Mastering Physics Ch 02 HW College Physics I Brian Uzpen LCCC
Short Description
Mastering Physics...
Description
5/6/2016
Ch 02 HW
Ch 02 HW Due: 9:30am on Monday, February 8, 2016 You will receive no credit for items you complete after the assignment is due. Grading Policy
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 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 x 0 at t = 0.0 s and the position x at t = 50.0 s. In symbols, Δx = x − x 0
.
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
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
1/33
5/6/2016
Ch 02 HW
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 v av =
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 (t A , x A ) , to point B, with coordinates (t B , x B ) , 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: https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
2/33
5/6/2016
Ch 02 HW 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.
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?
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
3/33
5/6/2016
Ch 02 HW
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? 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
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
4/33
5/6/2016
Ch 02 HW
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.
PSS 2.1 Kinematics with Constant Acceleration Learning Goal: To practice ProblemSolving Strategy 2.1 for constant acceleration problems. A car is traveling at a constant velocity of magnitude v 0 when the driver notices a garbage can on the road in front of him. At that moment, the distance between the garbage can and the front of the car is d . A time t after noticing the garbage can, the driver applies the brakes and slows down at a constant rate before coming to a halt just before the garbage can. What is the magnitude of the car's acceleration after the brakes are applied? PROBLEMSOLVING STRATEGY 2.1 Kinematics with constant acceleration MODEL: Use the particle model. Make simplifying assumptions. VISUALIZE: Use different representations of the information in the problem.
Draw a pictorial representation. This helps you assess the information you are given and starts the process of translating the problem into symbols. Use a graphical representation if it is appropriate for the problem. Go back and forth between these two representations as needed. SOLVE: The mathematical representation is based on the three kinematic equations
,
v f s = v is + a s Δt
sf = si + v is Δt + v
2 fs
= v
2 is
1 2
as (Δt)
+ 2a s Δs
2
, and
.
Use x or y, as appropriate to the problem, rather than the generic s . Replace i and f with numerical subscripts defined in the pictorial representation. Uniform motion with constant velocity has a s = 0. ASSESS: Is your result believable? Does it have proper units? Does it make sense?
Model Start by making simplifying assumptions appropriate for the situation. In this problem, the object of interest, the car, should be modeled as a particle.
Visualize Part A Below is a sketch of the situation described in this problem, along with four different motion diagrams. Which of https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
5/33
5/6/2016
Ch 02 HW
these diagrams is the correct pictorial representation of the problem?
ANSWER: (a) (b) (c) (d)
Correct
Part B As an alternative to the pictorial representation shown above, you could consider using a graphical representation instead. Below are four velocityversustime graphs. Which graph correctly represents the situation described in this problem?
ANSWER:
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
6/33
5/6/2016
Ch 02 HW
A B C D
Correct Now that you have assessed the information given in the problem, you need to translate it into symbols. You know that when the driver hits the brakes at time t after seeing the obstacle, the magnitude of the car's acceleration changes from zero to a constant (unknown) value. Therefore, you will need to consider the two time intervals (before and after the brakes are pressed) separately, since they have different accelerations. Align the x axis with the direction of motion, and let t 0 be the moment when the driver sees the garbage can, x 0 be the car's position at that moment, and v 0x be the car's velocity at that moment. Let t 1 , x 1 , and v 1x be the time, position, and velocity, respectively, when the driver hits the brakes, and let t 2 , x 2 , and v 2x be the time, position, and velocity, respectively, when the car finally comes to rest. Note that it is most convenient to take the origin of the coordinate axis to be at x 0 . At this point, your pictorial representation should look like this:
Now, decide which quantities are known and which need to be found.
Solve Part C Find the magnitude of a x , the acceleration of the car after the brakes are applied. Express your answer in terms of some or all of the variables d , t , and v 0 .
