Formal Homework Assignment 2

10/26/14 10:37 PMFormal Homework Assignment 2

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Formal Homework Assignment 2Due: 11:15am on Tuesday, October 7, 2014

To understand how points are awarded, read the Grading Policy for this assignment.

A message from your instructor...

1. For the "Analyzing Position versus Time Graphs: Conceptual Question", for part E, there is a choice you can select; it should not be "none" or "Cannot bedetermined". 2. Problem 2.17 starts out as "A speed skater moving across frictionless ice at 8.20 m/s." Although a positive sign is not used, it is assumedthe object is moving in the +x direction. This is a good example for why drawing a picture and labeling an axis helps clarify a problem.

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. Thistype of graph is usually referred to as an x vs. t graph. To draw such a graph, choose an axis system in which time is plotted on the horizontal axis andposition on the vertical axis. Then, indicate the values of at various times . Mathematically, this corresponds to plotting the variable as a function of . 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 traveled by the particle?

Express your answer in meters.

Hint 1. Total distance

The total distance traveled by the particle is given by the difference between the initial position at and the position at . In symbols,

.

Hint 2. How to read an x vs. t graph

Remember that in an x vs. t graph, time is plotted on the horizontal axis and position on the vertical axis. For example, in the plot shown inthe figure, at .

ANSWER:

Correct

tx x t x

t

!x

!x x0 t = 0.0 s x

t = 50.0 s

!x = x " x0

t xx = 16.0 m t = 10.0 s

= 30 !x m



Part B

What is the average velocity of the particle over the time interval ?

Express your answer in meters per second.

Hint 1. Definition and graphical interpretation of average velocity

The average velocity of a particle that travels a distance along a straight line in a time interval is defined as

.

In an vs. 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 of a line from point A, with coordinates , to point B, with coordinates , is equal to the "rise" over the "run," or

.

ANSWER:

CorrectThe 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. tgraph.

Part C

What is the instantaneous velocity of the particle at ?

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'smotion, you can determine the instantaneous velocity of the particle at any point in the curve. The instantaneous velocity at any point is equal tothe slope of the line tangent to the curve at that point.

ANSWER:

CorrectThe 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 athand the curve is a straight line, the tangent line is the curve itself. Physically, this means that the instantaneous velocity of the particle is constantover 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 oftime. In this graph, time is plotted on the horizontal axis and velocity on the vertical axis. Note that by definition, velocity and acceleration are vectorquantities. In straight-line motion, however, these vectors have only one 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.

vav !t = 50.0 s

vav !x !t

=vav!x!t

x t

m ( , )tA xA ( , )tB xB

m = "xB xA"tB tA

= 0.600 vav m/s

v t = 10.0 s

= 0.600 v m/s

t vv

a



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 representsa variable that always has the same constant value at any time?

ANSWER:

CorrectWhenever 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 ahorizontal line.

Part E

Shown in the figure is the v vs. t curve selected in the previous part. What is the area of the shaded region under the curve?


Graph A

Graph B

Graph C

Graph D

A




The shaded region under the v vs. t curve is a rectangle whose horizontal and vertical sides lie on the axis and the axis, respectively. Sincethe 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 inseconds and another quantity expressed in meters per second. The area itself, then, will be in meters.

ANSWER:

CorrectCompare 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 traveledby the particle. This is true for any velocity curve and any time interval: The area of the region that extends over a time interval under the v vs. tcurve is always equal to the distance traveled in .

Analyzing Position versus Time Graphs: Conceptual Question

Two cars travel on the parallel lanes of a two-lane road. The cars’ motions are represented by theposition versus time graph shown in the figure. Answer the questions using the times from thegraph indicated by letters.

Part A

At which of the times do the two cars pass each other?

Hint 1. Two cars passing

t v

= 30 A m

!t!t



Two objects can pass each other only if they have the same position at the same time.

ANSWER:

Correct

Part B

Are the two cars traveling in the same direction when they pass each other?

ANSWER:

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 inposition over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object beinggraphed.

