HW02 - Kinematics in One Dimension

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10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 1/41 HW02 - Kinematics in One Dimension Due: 11:59pm on Friday, September 30, 2022 To understand how points are awarded, read the Grading Policy for this assignment. Position versus Time Graphs Conceptual Question The motions described in each of the questions take place at an intersection on a two-lane road with a stop sign in each direction. For each motion, select the correct position versus time graph. For all of the motions, the stop sign is at the position , and east is the positive x direction. Part A A driver ignores the stop sign and continues driving east at constant speed. Hint 1. Determining velocity from a position versus time graph The slope on a position versus time graph is the rise (change in position) over the run (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed. Hint 2. Driving east Since east is defined as the positive x direction, a car traveling east must have a positive velocity. A positive velocity is represented as a positive slope on a position versus time graph. Hint 3. Constant speed Since velocity is represented by the slope on a position versus time graph, a car moving at constant speed must be represented by a line of constant slope. ANSWER:

10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 2/41 Correct Part B A driver ignores the stop sign and continues driving west at constant speed. Hint 1. Determining velocity from a position versus time graph The slope on a position versus time graph is the rise (change in position) over the run (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed. Hint 2. Driving west Since east is defined as the positive x direction, a car traveling west must have a negative velocity. A negative velocity is represented as a negative slope on a position versus time graph. Hint 3. Constant speed Since velocity is represented by the slope on a position versus time graph, a car moving at constant speed must be represented by a line of constant slope. ANSWER: Correct Part C A driver, traveling west, slows and stops at the stop sign. Hint 1. Determining velocity from a position versus time graph The slope on a position versus time graph is the rise (change in position) over the run (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time A B C D E F A B C D E F

10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 3/41 graph is the velocity of the object being graphed. Hint 2. Driving west Since east is defined as the positive x direction, a car traveling west must have a negative velocity. A negative velocity is represented as a negative slope on a position versus time graph. Hint 3. Acceleration Since velocity is represented by the slope on a position versus time graph, a car that accelerates must be represented as a curve with changing slope. If a car slows, then the slope of the graph must approach zero. If a car's speed increases, the slope must become more positive or more negative (depending upon which direction it is moving). ANSWER: Correct Part D A driver, after stopping at the stop sign, travels east with a positive acceleration. Hint 1. Determining velocity from a position versus time graph The slope on a position versus time graph is the rise (change in position) over the run (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed. Hint 2. Driving east Since east is defined as the positive x direction, a car traveling east must have a positive velocity. A positive velocity is represented as a positive slope on a position versus time graph. Hint 3. Acceleration Since velocity is represented by the slope on a position versus time graph, a car that accelerates must be represented as a curve with changing slope. If a car slows, then the slope of the graph must approach zero. If a car's speed increases, the slope must become more positive or more negative (depending upon which direction it is moving). ANSWER: A B C D E F

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10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 4/41 Correct ± Average Velocity from a Position vs. Time Graph Learning Goal: To learn to read a graph of position versus time and to calculate average velocity. In this problem you will determine the average velocity of a moving object from the graph of its position as a function of time . A traveling object might move at different speeds and in different directions during an interval of time, but if we ask at what constant velocity the object would have to travel to achieve the same displacement over the given time interval, that is what we call the object's average velocity . We will use the notation to indicate average velocity over the time interval from to . For instance, is the average velocity over the time interval from to . Part A Consulting the graph shown in the figure, find the object's average velocity over the time interval from 0 to 1 second. Answer to the nearest integer. Hint 1. Definition of average velocity Average velocity is defined as the constant velocity at which an object would have to travel to achieve a given displacement (difference between final and initial positions, which can be negative) over a given time interval, from the initial time to the final time . The average velocity is therefore equal to the displacement divided by the given time interval. In symbolic form, average velocity is given by . ANSWER: A B C D E F = 0

10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 5/41 Correct Part B Find the average velocity over the time interval from 1 to 3 seconds. Express your answer in meters per second to the nearest integer. Hint 1. Find the change in position The final and initial positions can be read off the y axis of the graph. What is the displacement during the time interval from 1 to 3 seconds? Express your answer numerically, in meters ANSWER: Hint 2. Definition of average velocity Average velocity is defined as the constant velocity at which an object would have to travel to achieve a given displacement (difference between final and initial positions, which can be negative) over a given time interval, from the initial time to the final time . The average velocity is therefore equal to the displacement divided by the given time interval. In symbolic form, average velocity is given by . ANSWER: Correct A note about instantaneous velocity. The instantaneous velocity at a certain moment in time is represented by the slope of the graph at that moment. For straight-line graphs, the (instantaneous) velocity remains constant over the interval, so the instantaneous velocity at any time during an interval is the same as the average velocity over that interval. For instance, in this case, the instantaneous velocity at any time from 1 to 3 seconds is the same as the average velocity of . Part C Now find . Give your answer to three significant figures. Hint 1. A note on the displacement Since the object's position remains constant from time 0 to time 1, the object's displacement from 0 to 3 is the same as in Part B. However, the time interval has changed. ANSWER: = 40 = 20 = 13.3