Hint 1. How to approach the problem The relevant kinematic equations needed to solve this problem are given in the strategy above. You might find, though, that you will not need them all. In fact, even though you know the delay occurring from when the driver sees the garbage can and when the brakes are applied, you don't know how long it takes the car to come to a rest. Thus, the first equation might not be very useful. What you do know, instead, is the total distance the car travels before stopping. You also know that a portion of that distance is covered at a constant velocity v 0 , so using the appropriate equation, you could calculate what distance the car still has to travel before stopping. That information, when used with the appropriate equation, will allow you to find the magnitude of the car's acceleration after the brakes are applied.
Hint 2. Find the distance traveled in the first time interval After the driver first notices the obstacle, the car moves uniformly for a time interval t 1 − t 0 = t before the brakes are applied. Find the distance x 1 − x 0 covered by the car in the time interval before the brakes are https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
7/33
5/6/2016
Ch 02 HW
applied and after the driver sees the obstacle. Express your answer in terms of some or all of the variables d , v 0 , t , and a x . ANSWER: x1 − x0
=
v0 t
Incorrect; Try Again; 5 attempts remaining Hint 3. Find the distance traveled in the second time interval When the driver first notices the garbage can, the distance between the car and the can is x 2 − x 0 = d. You have found that the distance traveled by the car before the brakes are applied is v 0 t. Find the distance x 2 − x 1 traveled by the car after the brakes are applied. Express your answer in terms of some or all of the variables d , v 0 , t , and a x . ANSWER: x2 − x1
=
d − v0 t
Hint 4. Find which equation to use When you apply each of the three kinematics equations listed in the strategy above to the car's motion after the brakes are applied, which equation will allow you to find the magnitude of the car's acceleration, given only the initial and final velocity and the distance traveled in that time interval? ANSWER: v f s = v is + a s Δt
sf = si + v is Δt + v
2 fs
= v
2 is
1 2
as (Δt)
2
+ 2a s Δs
ANSWER:
|a x |
v0
=
2
2(d−v 0 t)
Correct
Assess Part D https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
8/33
5/6/2016
Ch 02 HW
In Part C, you found that the magnitude of the car's acceleration after the brakes are applied is |ax | =
v
2 0
2(d−v 0 t)
.
Based on this expression, what happens to |a x | if t increases and all the other variables remain constant? ANSWER: |a x |
decreases because it is inversely proportional to a linear function of t that increases as t increases.
|a x |
increases because it is inversely proportional to a linear function of t that increases as t increases.
|a x |
increases because it is a linear function of t .
|a x |
decreases because it is inversely proportional to a linear function of t that decreases as t increases.
|a x |
increases because it is inversely proportional to a linear function of t that decreases as t increases.
Correct Your answer does make physical sense! If the driver takes longer to react (i.e., if t increases), the acceleration required to stop the car just before the garbage can must be larger in magnitude because now the distance traveled after the brakes are applied is smaller.
Clear the Runway To take off from the ground, an airplane must reach a sufficiently high speed. The velocity required for the takeoff, the takeoff velocity, depends on several factors, including the weight of the aircraft and the wind velocity.
Part A A plane accelerates from rest at a constant rate of 5.00 m/s 2 along a runway that is 1800 m long. Assume that the plane reaches the required takeoff velocity at the end of the runway. What is the time t TO needed to take off? Express your answer in seconds using three significant figures.
Hint 1. How to approach the problem As the plane travels along the runway, it has constant acceleration. To solve the problem, you'll need to use the kinematics equations for such motion. In particular, you need to use the equation relating the distance traveled and time.
Hint 2. Find the equation for the distance traveled by the plane Which expression best describes the distance d traveled by the plane during a certain interval of time t ? Let s 0 and v 0 be, respectively, the initial position and speed of the plane, and use a for the acceleration of the plane. Remember that the plane accelerates from rest. ANSWER:
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
9/33
5/6/2016
Ch 02 HW
1
v0 +
2
at
2
s0 + v 0 t + d
=
1 2
at
1 2
at
2
2
s0 + v 0 + s0 +
1 2
at
1 2
at
2
2
ANSWER: t TO
= 26.8 s
Correct If you need to use the answer from this part in subsequent parts, use the unrounded value you calculated before you rounded the answer to three significant figures. Recall that you should only round as a final step before submitting your answer.
