t. Time t (min) Velocity v(t) (meters/min) At t = 0, is the particle moving to the right or to the left? Explain your answer. 2. Is there a time during the time interval 0 t 12 minutes when the particle is at rest? Explain your answer. 3. Use data from the table to find an approximation for v (10) and explain the meaning of v (10) in terms of the motion of the particle. Show the computations that lead to your answer and indicate units of measure. 4. Let a(t) denote the acceleration of the particle at time t. Is there guaranteed to be a time t c in the interval 0 t 12 such that a(c) 0? Justify your answer. 7

5 6. At what time t in the given interval is the particle farthest to the right? Explain your answer. Example 3 (analytic). A particle moves along the x-axis so that at time t its position is given by: x(t) t 3 6t 2 9t At t = 0, is the particle moving to the right or to the left? Explain your answer. 2. At t = 1, is the velocity of the particle increasing or decreasing? Explain your answer. 3. Find all values of t for which the particle is moving to the left. 4. Find the total distance traveled by the particle over the time interval 0 t 5. Dixie Ross, Pflugerville High School, Pflugerville, Texas 9

6 Worksheet 2. Solutions Example 1 (numerical) 1. At t = 0, the particle is moving to the left because the velocity is negative. 2. Yes, there is a time when the particle is at rest during the time interval 0 < t < 12 minutes. Since the velocity function is differentiable, it also is continuous. Hence, by the Intermediate Value Theorem, since velocity goes from negative to positive, it must go through zero and v(t) = 0 means the particle is at rest. 3. Since t = 10 is not one of the times given in the table, we should approximate the derivative by using a difference quotient with the best (closest) data available. Because 10 lies between 8 and 12, the best approximation is given by: Here, v(12) v(8) v (10) m / min min v (10) is the acceleration of the particle at t = 10 minutes. m min Yes, such a point is guaranteed by the Mean Value Theorem or Rolle s Theorem. Since velocity is differentiable (and therefore also continuous) over the interval 6 t 12 and: v(12) v(6) , then there must exist a point c in the interval such that v (c) a(c) 0. Note: If we add the hypothesis that v is continuous, then we may use the Intermediate Value Theorem to establish the result. In this case, since the values in the table indicate that velocity increases and then decreases on the interval 0 t 12, then v (t) a(t) must go from positive to negative and by the Intermediate Value Theorem must therefore pass through zero somewhere in that interval. It is the Mean Value Theorem, applied to the differentiable function v, that guarantees v takes on at least one positive value in the interval 0 t 8 (note that 7 ( 3) is one such 4 value), and at least one negative value in the interval 8 t 12 (note that is one such value). 2 10

7 Example 2 (graphical) 1. At t = 4 seconds, the particle is moving to the right because the velocity is positive. 2. The particle is moving to the left over the interval 5 t 9 seconds because the velocity is negative. 3. The acceleration of the particle is negative because the velocity is decreasing, OR the acceleration is the slope of the velocity graph and the slope of the velocity graph at t = 4 is negative. 4. Average acceleration over the time interval can be found by dividing the change in velocity by the change in time: v(4) v(2) ft / sec sec ft sec 2 5. No such time is guaranteed. The Mean Value Theorem does not apply since the function is not differentiable at t = 3 due to the sharp turn in the graph. If students have not yet learned the MVT, you can slide a tangent line (toothpick or stick) along the graph to show that no such point exists where the slope of the tangent line would be equal to the slope of a secant line between t = 2 and t = The particle is farthest to the right at t = 5 seconds. Since the velocity is positive during the time interval 0 t 5 seconds and negative during the time interval 5 t 9 seconds, the particle moves to the right before time t = 5 seconds and moves to the left after that time. Therefore, it is farthest to the right at t = 5 seconds. Example 3 (analytic) 1. The particle is moving to the right because x (0) v(0) 9 which is positive. 2. At t = 1, the velocity of the particle is decreasing because x (1) v (1) a(1) 6, and if the acceleration is negative then the velocity is decreasing. 3. The particle is moving to the left for all values of t where v(t) < 0. We have: v(t) x (t) 3t 2 12t 9 0 for 1 < t < 3. 11