10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 6/41 Correct Note that is not equal to the simple arithmetic average of and , i.e., , because they are averages for time intervals of different lengths. Part D Find the average velocity over the time interval from 3 to 6 seconds. Express your answer to three significant figures. Hint 1. Determine the displacement What is the displacement? Answer to the nearest integer. ANSWER: Hint 2. Determine the time interval What is the time interval? Answer to two significant figures. ANSWER: ANSWER: Correct Part E Finally, find the average velocity over the whole time interval shown in the graph. Express your answer to three significant figures. Hint 1. Determine the displacement What is the displacement? Answer to the nearest integer. ANSWER: ANSWER: = -40 = 3.0 = -13.3 = 0

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10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 7/41 Correct Note that though the average velocity is zero for this time interval, the instantaneous velocity (i.e., the slope of the graph) has several different values (positive, negative, zero) during this time interval. Note as well that since average velocity over a time interval is defined as the change in position (displacement) in the given interval divided by the time, the object can travel a great distance (here 80 meters) and still have zero average velocity, since it ended up exactly where it started. Therefore, zero average velocity does not necessarily mean that the object was standing still the entire time! Problem 2.1 - Enhanced - with Hints and Feedback Larry leaves home at 9:06 and runs at constant speed to the lamppost seen in . He reaches the lamppost at 9:09, immediately turns, and runs to the tree. Larry arrives at the tree at 9:15. Part A What is Larry's average velocity, in , during each of these two intervals. Express your answers in meters per minute and separated by a comma. Hint 1. How to approach the problem Recall the relationship between the average velocity, displacement, and time interval in a case of constant speed. Pay attention to the direction of motion during each time interval to determine the sign of the answer correctly. ANSWER: Correct Part B What is Larry's average velocity for the entire run? xpress your answer in meters per minute. = 0 , = -133,167

10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 8/41 Hint 1. How to approach the question Recall the relationship between the average velocity, displacement, and time interval in a case of constant speed. Also recall that displacement is not the same as the total distance traveled and depends only on the initial and final positions. ANSWER: Correct Velocity from Graphs of Position versus Time An object moves along the x axis during four separate trials. Graphs of position versus time for each trial (with the same scales on each axis) are shown in the figure. Part A During which trial or trials is the object's velocity not constant? Check all that apply. Hint 1. Finding velocity from a position versus time graph On a graph of coordinate x as a function of time , the instantaneous velocity at any point is equal to the slope of the curve at that point. Hint 2. Equation for slope The slope of a line is its rise divided by the run: . ANSWER: = 66.7

10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 9/41 Correct The graph of the motion during Trial B has a changing slope and therefore is not constant. The other trials all have graphs with constant slope and thus correspond to motion with constant velocity. Part B During which trial or trials is the magnitude of the average velocity the largest? Check all that apply. Hint 1. Definition of average velocity Recall that . Then note that the question asks only about the magnitude of the velocity. ANSWER: Correct While Trial B and Trial D do not have the same average velocity, the only difference is the direction! The magnitudes are the same. Neither one is "larger" than the other, and it is only because of how we chose our axes that Trial B has a positive average velocity while Trial D has a negative average velocity. In Trial C the object does not move, so it has an average velocity of zero. During Trial A the object has a positive average velocity but its magnitude is less than that in Trial B and Trial D. Derivatives Learning Goal: To understand the concept of derivatives as the slope of a function graph. There are two fundamental tools of calculus—the derivative and the integral. The derivative is a measure of the rate of change of a function. You’ll see derivatives often throughout your study of physics because much of physics involves describing rates of change. Part A Trial A Trial B Trial C Trial D Trial A Trial B Trial C Trial D

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10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 10/41 What does “rate of change” mean, exactly? A function describes how one quantity is related to another – how y is related to x . As an example, y could be your height and x your age. The three graphs below show three cases of how your height could change with age. Sort the graphs according to which best represents the graph of your height against age for the following periods: During your childhood, as you grew up Between age 30 and age 50 During old age, when your spine compacts a little Drag the appropriate items to their respective bins. Hint 1. Identify the graph that illustrates growth Age is plotted on the horizontal axis in each graph, with age increasing from left to right. Height is plotted on the vertical axis. Think about the childhood period, when you are growing. What kind of graph illustrates growth? ANSWER: ANSWER: A graph that falls as you move from left to right. A graph that is flat as you move from left to right. A graph that rises as you move from left to right. Reset Help Childhood Age 30 to age 50 Old age