Part B What is the speed v TO of the plane as it takes off? Express your answer numerically in meters per second using three significant figures.
Hint 1. How to approach the problem Since you are given the constant acceleration of the plane, and you have also found the time it takes to take off, you can calculate the speed of the plane as it ascends into the air using the equation for the velocity of an object in motion at constant acceleration.
Hint 2. Find the equation for the velocity of the plane Which expression best describes the velocity v of the plane after a certain interval of time t ? Let v 0 be the initial velocity of the plane, and use a for the acceleration of the plane. Remember that the plane starts from rest. ANSWER: v0 t
v
=
v 0 + at
v0 t +
1 2
at
2
at
ANSWER: https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
10/33
5/6/2016
Ch 02 HW v TO
= 134 m/s
Correct If you need to use the answer from this part in subsequent parts, use the unrounded value you calculated before you rounded the answer to three significant figures. Recall that you should only round as a final step before submitting your answer.
Part C What is the distance d f irst traveled by the plane in the first second of its run? Express your answer numerically in meters using three significant figures.
Hint 1. How to approach the problem Apply the same equation that you used to solve Part A. ANSWER: d f irst
= 2.50 m
Correct If you need to use the answer from this part in subsequent parts, use the unrounded value you calculated before you rounded the answer to three significant figures. Recall that you should only round as a final step before submitting your answer.
Part D What is the distance d last traveled by the plane in the last second before taking off? Express your answer numerically in meters using three significant figures.
Hint 1. How to approach the problem Use the equation that gives the distance traveled as a function of time. Note that you are looking for the distance traveled in the last second before the plane takes off, which can be expressed as the length of the runway minus the distance traveled by the plane up to that last second. ANSWER: d last
= 132 m
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
11/33
5/6/2016
Ch 02 HW
Correct Since the plane is accelerating, the average speed of the plane during the last second of its run is greater than its average speed during the first second of the run. Not surprisingly, so is the distance traveled. If you need to use the answer from this part in subsequent parts, use the unrounded value you calculated before you rounded the answer to three significant figures. Recall that you should only round as a final step before submitting your answer.
Part E What percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway? Express your answer numerically to the nearest percent.
Hint 1. How to approach the problem You need to find the velocity of the plane by the time it covers half the length of the runway and compare it with the takeoff velocity. Apply the same method that you used to determine the takeoff velocity. ANSWER: 70 %
Correct This is a "rule of thumb" generally used by pilots. Since the takeoff velocity for a particular aircraft can be computed before the flight, a pilot can determine whether the plane will successfully take off before the end of the runway by verifying that the plane has gained 70% of the takeoff velocity by the time it reaches half the length of the runway. If the plane hasn't reached that velocity, the pilot knows that there isn't enough time to reach the needed takeoff velocity before the plane reaches the end of the runaway. At that point, applying the brakes and aborting the takeoff is the safest course of action.
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 DA beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed v A . Car B starts at the starting line but has a better engine than Car A, and thus Car B travels at a constant speed v B , which is greater than v A .
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 t after Car B starts. (Note that we are taking this time to be t = 0.) D t https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
x = 0
12/33
5/6/2016
Ch 02 HW
Answer in terms of v A , v B , DA , and t for time, and take x
at the starting line.
= 0
Hint 1. What is the acceleration of Car A? The acceleration of Car A is zero, so the general formula x(t) one term equal to zero.
= x 0 + v 0 t + (1/2)at
2
has at least
ANSWER: x A (t)
=
DA + v A t
Hint 2. What is the relation between the positions of the two cars? The positions of the two cars are equal at time t catch .
Hint 3. Consider Car B's position as a function of time Write down an expression for the position of Car B at time t after starting. Give your answer in terms of any variables needed (use t for time). ANSWER: x B (t)
=
vB t
ANSWER:
t catch
=
DA v B −v A
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 t catch as well.)
