2017.02.02 Homework 2_ Kinematics in One Dimension.pdf - Homework 2 Kinematics in One Dimension Due 11:59pm on Tuesday To understand how points are

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Unformatted text preview: Homework 2: Kinematics in One Dimension Due: 11:59pm on Tuesday, January 31, 2017 To understand how points are awarded, read the Grading Policy for this assignment. Analyzing Position versus Time Graphs: Conceptual Question Two cars travel on the parallel lanes of a two­lane road. The cars’ motions are represented by the position versus time graph shown in the figure. Answer the questions using the times from the graph indicated by letters. Part A At which of the times do the two cars pass each other? Hint 1. Two cars passing Two objects can pass each other only if they have the same position at the same time. ANSWER: A B C D E None Cannot be determined Correct Part B Typesetting math: 88% Are the two cars traveling in the same direction when they pass each other? ANSWER: yes no Correct Part C At which of the lettered times, if any, does car #1 momentarily stop? 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. ANSWER: A B C D E none cannot be determined Correct Part D At which of the lettered times, if any, does car #2 momentarily stop? 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. ANSWER: Typesetting math: 88% A B C D E none cannot be determined Correct Part E At which of the lettered times are the cars moving with nearly identical velocity? 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. ANSWER: A B C D E None Cannot be determined Correct What x vs. t Graphs Can Tell You To describe the motion of a particle along a straight line, it is often convenient to draw a graph representing the position of the particle at different times. This type of graph is usually referred to as an x vs. t graph. To draw such a graph, choose an axis system in which time t is plotted on the horizontal axis and position x on the vertical axis. Then, indicate the values of x at Typesetting math: 88% various times t . Mathematically, this corresponds to plotting the variable x as a function of t . An example of a graph of position as a function of time for a particle traveling along a straight line is shown below. Note that an x vs. t graph like this does not represent the path of the particle in space. Now let's study the graph shown in the figure in more detail. Refer to this graph to answer Parts A, B, and C. Part A What is the total distance Δx traveled by the particle? Express your answer in meters. Hint 1. Total distance The total distance Δx traveled by the particle is given by the difference between the initial position x0 at t and the position x at t = 50.0 s. In symbols, Δx = x − x0 = 0.0 s . Hint 2. How to read an x vs. t graph Remember that in an x vs. t graph, time t is plotted on the horizontal axis and position x on the vertical axis. For example, in the plot shown in the figure, x = 16.0 m at t = 10.0 s. ANSWER: Δx = 30 m Correct Part B What is the average velocity v av of the particle over the time interval Δt = 50.0 s ? Express your answer in meters per second. Hint 1. Definition and graphical interpretation of average velocity The average velocity v av of a particle that travels a distance Δx along a straight line in a time interval Δt is defined as Typesetting math: 88% = Δx vav = Δx Δt . In an x vs. t graph, then, the average velocity equals the slope of the line connecting the initial and final positions. Hint 2. Slope of a line The slope m of a line from point A, with coordinates (tA , xA ), to point B, with coordinates (tB , xB ), is equal to the "rise" over the "run," or m= x B −x A t B −t A . ANSWER: v av = 0.600 m/s Correct The average velocity of a particle between two positions is equal to the slope of the line connecting the two corresponding points in an x vs. t graph. Part C What is the instantaneous velocity v of the particle at t = 10.0 s ? Express your answer in meters per second. Hint 1. Graphical interpretation of instantaneous velocity The velocity of a particle at any given instant of time or at any point in its path is called instantaneous velocity. In an x vs. t graph of the particle's motion, you can determine the instantaneous velocity of the particle at any point in the curve. The instantaneous velocity at any point is equal to the slope of the line tangent to the curve at that point. ANSWER: v = 0.600 m/s Correct The instantaneous velocity of a particle at any point on its x vs. t graph is the slope of the line tangent to the curve at that point. Since in the case at hand the curve is a straight line, the tangent