10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 11/41 Correct The positive (upward) slope on Graph 2 corresponds to your increasing height during childhood. Your height won’t change between ages 30 and 50, so Graph 1 is correct: It is horizontal, and its slope is zero. And finally, your decreasing height during old age corresponds to the negative (downward) slope on Graph 3. These three graphs of height vs. age are three important special cases of functions: Graph 1 shows a quantity that remains constant with respect to another quantity. Graph 2 shows a quantity that increases with respect to another quantity. Graph 3 shows a quantity that decreases with respect to another quantity. These graphs also give three important special cases in our understanding of rate of change . Part B Graphs 4 and 5 show the same information as Graphs 2 and 3, but this time with specific information about height and age on the graph axes. From the graphs, you can find the exact rate of change of the person’s height in each case. Find the rate of change of the height from age 8 to 16 and from age 60 to 80. Express both answers in units of cm per year, to two significant figures. Separate your answers with a comma. Express your answers numerically to two significant figures separated by a comma. Hint 1. How to approach the problem The rate of change corresponds to the slope on the graph. The slope is commonly thought of as rise/run. In terms of the x and y values on the graph, the slope is Read the values of and from the graph, and from these compute the slope as given by the formula above. In this problem, is the change in age and is the change in height. Hint 2. Calculate the slope of Graph 4 It’s often easiest to pick points on the line that lie on the intersection of gridlines: this makes reading off the values on the axes straightforward. On Graph 4, we have two such points: and . What are the rise and run values corresponding to these two points? Express your answers numerically separated by a comma. ANSWER: Hint 3. Calculate the slope of Graph 5 It’s often easiest to pick points on the line that lie on the intersection of gridlines: This makes it straightforward to read off the values on the axes. On Graph 5, we have two such points: and . What are the rise and run rise, run = 50,8

10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 12/41 values corresponding to these two points? Express your answers numerically separated by a comma. ANSWER: ANSWER: Correct What you’ve just done is calculate the derivative of a function, in this case a person’s height vs. time, because the derivative measures a function’s rate of change. The answers here illustrate that a derivative can have a positive value, when the function is increasing; a negative value, when the function is decreasing; a value of zero, when the function isn’t changing. These are the three special cases of rates of change we mentioned above. Part C The next graph shows a different function in . Which of the graphs below is the derivative of this function? Hint 1. How to approach the problem Notice that the slope of the graph in Part C is constant. So you’re looking for an answer graph that’s constant - that is, a graph with a horizontal slope. Hint 2. Narrowing down the options Which of the two graphs with horizontal slopes is correct? To determine this, measure carefully the slope of your graph, using the numbers provided on the axes. Then, determine the slope as you did in Part B. ANSWER: rise, run = -5,20 6.3,-0.25

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10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 13/41 Correct The original function is increasing at a constant rate throughout the interval, so the answer must be a constant. Measuring the slope of the graph carefully leads you to the correct answer, 0.6. Part D So far we have only looked at functions with a constant slope (positive, negative, or zero). But the slope of a function, and thus its derivative, can change. Consider the next function shown in . Rank, in increasing order, the derivatives of the function at each of the points marked A through D . Express your answer in order of increasing slopes as a string without spaces between points. For example, enter ABCD if you think the derivative at point A is the smallest, B is the next larger, and so on.

10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 14/41 Hint 1. How to approach the problem Look carefully at the graph’s slope at each of the four points. The slope may be positive, zero, or negative. If you are ranking in the order of increasing slopes, then negative slopes will be listed first, then zero, and then positive slopes. For a positive slope, the more sharply the graph slopes upward, the larger the slope. ANSWER: Correct The slope at point C is negative, so the derivative there is the smallest. At B , the graph is horizontal, so the derivative is zero. At points A and D , the steeper slope corresponds to the larger derivative. Part E Functions are not all straight lines. If a function is curved, the derivative is not constant but changing. To find the derivative at any point, draw a line tangent to the graph at that point. The slope of the tangent line is the derivative. As you move along the graph, the slope of the tangent line changes, and so does the derivative. The graph in shows a function ( ). The derivative of the function at is closest to: Hint 1. Narrowing down the choices The derivate at is the slope of the curve there. Is the function increasing, decreasing, or zero at ? ANSWER: CBDA