8 4. t x(t) The particle moves right 4 spaces, left 4 spaces and then right 20 spaces. Therefore, the particle has traveled a total of 28 units. The common error that students make is to calculate x(5) x(0) 20, which gives the displacement or net change in position, rather than the total distance traveled. Teachers should reinforce the difference between these two concepts every chance they get. Dixie Ross, Pflugerville High School, Pflugerville, Texas 12

9 Worksheet 3. Understanding the Relationships Among Velocity, Speed, and Acceleration Speed is the absolute value of velocity. It tells you how fast something is moving without regard to the direction of movement. 1. What effect does absolute value have on numbers? 2. What effect does taking the absolute value of a function have on its graph? For each situation below, the graph of a differentiable function giving velocity as a function of time t is shown for 1 t 5, along with selected values of the velocity function. In the graph, each horizontal grid mark represents 1 unit of time and each vertical grid mark represents 4 units of velocity. For each situation, plot the speed graph on the same coordinate plane as the velocity graph and fill in the speed values in the table. Then, answer the questions below based on both the graph and the table of values. Situation 1: Velocity graph time velocity speed In this situation, the velocity is and. Positive or negative? Because velocity is, we know acceleration is. Positive or negative? By examining the graph of speed and the table of values, we can conclude that speed is. 13

10 Situation 2: Velocity graph time velocity speed In this situation, the velocity is and. Positive or negative? Because velocity is, we know acceleration is. Positive or negative? By examining the graph of speed and the table of values, we can conclude that speed is. Situation 3: Velocity graph time velocity speed In this situation, the velocity is and. Positive or negative? Because velocity is, we know acceleration is. Positive or negative? By examining the graph of speed and the table of values, we can conclude that speed is. 14

11 Situation 4: Velocity graph time velocity speed In this situation, the velocity is and. Positive or negative? Because velocity is, we know acceleration is. Positive or negative? By examining the graph of speed and the table of values, we can conclude that speed is. Conclusion: In which situations was the speed increasing? When the speed is increasing, the velocity and acceleration have signs. Same or opposite? In which situations was the speed decreasing? When the speed is decreasing, the velocity and acceleration have signs. Same or opposite? Dixie Ross, Pflugerville High School, Pflugerville, Texas 15

12 Assessing Students Understanding (A Short Quiz): 1. If velocity is negative and acceleration is positive, then speed is. 2. If velocity is positive and speed is decreasing, then acceleration is. 3. If velocity is positive and decreasing, then speed is. 4. If speed is increasing and acceleration is negative, then velocity is. 5. If velocity is negative and increasing, then speed is. 6. If the particle is moving to the left and speed is decreasing, then acceleration is. Dixie Ross, Pflugerville High School, Pflugerville, Texas 16

13 Worksheet 3. Solutions Speed is the absolute value of velocity. It tells you how fast something is moving without regard to the direction of movement. 1. What effect does absolute value have on numbers? Absolute value makes all numbers non-negative. 2. What effect does taking the absolute value of a function have on its graph? Taking the absolute value of a function will cause any portion of its graph that is below the x- axis to be reflected across the x-axis. For each situation below, the graph of a differentiable function giving velocity as a function of time t is shown for 1 t 5, along with selected values of the velocity function. In the graph, each horizontal grid mark represents 1 unit of time and each vertical grid mark represents 4 units of velocity. For each situation, plot the speed graph on the same coordinate plane as the velocity graph and fill in the speed values in the table. Then, answer the questions below based on both the graph and the table of values. Situation 1: Speed graph time velocity speed In this situation, the velocity is positive and increasing. Positive or negative? Because velocity is increasing, we know acceleration is positive. Positive or negative? By examining the graph of speed and the table of values, we can conclude that speed is increasing. 17