line is the curve itself. Physically, this means that the instantaneous velocity of the particle is constant over the entire time interval of motion. This is true for any motion where distance increases linearly with time. Another common graphical representation of motion along a straight line is the v vs. t graph, that is, the graph of (instantaneous) velocity as a function of time. In this graph, time t is plotted on the horizontal axis and velocity v on the vertical axis. Note that by definition, velocity and acceleration are vector quantities. In straight­line motion, however, these vectors have only one nonzero component in the direction of motion. Thus, in this problem, we will call v the velocity and a the acceleration, even though they are really the components of the velocity and acceleration vectors in the direction of motion. Typesetting math: 88% Part D Which of the graphs shown is the correct v vs. t plot for the motion described in the previous parts? Hint 1. How to approach the problem Recall your results found in the previous parts, namely the fact that the instantaneous velocity of the particle is constant. Which graph represents a variable that always has the same constant value at any time? ANSWER: Graph A Graph B Graph C Graph D Correct Whenever a particle moves with constant nonzero velocity, its x vs. t graph is a straight line with a nonzero slope, and its v vs. t curve is a horizontal line. Part E Shown in the figure is the v vs. t curve selected in the previous part. What is the area A of the shaded region under the curve? Typesetting math: 88% Express your answer in meters. Hint 1. How to approach the problem The shaded region under the v vs. t curve is a rectangle whose horizontal and vertical sides lie on the t axis and the v axis, respectively. Since the area of a rectangle is the product of its sides, in this case the area of the shaded region is the product of a certain quantity expressed in seconds and another quantity expressed in meters per second. The area itself, then, will be in meters. ANSWER: A = 30 m Correct Compare this result with what you found in Part A. As you can see, the area of the region under the v vs. t curve equals the total distance traveled by the particle. This is true for any velocity curve and any time interval: The area of the region that extends over a time interval Δt under the v vs. t curve is always equal to the distance traveled in Δt. Position versus Time Graphs Conceptual Question The motions described in each of the questions take place at an intersection on a two­lane 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 x = 0, and east is the positive x direction. Typesetting math: 88% 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: A B C D E F Correct Typesetting math: 88% 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: A B C D E F 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 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 Typesetting math: 88% 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 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: Typesetting math: 88% A B C D E F Correct What Velocity vs. Time Graphs Can Tell You A common graphical representation of motion along a straight line is the v vs. t graph, that is, the graph of (instantaneous) velocity as a function of time. In this graph, time t is plotted on the horizontal axis and velocity v on the vertical axis. Note that by definition, velocity and acceleration are vector quantities. In straight­line motion, however, these vectors have only a single nonzero component in the direction of motion. Thus, in this problem, we will call v the velocity and a the acceleration, even though they are really the components of the velocity and acceleration vectors in the direction of motion, respectively. Here 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, v 0 ? Express your answer in meters per second. Hint 1. Initial velocity The initial velocity is the velocity at t = 0 s . Hint 2. How to read a v vs. t graph Recall that in a graph of velocity versus time, time is plotted on the horizontal axis and velocity on the vertical axis. = 2.00 m/s at t = 30.0 s. Typesetting math: 88% For example, in the plot shown in the figure, v ANSWER: v0 = 0.5 m/s Correct Part B What is the total distance Δx traveled by 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 Δt under the v vs. t curve is always equal to the distance traveled in Δt. Thus, to calculate the total distance, you need to find the area of the entire region under the v vs. t curve. In the case at hand, the entire region under the v vs. t curve is not an elementary