Hint 1. Which expression should you use? Just use your expression for the position of either car after time t t catch (found in the previous part).
, and substitute in the correct value for
= 0
ANSWER:
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
13/33
5/6/2016
Ch 02 HW d pass
=
t catch v B
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 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 t , 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: slope =
Δx Δt
.
ANSWER:
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
14/33
5/6/2016
Ch 02 HW
Trial A Trial B Trial C Trial D
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 average velocity =
Δ(position) Δ(time)
=
Δx Δt
.
Then note that the question asks only about the magnitude of the velocity. ANSWER: Trial A Trial B Trial C Trial D
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.
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.
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
15/33
5/6/2016
Ch 02 HW
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 Are the two cars traveling in the same direction when they pass each other? ANSWER: yes no
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
16/33
5/6/2016
Ch 02 HW
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:
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
17/33
5/6/2016
Ch 02 HW
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
PhET Tutorial: Moving Man Learning Goal: To understand the relationships between position, velocity, and acceleration. For this tutorial, use the PhET simulation The Moving Man. This simulation allows you to drag a person back and forth and look at the resulting position, velocity, and acceleration. You can also enter a position as a function of time mathematically and look at the resulting motion. Start the simulation. When you click the simulation link, you may be asked whether to run, open, or save the file. https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
18/33
5/6/2016
Ch 02 HW
Choose to run or open it.
Under the Charts tab you can click and drag the person left and right, or enter a numeric value in the boxes on the left panel to see plots for the person’s position, velocity, and acceleration as a function of time. Click the Play button to start a simulation and the Pause button to stop a simulation. You can also watch a playback by selecting the Playback radio button instead of the default Record radio button. You can click Clear to remove the current plot while maintaining your settings for position, velocity and acceleration or click Reset All to start over. In the Playback mode, the grey bar can be dragged over the plot to any value in time, and the digital readouts will show the corresponding values of the position, velocity, and acceleration. Under the Special Features menu, the Expression Evaluator option produces a second window in which you can mathematically type in any function for the position as a function of time, x(t). After typing in a function, click the Play button to start the simulation. To zoom in vertically, click any of the three + buttons to the top right of each plot. To zoom in horizontally, click the + button to the bottom right of the acceleration plot. Feel free to play around with the simulation. When you are done, click Reset All on the Charts tab before beginning Part A.
Part A First, you will focus on the relationship between velocity and position. Recall that velocity is the rate of change of position (v x = dx/dt). This means that the velocity is equal to the slope of the Position vs. Time graph. Move the person to the position x = −6 m or enter –6.00 in the position box. If you dragged the person to position, click the Pause button and then the Clear button. Next, drag the person to the right to roughly x = 6 m and reverse his direction, returning him to the original position, at x = −6 m. Move the person relatively quickly, about a few seconds for the round trip. Your plots should look something like those shown below.
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
19/33
5/6/2016
Ch 02 HW
Look at the Position vs. Time and Velocity vs. Time plots. What is the person's velocity when his position is at its maximum value (around 6 m )? ANSWER: positive. The person's velocity is
negative. zero.
Correct When the person’s position is a maximum, the slope of the position with respect to time is zero, so dx/dt = . However, due to the person’s acceleration, the velocity does not remain zero; he eventually moves to the left.
0
Part B Acceleration is the rate of change of the velocity, a x
, so it is the slope of the Velocity vs. Time graph.
= dv x /dt
Because it is difficult to drag the person in a consistent and reproducible way, use the Expression Evaluator under the Special Features menu for this question. Click Reset All and type in the function x(t) = 8 ∗ t − 2 ∗ t ∗ t in the Expression Evaluator. Click the Play button and let the simulation run roughly 5 simulation seconds before pressing the Pause button. Use the zoom buttons to adjust the plots so they fit in the screen. You should see a plot similar to what you got in the previous question, but much smoother. Look at the Position vs. Time, Velocity vs. Time, and Acceleration vs. Time plots.