14 Situation 2: Speed graph time velocity speed In this situation, the velocity is negative and decreasing. Positive or negative? Because velocity is decreasing, we know acceleration is negative. Positive or negative? By examining the graph of speed and the table of values, we can conclude that speed is increasing. Situation 3: Speed graph time velocity speed In this situation, the velocity is negative and increasing. Positive or negative? Because velocity is increasing, we know acceleration is positive. Positive or negative? By examining the graph of speed and the table of values, we can conclude that speed is decreasing. 18

15 Situation 4: Speed graph time velocity speed In this situation, the velocity is positive and decreasing. Positive or negative? Because velocity is decreasing, we know acceleration is negative. Positive or negative? By examining the graph of speed and the table of values, we can conclude that speed is decreasing. Conclusion: In which situations was the speed increasing? Situations 1 and 2 When the speed is increasing, the velocity and acceleration have the same signs. Same or opposite? In which situations was the speed decreasing? Situations 3 and 4 When the speed is decreasing, the velocity and acceleration have _ opposite signs. Same or opposite? Dixie Ross, Pflugerville High School, Pflugerville, Texas 19

16 Assessing Students Understanding (A Short Quiz): 1. If velocity is negative and acceleration is positive, then speed is decreasing. 2. If velocity is positive and speed is decreasing, then acceleration is negative. 3. If velocity is positive and decreasing, then speed is decreasing. 4. If speed is increasing and acceleration is negative, then velocity is negative. 5. If velocity is negative and increasing, then speed is decreasing. 6. If the particle is moving to the left and speed is decreasing, then acceleration is positive. Dixie Ross, Pflugerville High School, Pflugerville, Texas 20

17 Worksheet 4. What You Need to Know About Motion Along the x-axis (Part 2) 1. Speed is the absolute value of. 2. If the velocity and acceleration have the same sign (both positive or both negative), then speed is. 3. If the velocity and acceleration are in sign (one is positive and the other is negative), then speed is decreasing. There are three ways to use an integral in the study of motion that are easily confused. Watch out! 4. v(t)dt is an integral. It will give you an expression for at time t. Don t forget that you will have a, the value of which can be determined if you know a position value at a particular time. 5. t 2 v(t)dt t 1 is a integral and so the answer will be a. The number represents the change in over the time interval. By the Fundamental Theorem of Calculus, since v(t) x (t), the integral will yield x(t ) x(t ) 2 1. This is also known as displacement. The answer can be positive or depending upon if the particle lands to the or left of its original starting position. 6. t 2 v(t) dt t 1 is also a integral and so the answer will be a number. The number represents the traveled by the particle over the time interval. The answer should always be. Dixie Ross, Pflugerville High School, Pflugerville, Texas 21

18 Worksheet 4. Solutions 1. Speed is the absolute value of velocity. 2. If the velocity and acceleration have the same sign (both positive or both negative), then speed is increasing. 3. If the velocity and acceleration are opposite in sign (one is positive and the other is negative), then speed is decreasing. There are three ways to use an integral in the study of motion that are easily confused. Watch out! v(t)dt is an indefinite integral. It will give you an expression for position at time t. Don t forget that you will have a + C, or constant of integration, the value of which can be determined if you know a position value at a particular time. t 2 v(t)dt t 1 is a definite integral and so the answer will be a number. The number represents the change in position over the time interval. By the Fundamental Theorem of Calculus, since v(t) x (t), the integral will yield x(t 2 ) x(t 1 ). This is also known as displacement. The answer can be positive or negative depending upon if the particle lands to the right or left of its original starting position. t 2 v(t) dt t 1 is also a definite integral and so the answer will be a number. The number represents the total distance traveled by the particle over the time interval. The answer should always be non-negative. Note: One way to see that: t 2 t 1 v(t) dt gives the total distance traveled is to note that the speed function v(t), in transforming all negative velocities to positive velocities, also transforms all backward motion (motion to the left) to forward motion (motion to the right). The change in position from start to finish of the resulting forward-only journey is the total distance traveled during the original forward-and-backward journey. Dixie Ross, Pflugerville High School, Pflugerville, Texas 22