10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 15/41 Hint 2. Finding the answer To determine the exact derivative of the function at , draw a line tangent to the graph at that point. The slope of the tangent line tells you the function’s derivative. ANSWER: Correct At , the function is increasing, so that tells you that the derivative (the slope of the tangent line) is positive. Here, you needed to measure the slope of the tangent line carefully to find the numerical answer. Part F Referring again to the graph in Part E, rank, in increasing order, the derivatives of the function at each of the points A through E . If two of the values are equal, you may list them in either order. Express your answer in order of increasing slopes as a string without spaces between points. For example, enter ABCDE if you think the derivative at point A is the smallest, B is the next larger, and so on. Hint 1. How to approach the problem Assess the values of the derivatives by looking carefully at the graph’s slope at each of the five points. The slope may be positive, zero, or negative. To rank the derivatives in increasing order, the negative derivatives (negative slopes) will be listed first, then zero, and then the positive derivatives (positive slopes). For a positive slope, the more sharply the graph slopes upward, the larger the slope and hence the larger the derivative. ANSWER: Correct At points B and D , the tangent line is horizontal, so the derivative is zero. At point C , the tangent line slopes downward, and the derivative is negative. Thus, point C has the smallest derivative, followed by B and D , which are equal. Measuring the two positive derivatives (the slopes of the tangent lines at A and E ) reveals that the slope (derivative) at E is smaller than at A . In physics, ( ) might represent an object’s position in one dimension versus time. In that case, the derivative represents velocity. Positive and negative velocity correspond to motion in opposite directions, and zero velocity means the object is at rest. This is just one of the important ways that derivatives are used in physics. decreasing increasing zero 10 5 –5 –10 0 CDBEA

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10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 16/41 Problem 2.5 shows the position graph of a particle. Part A Select the correct graph of the particle’s velocity for the interval 0 4 . ANSWER:

10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 17/41 Correct Part B Does this particle have a turning point or points? If so, at what time or times? Check all that apply. ANSWER: Correct = 0 = 1 = 2 = 3 = 4 This particle has no turning point.

10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 18/41 Overcoming a Head Start Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance beyond the starting line at . The starting line is at . Car A travels at a constant speed . Car B starts at the starting line but has a better engine than Car A, and thus Car B travels at a constant speed , which is greater than . Part A How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities. Hint 1. Consider the kinematics relation Write an expression for the displacement of Car A from the starting line at a time after Car B starts. (Note that we are taking this time to be .) Answer in terms of , , , and for time, and take at the starting line. Hint 1. What is the acceleration of Car A? The acceleration of Car A is zero, so the general formula has at least one term equal to zero. ANSWER: Hint 2. What is the relation between the positions of the two cars? The positions of the two cars are equal at time . Hint 3. Consider Car B's position as a function of time Write down an expression for the position of Car B at time after starting. Give your answer in terms of any variables needed (use for time). ANSWER: ANSWER: Correct Part B How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities. (You may use as well.) = = =

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10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 19/41 Hint 1. Which expression should you use? Just use your expression for the position of either car after time , and substitute in the correct value for (found in the previous part). ANSWER: Correct Integrals Learning Goal: To understand the concept of an integral as the area under a function graph. There are two fundamental tools of calculus—the derivative and the integral. Just as the derivative can be understood in a visual way as the slope of a function graph, what is called a definite integral can be understood as the area under a function graph. Definite integrals have many applications in physics, such as finding an object’s displacement from a known velocity and finding work done by a known force. A definite integral gives you a numerical answer. There is a second kind of integral, called the indefinite integral (or antiderivative), which is itself another function, but we won’t discuss that kind in this tutorial. Part A Many practical applications require calculations of area. Some of these calculations are straightforward, but others are more difficult, and it turns out that the idea of the definite integral can help us with the more difficult cases. Let’s start with some familiar shapes. What is the area of a rectangle of length and width ? Express your answer in terms of and . ANSWER: Correct The area of a rectangle is found by simply multiplying the length by the width. You also know how to find the area of a circle. But if you didn’t already know the formula, how would you go about figuring it out? In Part B, we’ll look at one method in detail to introduce key ideas about integrals. Part B Take a circle of radius as shown in , and imagine peeling it like an apple: Cut off a thin strip from all the way around the edge (the green strip), and straighten that strip out to a rectangle, as shown in . Call the tiny width of the strip . The length of the strip is because that’s the circumference of the circle. The bit of “peel” is the circumference of the skin on the outside of the circle. What is the area of the green strip? Express your answer in terms of , , and . =

10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 20/41 Hint 1. How to approach the problem Notice that the strip in question becomes a rectangle in the figure Part B.2. From Part A, the area of a rectangle is its length times its width. ANSWER: Correct The area of this rectangle is its length multiplied by its width . Part C Now, imagine peeling more and more strips of width from our circle, and laying them all out as rectangles side by side, as shown in – starting with the first (outermost) strip at the far right. As you peel more and more, the strips get shorter and shorter because the circle is getting smaller and smaller, until you have the smallest strip laid at far left – this strip, if is really tiny, is essentially the center of the circle! What is the area of the circle? You’ve now chopped it into little strips, but notice that the area of the circle is the same as the sum of the areas of all our strips. If is really small, you can see that all the strips approximate a triangle in the figure Part C (the area under the diagonal line). So the sum of the areas of all our strips is the area of this triangle. What is the area of the triangle? Express your answer in terms of and . Hint 1. The formula for the area of a triangle Keep in mind that the area of a triangle is one half its base times its height. ANSWER:

10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 21/41 Correct The area of the triangle is half the base (which is ) times the height (which is ). The area of the triangle in the figure Part C is , the formula you know for the area of a circle. We’ve proved it! We didn’t do it entirely rigorously, but the essence of our method (dividing something into tiny pieces and adding those up) lies at the heart of understanding definite integrals. Part D Now, look at a general function , as shown in . Suppose you want to find the area under the curve, between and —that is, the blue shaded area. Using our method above, we can add up the areas of lots of skinny rectangles like the green one. Each rectangle has area , and we want to sum these areas between and , letting get very small. Mathematically, you may recognize this sum as a definite integral: . The notation may look intimidating, but just keep in mind that the refers to the small rectangle areas, the sign stands, in a sense, for the S in “sum” (that’s actually how the symbol for integration was chosen), and the and at the bottom and top of give you the start and end of the region you’re interested in. Thus, the definite integral means the total area under the curve between and . For the function shown below, find the definite integral . Express your answer numerically. Hint 1. How to approach the problem is the area under the graph between and . Because this function graph is horizontal between and , the area you need to find is that of a rectangle. ANSWER: Correct The area under this function curve is a rectangle, so its area is just its length times its width. Part E 15

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10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 22/41 For the function shown below, find the definite integral . Express your answer numerically. Hint 1. The areas below the x -axis Again, in this part you are finding the area of a rectangle, but when the function graph is below the x -axis, the area is counted as negative . ANSWER: Correct Just as in Part D, the area is that of a rectangle, so it’s straightforward to find. The twist here is that the area lies below the x -axis, so that area is counted as negative. Part F For the function shown below, find the definite integral . Express your answer numerically. Hint 1. How to approach the problem This time you have several rectangles. Find the area of each and add them, keeping in mind that area below the x - axis is negative. ANSWER: -6

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10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 23/41 Correct In this case, the net area is the sum of positive area (above the x -axis) and negative area (below the x -axis). Part G Find the definite integral of the function shown in the graph below, over the limits from 0 to 3. Express your answer numerically. Hint 1. How to approach the problem In this case the function varies, but you can still find the definite integral by finding the area under the graph. That area is a triangle, so its area is half the base multiplied by the height. ANSWER: Correct In this case, the function varies, but you can still find the definite integral by finding the area under the graph. That area is a triangle, so its area is ½(base)(height) = ½(3)(12) = 18. One application of definite integrals in physics is in calculating the work done by a force acting through a displacement. If the function shown in the figure represents force, then the definite integral is the work done by that force. In this example, if the force is measured in newtons ( ) and position is in meters ( ), then the work done, measured in joules ( ), is Part H For the function shown below, the definite integral is closest to which of the following? 8 18

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10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 24/41 Hint 1. How to approach the problem This problem involves adding different kinds of area. Break the entire interval (from to ) into smaller intervals to find shapes that you recognize. Find the area of each, and then add your results to find the net area (the definite integral). Hint 2. What are the shapes? As shown below, this function can be broken into rectangles (yellow) and triangles (green). The area of each rectangle is length times width, and the area of each triangle is half its base times height. Read the lengths you need from the values on the graph. Hint 3. The positive and negative areas Recall that area above the x -axis is counted as positive, and area below the x -axis is negative, as shown below.

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10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 25/41 ANSWER: Correct In this case, you had to break the larger interval into smaller ones to identify the different shapes and find their areas. The definite integral always corresponds to the net area over the entire interval. Problem 2.6 - Enhanced - with Expanded Hints A particle starts from = 11 at = 0 and moves with the velocity graph shown in . Part A Does this particle have a turning point? If so, at what time? Hint 1. How to approach the problem A turning point is a point in the motion where a particle reverses direction. Examine the graph given to determine if there are turning points. Hint 2. Simplify: direction of motion How can you say in what direction the particle moves from the velocity-versus-time graph? ANSWER: 4 –8 0 –4 8

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10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 26/41 ANSWER: Correct VISUALIZE: The particle starts at = 11 at = 0. Its velocity initially points in the direction. The speed decreases as time increases during the first second, is zero at = 1 , and then increases after the particle reverses direction. SOLVE: The particle reverses direction at = 1 , when changes sign. Part B What is the object's position at = 2 ? Express your answer with the appropriate units. Hint 1. How to approach the problem The particle's position at = 0 is known. To determine the position at = 2 , you need to find the displacement of the particle in the time interval from 0 to 2 . Hint 2. Simplify: equation for displacement How can you determine the particle's displacement in the time interval from to from the velocity-versus-time graph? The velocity is at and at . ANSWER: The particle moves in the direction when the slope of the graph is positive and in the direction when the slope of the graph is negative. The particle moves in the direction when the slope of the graph is negative and in the direction when the slope of the graph is positive. The particle moves in the direction when the velocity is negative and in the direction when the velocity is positive. The particle moves in the direction when the velocity is positive and in the direction when the velocity is negative. = 0 = 1 = 2 = 3 = 4 This particle has no turning point.