geometrical figure, but rather a combination of triangles and rectangles. Hint 2. Find the distance traveled in the first 20.0 seconds What is the distance Δx1 traveled in the first 20 seconds of motion, between t and t = 0.0 s = 20.0 s ? Express your answer in meters. Hint 1. Area of the region under the v vs. t curve The region under the v vs. t curve between t = 0.0 s and t = 20.0 s can be divided into a rectangle of dimensions 20.0 s by 0.50 m/s, and a triangle of base 20.0 s and height 1.50 m/s, as shown in the figure. ANSWER: Δx1 = 25 m Typesetting math: 88% Hint 3. Find the distance traveled in the second 20.0 seconds Δ = 40.0 s What is the distance Δx2 traveled in the second 20 seconds of motion, from t = 20.0 s to t = 40.0 s ? Express your answer in meters. Hint 1. Area of the region under the v vs. t curve The region under the v vs. t curve between t by 2.00 m/s, as shown in the figure. = 20.0 s and t = 40.0 s is a rectangle of dimensions 20.0 ANSWER: Δx2 = 40 m Hint 4. Find the distance traveled in the last 10.0 seconds What is the distance Δx3 traveled in the last 10 seconds of motion, from t = 40.0 s to t = 50.0 s ? Express your answer in meters. Hint 1. Area of the region under the v vs. t curve The region under the v vs. t curve between t height 2.00 m/s, as shown in the figure. Typesetting math: 88% = 40.0 s and t = 50.0 s is a triangle of base 10.0 s and s ANSWER: Δx3 = 10 m ANSWER: Δx = 75 m Correct Part C What is the average acceleration aav 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 aav of a particle that travels along a straight line in a time interval Δt is the ratio of the change in velocity Δv experienced by the particle to the time interval Δt, or aav = Δv Δt . In a v vs. t 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 m of a line from point A, of coordinates (xA , yA ), to point B, of coordinates (xB , yB ), is equal to the "rise" over the "run," or m= y B −y A x B −x A . ANSWER: aav = 0.075 m/s2 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 v vs. t graph. Part D What is the instantaneous acceleration a of the particle at t = 45.0 s ? Hint 1. Graphical interpretation of instantaneous acceleration Typesetting math: 88% 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 v vs. t 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 m of a line from point A, of coordinates (xA , yA ), to point B, of coordinates (xB , yB ), is equal to the "rise" over the "run," or m= y B −y A x B −x A . ANSWER: 1 m/s2 0.20 m/s2 = a ­0.20 m/s2 0.022 m/s2 ­0.022 m/s2 Correct The instantaneous acceleration of a particle at any point on a v vs. t graph is the slope of the line tangent to the curve at that point. Since in the last 10 seconds of motion, between t = 40.0 s and t = 50.0 s, 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 t is plotted on the horizontal axis, while the vertical axis is used to indicate acceleration a. Part E Which of the graphs shown below is the correct acceleration vs. time plot for the motion described in the previous parts? Typesetting math: 88% 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 s of motion. In the second 20.0 s , the particle's velocity is constant, and then it decreases linearly with time in the last 10 s. 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 \rm s What is \texttip{a_{\rm 1}}{a_1}, the particle's acceleration in the first 20 \rm s of motion, between t=0.0\;\rm s and t=20.0\;\rm s? Express your answer in meters per second per second. Hint 1. Constant acceleration Since we have already determined that in the first 20 \rm s 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: \texttip{a_{\rm 1}}{a_1} = 0.075 \rm m/s^2 Hint 3. Find the acceleration in the second 20 \rm s What is \texttip{a_{\rm 2}}{a_2}, the particle's acceleration in the second 20 \rm s of motion, between t=20.0\;\rm s and t=40.0\;\rm s? Express your answer in meters per second per second. Typesetting math: 88% Hint 1. Constant velocity In the second 20 \rm s of motion, the particle's velocity remains unchanged. This means that in this time interval, the particle does not accelerate. ANSWER: \texttip{a_{\rm 2}}{a_2} = 0 \rm m/s^2 Hint 4. Find the acceleration in the last 10 \rm s What is \texttip{a_{\rm 3}}{a_3}, the particle's acceleration in the last 10 \rm s of motion, between t=40.0\;\rm s and t=50.0\;\rm s? Express your answer in meters per second per second. Hint 1. Constant acceleration Since we have already determined that in the last 10 \rm s of motion the particle's acceleration is constant, its constant value will be equal to the instantaneou...
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