Hint 1. How to approach the problem In Playback mode, use the grey vertical bar. Slide the bar until the value x https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
is displayed in the
= 8m
x = 8m
20/33
5/6/2016
Ch 02 HW
position box on the left panel. What are the values of velocity and acceleration when x
?
= 8m
ANSWER: both the velocity and the acceleration are nonzero. When the person is 8 m to the right of the origin,
the velocity is zero but the acceleration is negative.
the velocity is zero but the acceleration is positive. both the velocity and the acceleration are zero.
Correct At x = 8 m, the person turns to go back in the opposite direction. His velocity is zero, but his acceleration is negative since the velocity is decreasing with time. This is similar to throwing a ball straight up into the air; at its highest point, the velocity is zero but the acceleration is still directed downward.
Part C Keep the function x(t) = 8 ∗ t − 2 ∗ t ∗ t in the Expression Evaluator. What is the value of the person’s acceleration a at t = 2 s?
Hint 1. How to approach the problem Use the grey vertical bar. Slide the bar until it coincides with t value displayed in the acceleration box on the left panel?
on the horizontal axis. What is the
= 2s
ANSWER: 4 m/s
ax
=
2
−4 m/s
2
0 −2 m/s
2
Correct This is an example of onedimensional motion with constant acceleration. The position of an object undergoing this type of motion obeys the kinematic equation x(t) = x 0 + v x,0 t + 1/2 a x t 2 . In this case, the initial velocity is v x,0
and the acceleration is a x
= 8 m/s
= −4 m/s
2
(since 1/2 a x
= −2 m/s
2
).
Part D In the previous question, the person had an initial velocity of 8 m/s and a constant acceleration of −4 m/s 2 . How would the maximum distance he travels to the right of the origin change if instead his initial velocity were doubled ( v x,0 = 16 m/s)? https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
21/33
5/6/2016
Ch 02 HW
Hint 1. How to approach the problem Go to the Introduction tab to run the simulation using the new initial velocity v x,0 = 16 m/s and the same acceleration of −4 m/s 2 , and read the value for position when the velocity equals zero. Remember to remove the walls from the simulation by clicking on the red close button on the walls. In Playback mode the simulation can be run slowly and paused when the velocity is zero. Or, mathematically, determine how long it takes for the person to stop and use this value of time in the equation for x(t).
ANSWER: The maximum distance would double. The maximum distance would not change. The maximum distance would increase by a factor of four.
Correct Because it takes twice as much time to momentarily stop, and because his average velocity will be twice as fast, the distance he travels will be four times greater. Using the kinematic equation, 2 2 x(4 s) = (16 m/s) × (4 s) − (1/2) × (4 m/s ) × (4 s) = 32 m .
Part E Now, assume that the position is given by the equation x(t)
= 4t
3
.
Enter this function in the Expression Evaluator as x(t) = 4 ∗ t ∗ t ∗ t . Run the simulation by clicking the Play button in the Record mode for roughly three simulated seconds and then click the Pause button. Now take a look at the graphs. You will have to zoom in horizontally (bottom right), so that your range covers two seconds. Which of the following statements is true? ANSWER: The position is increasing at a constant rate. The velocity is increasing at a constant rate. The acceleration is increasing at a constant rate. The acceleration is constant in time.
Correct The graph showing Acceleration vs. Time is a straight line that is not horizontal.
Part F What is the position of the person when t
?
= 1s
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
22/33
5/6/2016
Ch 02 HW
Express your answer numerically in meters to one significant figure.
Hint 1. How to approach the problem You can solve this mathematically using the expression input for position as a function of time, x(t) = 4 ∗ t ∗ t ∗ t . Alternatively, you can put the simulation in Playback mode and drag the grey box/bar to t the approximate value for position.
, then read
= 1s
ANSWER: 4 m
Correct Notice that since the position is given by x x = 4(1)
3
= 4t
3
, when the time is t
, the position is
= 1s
m = 4 m.
Part G What is the velocity of the person when t
?