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10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 27/41 Hint 3. Simplify: displacement between 0 and 2 seconds To calculate the area under the velocity curve between two time values, you can divide the entire region into simple geometric shapes, find the area of each shape, and sum them all. What is the particle's displacement between 0 and 2 seconds? Express your answer with the appropriate units. ANSWER: ANSWER: Correct SOLVE: Using the equation area of the velocity graph between and , Part C What is the object's position at = 4 ? Express your answer with the appropriate units. Hint 1. How to approach the problem You can use the position at = 2 found in the previous part and the displacement of the particle in the time interval from 2 to 4 to determine the position at = 4 . Hint 2. Simplify: equation for displacement How can you determine the particle's displacement in the time interval from to from the velocity-versus-time graph? The velocity is at and at . ANSWER: Displacement = Displacement = Displacement = Displacement = = 0 = 11

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10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 28/41 Hint 3. Simplify: displacement between 2 and 4 seconds To calculate the area under the velocity curve between two time values, you can divide the entire region into simple geometric shapes, find the area of each shape, and sum them all. What is the particle's displacement between 2 and 4 seconds? Express your answer with the appropriate units. ANSWER: ANSWER: Correct SOLVE: Using the equation area of the velocity graph between and , Problem 2.8 - Enhanced - with Hints and Feedback The following figure is a somewhat idealized graph of the velocity of blood in the ascending aorta during one beat of the heart. Displacement = Displacement = Displacement = Displacement = = 16 = 27

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10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 29/41 Part A Approximately how far, in , does the blood move during one beat? Express your answer in centimeters. Hint 1. How to approach the problem Recall the velocity versus time relationship in a case of constant acceleration. Pay attention to the units used in the problem. ANSWER: Correct What Velocity vs. Time Graphs Can Tell You A common graphical representation of motion along a straight line is the vs. graph , that is, the graph of (instantaneous) velocity as a function of time. In this graph, time is plotted on the horizontal axis and velocity on the vertical axis. Note that by definition, velocity and acceleration are vector quantities. In straight-line motion, however, these vectors have only a single nonzero component in the direction of motion. Thus, in this problem, we will call the velocity and the acceleration, even though they are really the components of the velocity and acceleration vectors in the direction of motion, respectively. is a plot of velocity versus time for a particle that travels along a straight line with a varying velocity. Refer to this plot to answer the following questions. Part A What is the initial velocity of the particle, ? Express your answer in meters per second. Hint 1. Initial velocity The initial velocity is the velocity at . Hint 2. How to read a vs. graph Recall that in a graph of velocity versus time, time is plotted on the horizontal axis and velocity on the vertical axis. For example, in the plot shown in the figure, at . = 8.0

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10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 30/41 ANSWER: Correct Part B What is the displacement of the particle? Express your answer in meters. Hint 1. How to approach the problem Recall that the area of the region that extends over a time interval under the vs. curve is always equal to the displacement in . Thus, to calculate the displacement, you need to find the area of the entire region under the vs. curve. In the case at hand, the entire region under the vs. curve is not an elementary geometrical figure, but rather a combination of triangles and rectangles. Hint 2. Find the displacement in the first 20.0 seconds What is the displacement traveled in the first 20 seconds of motion, between and ? Express your answer in meters. Hint 1. Area of the region under the vs. curve The region under the vs. curve between and can be divided into a rectangle of dimensions by , and a triangle of base and height , as shown in the figure. ANSWER: Hint 3. Find the displacement in the second 20.0 seconds What is the displacement in the second 20 seconds of motion, from to ? Express your answer in meters. Hint 1. Area of the region under the vs. curve = 0.5 = 25

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10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 31/41 The region under the vs. curve between and is a rectangle of dimensions by , as shown in the figure. ANSWER: Hint 4. Find the displacement in the last 10.0 seconds What is the displacement in the last 10 seconds of motion, from to ? Express your answer in meters. Hint 1. Area of the region under the vs. curve The region under the vs. curve between and is a triangle of base and height , as shown in the figure. ANSWER: ANSWER: = 40 = 10 = 75