19 Worksheet 5. Sample Practice Problems for the Topic of Motion (Part 2) Example 1 (graphical). The graph to the right shows the velocity, v(t), of a particle moving along the x-axis for 0 t 11. It consists of a semicircle and two line segments. Use the graph and your knowledge of motion to answer the following questions. 1. At what time t, 0 t 11, is the speed of the particle the greatest? 2. At which of the times, t = 2, t = 6 or t = 9, is the acceleration of the particle the greatest? Explain your answer. 3. Over what time intervals is the particle moving to the left? Explain your answer. 4. Over what time intervals is the speed of the particle decreasing? Explain your answer. 5. Find the total distance traveled by the particle over the time interval 0 t Find the value of the problem v(t) dt and explain the meaning of this integral in the context of 23

20 7. If the initial position of the particle is x(0) = 2, find the position of the particle at time t = 11. Example 2 (analytic/graphical/calculator active). The rate of change, in kilometers per hour, of the altitude of a hot air balloon is given by r(t) t 3 4t 2 6 for time 0 t 4, where t is measured in hours. Assume the balloon is initially at ground level. 1. For what values of t, 0 t 4, is the altitude of the balloon decreasing? 2. Find the value of r (2) and explain the meaning of the answer in the context of the problem. Indicate units of measure. 3. What is the altitude of the balloon when it is closest to the ground during the time interval 2 t 4? 4 r(t) dt 4. Find the value of 0 and explain the meaning of the answer in the context of the problem. Indicate units of measure. 4 r(t) dt 5. Find the value of 0 and explain the meaning of the answer in the context of the problem. Indicate units of measure. 24

21 6. What is the maximum altitude of the balloon during the time interval 0 t 4? Example 3 (numerical). The table below gives values for the velocity and acceleration of a particle moving along the x-axis for selected values of time t. Both velocity and acceleration are differentiable functions of time t. The velocity is decreasing for all values of t, 0 t 10. Use the data in the table to answer the questions that follow. Time, t Velocity, v(t) Acceleration, a(t) Is there a time t when the particle is at rest? Explain your answer. 2. At what time indicated in the table is the speed of the particle decreasing? Explain your answer. v(t) dt 3. Use a left Riemann sum to approximate 0. Show the computations you use to arrive at your answer. Explain the meaning of the definite integral in the context of the problem

22 4. Is the approximation found in part (3) greater than or less than the actual value of the definite integral shown below? Explain your reasoning v(t) dt 10 v(t) dt 5. Approximate the value of 0 using a trapezoidal approximation with the three sub-intervals indicated by the values in the table. Show the computations you use to arrive at your answer. Explain the meaning of the definite integral in the context of the problem. 6. Determine the value of context of the problem a(t) dt. Explain the meaning of the definite integral in the Dixie Ross, Pflugerville High School, Pflugerville, Texas 26