= 1s
Express your answer numerically in meters per second to two significant figures.
Hint 1. How to approach the problem The velocity is the first derivative of position with respect to time, v x of the expression of x(t) and evaluate when t = 1 s.
. You can take the derivative
= dx/dt
Alternatively, you can put the simulation in Playback mode and drag the grey box/bar to t the approximate value for velocity.
, then read
= 1s
ANSWER: 12 m/s
Correct Notice that since the position is given by x(t) = 4t 3 , the velocity, which is the first derivative of position with respect to time, is given by v x = dx/dt = 12t 2. So when t = 1 s, v = 12 m/s.
Part H What is the acceleration of the person when t
?
= 1s
Express your answer numerically in meters per second squared to two significant figures. https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
23/33
5/6/2016
Ch 02 HW
Hint 1. How to approach the problem The acceleration is the first derivative of velocity with respect to time, a x derivative of position with respect to time, a x v x (t) and evaluate when t = 1 s.
= d
2
x/dt
2
or the second
= dv x /dt
. You can take the derivative of the expression of
Alternatively, you can put the simulation in Playback mode and drag the grey box/bar to t the approximate value for acceleration.
, then read
= 1s
ANSWER: 24 m/s 2
Correct Notice that since the position is given by x = 4t 3 , the acceleration, which is the first derivative of velocity with respect to time, is given by a x = dv x /dt = d(12t 2 )/dx = 24t. So when t = 1 s, a x = 24 m/s 2 . Notice also that the acceleration is proportional to time, which explains why it is increasing at a constant rate (as discovered in part G).
PhET Interactive Simulations University of Colorado http://phet.colorado.edu
Problem 2.28 The figure shows the acceleration graph for a particle that starts from rest at t = 0s .
Part A Determine the object's velocity at times t
0 s , 2 s , 4 s , 6 s , and 8 s .
=
Express your answer using two significant figures. Enter your answers numerically separated by commas. https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
24/33
5/6/2016
Ch 02 HW
ANSWER: v(0), v(2), v(4), v(6), v(8)
= 0,5.0,20,30,30 m/s
Correct
Problem 2.3 Julie drives 100 mi to Grandmother's house. On the way to Grandmother's, Julie drives half the distance at 40.0 mph and half the distance at 60.0 mph . On her return trip, she drives half the time at 40.0 mph and half the time at 60.0 mph .
Part A What is Julie's average speed on the way to Grandmother's house? Express your answer with the appropriate units. ANSWER: 48.0 mph
Correct
Part B What is her average speed on the return trip? Express your answer with the appropriate units. ANSWER: 50.0 mph
Correct
Problem 2.11 The figure shows the velocity graph of a particle moving along the xaxis. Its initial position is x 0 = t = 2 s , what are the particle's (a) position, (b) velocity, and (c) acceleration?
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
at t 0 = 0 . At
2 m
25/33
5/6/2016
Ch 02 HW
Part A Express your answer to two significant figures and include the appropriate units. ANSWER: x
= 6.0 m
Correct
Part B Express your answer to two significant figures and include the appropriate units. ANSWER: vx
= 4.0
m s
Correct
Part C Express your answer to two significant figures and include the appropriate units. ANSWER: ax
= 2.0
m s
2
Correct
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
26/33
5/6/2016
Ch 02 HW
Problem 2.12 The velocityversustime graph is shown for a particle moving along the xaxis. Its initial position is x 0 = 3.0 m at t 0 = 0 s.
Part A What is the particle's position at t =
1.0 s ?
Express your answer to two significant figures and include the appropriate units. ANSWER: x
= 7.0 m
Correct
Part B What is the particle's velocity at t =
1.0 s ?
Express your answer to two significant figures and include the appropriate units. ANSWER: v
= 4.0
m s
Correct
Part C What is the particle's acceleration at t =
1.0 s ?
Express your answer to two significant figures and include the appropriate units. ANSWER:
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
27/33
5/6/2016
Ch 02 HW
a
m
= 0 s
2
Correct
Part D What is the particle's position at t =
3.0 s ?