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10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 32/41 Correct Here, the displacement of the particle also equals the distance traveled (75 ). Part C What is the average acceleration of the particle over the first 20.0 seconds? Express your answer in meters per second per second. Hint 1. Definition and graphical interpretation of average acceleration The average acceleration of a particle that travels along a straight line in a time interval is the ratio of the change in velocity experienced by the particle to the time interval , or . In a vs. graph, then, the average acceleration equals the slope of the line connecting the two points representing the initial and final velocities. Hint 2. Slope of a line The slope of a line from point A, of coordinates , to point B, of coordinates , is equal to the "rise" over the "run," or . ANSWER: Correct The average acceleration of a particle between two instants of time is the slope of the line connecting the two corresponding points in a vs. graph. Part D What is the instantaneous acceleration of the particle at ? Hint 1. Graphical interpretation of instantaneous acceleration The acceleration of a particle at any given instant of time or at any point in its path is called the instantaneous acceleration. If the vs. graph of the particle's motion is known, you can directly determine the instantaneous acceleration at any point on the curve. The instantaneous acceleration at any point is equal to the slope of the line tangent to the curve at that point. Hint 2. Slope of a line The slope of a line from point A, of coordinates , to point B, of coordinates , is equal to the "rise" over the "run," or . ANSWER: = 0.075

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10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 33/41 Correct The instantaneous acceleration of a particle at any point on a vs. graph is the slope of the line tangent to the curve at that point. Since in the last 10 seconds of motion, between and , the curve is a straight line, the tangent line is the curve itself. Physically, this means that the instantaneous acceleration of the particle is constant over that time interval. This is true for any motion where velocity increases linearly with time. In the case at hand, can you think of another time interval in which the acceleration of the particle is constant? Now that you have reviewed how to plot variables as a function of time, you can use the same technique and draw an acceleration vs. time graph, that is, the graph of (instantaneous) acceleration as a function of time. As usual in these types of graphs, time is plotted on the horizontal axis, while the vertical axis is used to indicate acceleration . Part E Which of the graphs shown below is the correct acceleration vs. time plot for the motion described in the previous parts? Hint 1. How to approach the problem Recall that whenever velocity increases linearly with time, acceleration is constant. In the example here, the particle's velocity increases linearly with time in the first 20.0 of motion. In the second 20.0 , the particle's velocity is constant, and then it decreases linearly with time in the last 10 . This means that the particle's acceleration is constant over each time interval, but its value is different in each interval. Hint 2. Find the acceleration in the first 20 What is , the particle's acceleration in the first 20 of motion, between and ? 1 0.20 -0.20 0.022 -0.022

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10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 34/41 Express your answer in meters per second per second. Hint 1. Constant acceleration Since we have already determined that in the first 20 of motion the particle's acceleration is constant, its constant value will be equal to the average acceleration that you calculated in Part C. ANSWER: Hint 3. Find the acceleration in the second 20 What is , the particle's acceleration in the second 20 of motion, between and ? Express your answer in meters per second per second. Hint 1. Constant velocity In the second 20 of motion, the particle's velocity remains unchanged. This means that in this time interval, the particle does not accelerate. ANSWER: Hint 4. Find the acceleration in the last 10 What is , the particle's acceleration in the last 10 of motion, between and ? Express your answer in meters per second per second. Hint 1. Constant acceleration Since we have already determined that in the last 10 of motion the particle's acceleration is constant, its constant value will be equal to the instantaneous acceleration that you calculated in Part D. ANSWER: ANSWER: Correct In conclusion, graphs of velocity as a function of time are a useful representation of straight-line motion. If read correctly, they can provide you with all the information you need to study the motion. = 0.075 = 0 = -0.20 Graph A Graph B Graph C Graph D

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10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 35/41 Problem 2.12 - Enhanced - with Expanded Hints The velocity-versus-time graph is shown for a particle moving along the -axis. Its initial position is 1.5 at 0 . Part A What is the particle's position at 1.0 ? Express your answer with the appropriate units. Hint 1. How to approach the problem When we have a velocity-versus-time graph, it is convenient to use a geometric method to find the position of a particle. The displacement of a particle can be found by the area under the velocity-versus-time graph. Hint 2. Simplify: find the area under the graph Find the area of the rectangle with sides 4 and 1 . Note: this way you will find the displacement of the particle, not its position! Express your answer with the appropriate units. ANSWER: ANSWER: Correct SOLVE: We use a graphical method to find the position of the particle at time 1.0 . Use the equation: area under the velocity-versus-time graph between and . The area between 0.0 and 1.0 is found by the formula for the area of a rectangle: . As a result, we get: . Part B Area = 4 = 5.5

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10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 36/41 What is the particle's velocity at 1.0 ? Express your answer with the appropriate units. Hint 1. How to approach the problem Find the moment of time 1.0 on the graph and find the velocity of the particle at this moment. You can also find the equation for the velocity function and calculate the velocity at any given time. Hint 2. Simplify: equation Choose the correct expression for the particle's velocity as a function of time. ANSWER: ANSWER: Correct SOLVE: Let's find the moment 1.0 on the graph, and find the speed value corresponding to this moment in time. As a result, we get: . Part C What is the particle's acceleration at 1.0 ? Express your answer with the appropriate units. Hint 1. How to approach the problem Remember that the slope of the velocity-versus-time graph at a certain time is the acceleration of a particle. Hint 2. Simplify: find the slope The slope of the graph can be calculated by the equation: ANSWER: = 4.0