23 Worksheet 5. Solutions Example 1 (graphical) 1. Speed is greatest at t = 8 since that is when the magnitude of the velocity is the greatest. 2. The acceleration is the derivative or slope of the velocity graph. Therefore: a(2) = 0, a(6) 3, and a(9) = 3. 2 The acceleration of the particle is greatest at t = The particle is moving to the left over the time interval 4 < t < 10 because that is where the velocity is negative (velocity graph lies below the horizontal axis). 4. Whenever velocity and acceleration are opposite in sign, the speed is decreasing. The velocity is positive and decreasing for 2 t 4, which means v (t) a(t) 0 for 2 t 4. This means the speed of the particle is decreasing for the time interval 2 t 4 and in fact for the time interval 2 t 4. The velocity is negative and increasing for 8 t 10, which means v (t) a(t) 0 for 8 t 10. This means the speed of the particle also is decreasing for the time interval 8 t 10 and in fact for the time interval 8 t 10. Or, sketch and label clearly the graph of the speed function, v(t), to see that the speed decreases for 2 t 4 and 8 t The total distance traveled by the particle is the sum of the areas of the regions between the velocity graph and the t-axis. Therefore the total distance traveled is given by the expression: Equivalently, the total distance traveled is given by: 11 v(t) dt = The definite integral shown below represents the net change in position, or the displacement, of the particle over the time interval 0 t v(t) dt Using the Fundamental Theorem of Calculus, x(11) x(0) v(t)dt

24 Example 2 (analytic/graphical/calculator active) 1. The altitude of the balloon is decreasing whenever r(t) 0. By graphing r(t) and using the zero feature of the calculator, the time interval will be: t hours. Students should be taught to store these values to use for subsequent calculations or to use more decimal places than the three that are required so that their subsequent answers will have three decimal place accuracy. 2. Students can use the numerical derivative feature of their calculator or compute by hand the value: r (2) 4 km/hr = km hr hr 2 This value is the rate of change of the altitude of the hot air balloon at time t = 2 hours. In this case, at time t = 2 hours, the rate of change of the altitude is decreasing at a rate of 4 km/hr 2. Note: The rate of change of the altitude at time t = 2 hours is r(2) 2 km/hr, indicating that the altitude itself also is decreasing at this time. Note also that if the balloon moved only vertically, then r(t) would be its velocity and r (t) its acceleration. In this case, students could report that r (2) 4 means the acceleration of the balloon is 4 km/hr 2 or, equivalently, that the velocity of the balloon is decreasing at a rate of 4 km/hr By the result of part (1), the balloon is closest to the ground at t = hours. If y(t) denotes the altitude of the balloon after t hours, then, by the Fundamental Theorem of Calculus, the minimum altitude of the balloon for 2 t 4 is: y(3.514) y(0) r(t)dt km. Students should use the definite integral feature of the calculator to evaluate this integral. 28

25 4. The definite integral shown below represents the net change in the altitude of the balloon over the time interval 0 t r(t) dt km. Since the balloon was initially at ground level, this definite integral also gives the altitude of the balloon at t = 4 hours. 5. The definite integral shown below represents the total vertical distance traveled by the balloon over the time interval 0 t 4. (Note that if the balloon moved only vertically, then r(t) would be its speed.) 0 4 r(t) dt km. 6. The maximum altitude of the balloon can occur either at the critical point where r(t) goes from positive to negative (t = 1.572) or at the right endpoint of the time interval (t = 4). Since: y(1.572) y(0) r(t)dt km. and since, by part (d), y(4) = km, then the maximum altitude of the balloon is km. 29

26 Example 3 (numerical) 1. Yes, there is a time when the particle is at rest. Since velocity is a differentiable function of time t, then velocity must also be continuous and the Intermediate Value Theorem can be applied. Since v(2) = 3 and v(6) 1, then there must be a value of t in the interval 2 t 6 such that v(t) = Speed is decreasing at t = 2 since at that time velocity and acceleration are opposite in sign. 3. Below, the definite integral represents the net change in position, or displacement, over the time interval 0 t v(t) dt ( 1)(4) Because velocity is given to be a decreasing function on the interval 0 t 10, the approximation in part (3) will be greater than the actual value of the integral. 5. The definite integral shown below represents the total distance traveled by the particle over the time interval 0 t v(t) dt (5 3) 1 2 4(3 1) 1 4(1 8) Using the Fundamental Theorem of Calculus, 10 0 a(t) dt v(10) v(0) The definite integral gives the net change in the velocity of the particle over the time interval 0 t 10. Dixie Ross, Pflugerville High School, Pflugerville, Texas 30

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