ANSWER:

Correct

Part D

A

B

C

D

E

None

Cannot be determined

yes

no

A

B

C

D

E

none

cannot be determined



At which of the lettered times, if any, does car #2 momentarily stop?


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 inposition over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object beinggraphed.

ANSWER:

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 inposition over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object beinggraphed.

ANSWER:

Correct

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, selectthe correct position versus time graph. For all of the motions, the stop sign is at the position , and east is the positive x direction.

A

B

C

D

E

none

cannot be determined

A

B

C

D

E

None

Cannot be determined

x = 0



Part A

A driver ignores the stop sign and continues driving east at constant speed.


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 positionover 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 positiveslope 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 ofconstant slope.

ANSWER:

Correct

Part B

A driver ignores the stop sign and continues driving west at constant speed.



Hint 2. Driving west

A

B

C

D

E

F



Since east is defined as the positive x direction, a car traveling west must have a negative velocity. A negative velocity is represented as anegative 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 ofconstant slope.

ANSWER:

Correct

Part C

A driver, traveling west, slows and stops at the stop sign.



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 anegative 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 changingslope. 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 morenegative (depending upon which direction it is moving).

ANSWER:

Correct

Part D

A driver, after stopping at the stop sign, accelerates to the east.

A

B

C

D

E

F

A

B

C

D

E

F





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 positiveslope 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 changingslope. 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 morenegative (depending upon which direction it is moving).

ANSWER:

Correct

± The Graph of a Sports Car's Velocity

The graph in the figure shows the velocity of a sports car as a function of time . Use the graph to answerthe following questions.

Part A

Find the maximum velocity of the car during the ten-second interval depicted in the graph.

Express your answer in meters per second to the nearest integer.


Because the graph displays the car's velocity at each moment in time, the maximum velocity of the car can be found simply by locating themaximum value of the velocity on the graph.

ANSWER:

A

B

C

D

E

F

v t

vmax



Correct

Part B

During which time interval is the acceleration positive?

Indicate the best answer.

Hint 1. Finding acceleration from the graph

Recall that acceleration is the rate of change of velocity with respect to time. Therefore, on this graph of velocity vs. time, acceleration is theslope of the graph. Recall that the slope is defined by for a graph of vs. , or in this case. If the graph isincreasing from left to right, then the slope is positive.

ANSWER:

Correct

Part C

Find the maximum acceleration of the car.

Express your answer in meters per second per second to the nearest integer.


The car's acceleration is the rate of change of the car's velocity with respect to time . In this problem, the car's velocity is given graphically, sothe car's acceleration at a given moment is found from the slope of the vs. curve at that moment. If the vs. curve over some time intervalis represented by a straight line, the instantaneous acceleration anywhere in that interval is equal to the slope of the line, that is, to the averageacceleration over that time interval. To find the maximum acceleration, find the value of the curve's greatest positive slope.

Hint 2. Find the final velocity on the interval with greatest acceleration

The slope of the curve is greatest during the first second of motion. The slope of the graph on this interval is given by the change in velocitydivided by the change in time over the interval from to . At time , the car's velocity is zero. Find the velocity of thecar at time .


ANSWER:

ANSWER:

= 55 vmax m/s

m m = !y/!x y x m = !v/!t

to

to

to

to

to

t = 0 s t = 6 st = 0 s t = 4 st = 0 s t = 10 st = 4 s t = 10 st = 2 s t = 6 s

amax

v tv t v t

t = 0 t = 1 t = 0 s v(0) v(1)t = 1 s

= 30 v(1) m/s



Correct

Part D

Find the minimum magnitude of the acceleration of the car.

Express your answer in meters per second per second to the nearest integer.


To find the minimum magnitude of the acceleration of the car, you must find the point where the absolute value of the slope is smallest.

ANSWER:

Correct

Part E

Find the distance traveled by the car between and .

Express your answer in meters to the nearest integer.


In this problem, the car's velocity as a function of time is given graphically, so the distance traveled is represented by the area under the vs. graph between and .