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10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 37/41 ANSWER: Correct SOLVE: The acceleration is found by the equation: . Let's use the geometric meaning of the derivative, namely the slope of the tangent to the graph at a point. The slope is found by the following equation , substitute convenient values into this formula, for example, 0.0 and 2.0 . As a result, we get: . Part D What is the particle's position at 3.0 ? Express your answer with the appropriate units. Hint 1. How to approach the problem When we have a velocity-versus-time graph, it is convenient to use a geometric method to find the position of a particle. The displacement of a particle can be found by the area under the velocity-versus-time graph. Hint 2. Simplify: find the area under the graph In this case, the area under the graph will consist of the areas of a rectangle with sides 4 and 2 , and a trapezoid with bases 4 and 2 and height 1 . What is it equal to? Note: this way you will find the displacement of the particle, not its position! Express your answer with the appropriate units. ANSWER: ANSWER: = 0 Area = 11 = 12.5

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10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 38/41 Correct SOLVE: We use a graphical method to find the position of the particle at time 3.0 . Use the equation: area under the velocity-versus-time graph between and . The area between 0.0 and 3.0 can be found as a sum of the areas between 0.0 and 2.0 and between 2.0 and 3.0 . So, . As a result, we get: . Part E What is the particle's velocity at 3.0 ? Express your answer with the appropriate units. Hint 1. How to approach the problem Find the moment of time 3.0 on the graph and find the velocity of the particle at this moment. You can also find the equation for the velocity function and calculate the velocity at any given time. Hint 2. Simplify: equation Choose the correct expression for the particle's velocity as a function of time. ANSWER: ANSWER: Correct SOLVE: Let's find the moment 3.0 on the graph, and find the speed value corresponding to this moment in time. As a result, we get: . Part F What is the particle's acceleration at 3.0 ? Express your answer with the appropriate units. Hint 1. How to approach the problem = 2.0

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10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 39/41 Remember that the slope of the velocity-versus-time graph at a certain time is the acceleration of a particle. Hint 2. Simplify: find the slope The slope of the graph can be calculated by the equation: ANSWER: ANSWER: Correct SOLVE: The acceleration is found by the equation: . Let's use the geometric meaning of the derivative, namely the slope of the tangent to the graph at a point. The slope is found by the following equation , substitute convenient values into this formula, for example, 2.0 and 4.0 . As a result, we get: . Problem 2.13 - Enhanced - with Hints and Feedback Part A What constant acceleration, in SI units, must a car have to go from zero to 70 in 14 ? Express your answer with the appropriate units. Hint 1. How to approach the question Recall the velocity versus time relationship in a case of constant acceleration. Pay attention to the fact that the final velocity in the problem is given in units of miles per hour, so it should be converted to SI units before the calculations. ANSWER: Correct Part B How far has the car traveled when it reaches 70 ? Give your answer in SI units. = -2.0 = 2.2

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10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 40/41 Express your answer with the appropriate units. Hint 1. How to approach the question After you calculate acceleration in the previous part, use the expression for distance traveled in a case of constant acceleration. ANSWER: Correct Part C How far has the car traveled when it reaches 70 ? Give your answer in feet. Express your answer in feet. Hint 1. How to approach the question Simply convert the answer from the previous part to feet. Alternatively, you can convert the velocity given in part A to feet per second, find the requested acceleration in feet per second squared, and apply the expression for distance traveled in a case of constant acceleration. ANSWER: Correct Problem 2.16 When you sneeze, the air in your lungs accelerates from rest to 160 in approximately 0.70 . Part A What is the acceleration of the air in ? Express your answer in meters per second squared. ANSWER: Correct Problem 2.18 - Enhanced - with Hints and Feedback = 220 = 720 = 63

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10/3/22, 1:49 PM HW02 - Kinematics in One Dimension https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10072419 41/41 A Porsche challenges a Honda to a 500- race. Because the Porsche's acceleration of 3.5 is larger than the Honda's 3.0 , the Honda gets a 1.0 head start. Part A Who wins? Hint 1. How to approach the problem Treat both cars as particles and use the position equation to calculate the time required for each to finish the race. Note that Porsche starts race 1 second later than Honda. ANSWER: Correct Part B By how many seconds? Express your answer in seconds. Hint 1. How to approach the problem Treat both cars as particles and use the position equation to calculate the time required for each to finish the race. Note that Porsche starts race 1 second later than Honda. ANSWER: Correct Score Summary: Your score on this assignment is 106%. You received 15.97 out of a possible total of 16 points, plus 1 points of extra credit. Honda Porsche = 0.35

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