Express your answer to three significant figures and include the appropriate units. ANSWER: x
= 14.0 m
Correct
Part E What is the particle's velocity at t =
3.0 s ?
Express your answer to two significant figures and include the appropriate units. ANSWER: v
m = 2.0 s
Correct
Part F What is the particle's acceleration at t =
3.0 s ?
Express your answer to two significant figures and include the appropriate units. ANSWER: a
= 2.0
m s
2
Correct
Problem 2.17 Ball bearings are made by letting spherical drops of molten metal fall inside a tall tower − called a shot tower − and https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
28/33
5/6/2016
Ch 02 HW
solidify as they fall.
Part A If a bearing needs 4.20 s to solidify enough for impact, how high must the tower be? Express your answer with the appropriate units. ANSWER: 86.4 m
Correct
Part B What is the bearing's impact speed? Express your answer with the appropriate units. ANSWER: 41.2
m s
Correct
Problem 2.24 A particle moving along the xaxis has its velocity described by the function v x = position is x 0 = 1.7 m at t 0 = 0 s .
2t
2
, where t is in s . Its initial
m/s
Part A At 1.5 s , what is the particle's position? Express your answer with the appropriate units. ANSWER: x
= 4.0 m
Correct
Part B At 1.5 s , what is the particle's velocity? Express your answer with the appropriate units. https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
29/33
5/6/2016
Ch 02 HW
ANSWER: v
m = 4.5 s
Correct
Part C At 1.5 s , what is the particle's acceleration? Express your answer with the appropriate units. ANSWER: a
= 6.0
m s
2
Correct
Problem 2.43 You are driving to the grocery store at 20 m/s . You are 110 m from an intersection when the traffic light turns red. Assume that your reaction time is 0.50 s and that your car brakes with constant acceleration.
Part A How far are you from the intersection when you begin to apply the brakes? Express your answer to two significant figures and include the appropriate units. ANSWER: s
= 100 m
Correct
Part B What acceleration will bring you to rest right at the intersection? Express your answer to two significant figures and include the appropriate units. ANSWER: a
m
= 2.0 s
2
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
30/33
5/6/2016
Ch 02 HW
Correct
Part C How long does it take you to stop after the light turns red? Express your answer to two significant figures and include the appropriate units. ANSWER: t
= 11 s
Correct
Problem 2.52 A hotel elevator ascends 200 magnitude of 1.0 m/s 2 .
m
with maximum speed of 5
m/s
. Its acceleration and deceleration both have a
Part A How far does the elevator move while accelerating to full speed from rest? Express your answer with the appropriate units. ANSWER: 12.5 m
Correct
Part B How long does it take to make the complete trip from bottom to top? Express your answer with the appropriate units. ANSWER: 45.0 s
Correct
Problem 2.55 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
31/33
5/6/2016
Ch 02 HW
Santa loses his footing and slides down a frictionless, snowy roof that is tilted at an angle of 30.0 ∘ .
Part A If Santa slides 10.0 m before reaching the edge, what is his speed as he leaves the roof? Express your answer with the appropriate units. ANSWER: 9.90 m s
Correct
Problem 2.65 David is driving a steady 21.0 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2.70 m/s 2 at the instant when David passes.
Part A How far does Tina drive before passing David? Express your answer with the appropriate units. ANSWER: 327 m
Correct
Part B What is her speed as she passes him? Express your answer with the appropriate units. ANSWER: 42.0
m s
Correct
Problem 2.7 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
32/33
5/6/2016
Ch 02 HW
The following figure is a somewhat idealized graph of the velocity of blood in the ascending aorta during one beat of the heart.
Part A Approximately how far, in cm, does the blood move during one beat? Express your answer to two significant figures and include the appropriate units. ANSWER: l
= 8.0 cm
Correct Score Summary: Your score on this assignment is 114%. You received 85.07 out of a possible total of 85 points, plus 12.21 points of extra credit.
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057468
33/33
View more...
Comments