Hint 2. Find the distance traveled in the first second

What is the distance traveled between and ?


Hint 1. The area of a triangle

Observe that the region in question is a triangle , whose area is therefore one-half the product of the base and the height.

= 30 amax m/s2

amin

= 0 amin m/s2

d0,2 t = 0 s t = 2 s

v tt = 0 s t = 2 s

d0,1 t = 0 s t = 1 s



ANSWER:

Hint 3. Find the distance traveled in the second second

What is the distance traveled between and ?


Hint 1. The shape of the region

The region under the graph between 1 and 2 seconds can be seen as consisting of a rectangle and a triangle.

ANSWER:

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 are shown in the figure.

d_0,1 = 15 m

d1,2 t = 1 s t = 2 s

= 40 d1,2 m

= 55 d m



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:

CorrectThe 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 andthus 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.

t

slope = !x!t

Trial A

Trial B

Trial C

Trial D

average velocity = =!(position)!(time)

!x!t



ANSWER:

CorrectWhile 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 negativeaverage 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 velocitybut its magnitude is less than that in Trial B and Trial D.

Reading Quiz 2.1

Part A

The slope at a point on a position-versus-time graph of an object is

ANSWER:

Correct

Problem 2.11

The figure shows the velocity graph of a particle moving along the x-axis. Its initial position is at . At = 1 , what are the particle's (a) position, (b) velocity, and (c) acceleration?

Part A

Express your answer to two significant figures and include the appropriate units.

ANSWER:

Trial A

Trial B

Trial C

Trial D

the object’s speed at that point.

the object’s average velocity at that point.

the object’s instantaneous velocity at that point.

the object’s acceleration at that point.

the distance traveled by the object to that point.

x0= 2 m t0 = 0 t s



Correct

Part B


ANSWER:

Correct

Part C


ANSWER:

Correct

Problem 2.12

The velocity-versus-time graph is shown for a particle moving along the x-axis. Its initial position is = 2.1 at .

Part A

What is the particle's position at


ANSWER:

Correct

= 3.0 x m

= 2.0 vxms

= 2.0 axms2

x0 m t0 = 0 s

t = 1.0 s ?

= 6.1 x m



Part B

What is the particle's velocity at


ANSWER:

Correct

Part C

What is the particle's acceleration at


ANSWER:

Correct

Part D

What is the particle's position at

Express your answer to three significant figures and include the appropriate units.

ANSWER:

Correct

Part E

What is the particle's velocity at


ANSWER:

Correct

Part F

What is the particle's acceleration at


ANSWER:

t = 1.0 s ?

= 4.0 v ms

t = 1.0 s ?

= 0 a ms2

t = 3.0 s ?

= 13.1 x m

t = 3.0 s ?

= 2.0 v ms

t = 3.0 s ?



Correct

Problem 2.24

A particle moving along the x-axis has its velocity described by the function , where is in . Its initial position is = 2.1 at = 0 .

Part A

At 2.1 , what is the particle's position?

Express your answer with the appropriate units.

ANSWER:

Correct

Part B

At 2.1 , what is the particle's velocity?


ANSWER:

Correct

Part C

At 2.1 , what is the particle's acceleration?


ANSWER:

Correct

Problem 2.28

The figure shows the acceleration graph for a particle that starts from rest at = 0 .

= -2.0 a ms2

vx = 2 m/st2 t s x0 m t0 s

s

= 8.3 x m

s

= 8.8 v ms

s

= 8.4 a ms2

t s



Part A

Determine the object's velocity at times 0 , 2 , 4 , 6 , and 8 .

Express your answer using two significant figures. Enter your answers numerically separated by commas.

ANSWER:

Correct

PSS 2.1 Kinematics with Constant Acceleration - Copy

Learning Goal:

To practice Problem-Solving Strategy 2.1 for constant acceleration problems.

A car is traveling at a constant velocity of magnitude when the driver notices a garbage can on the road in front of him. At that moment, the distancebetween the garbage can and the front of the car is . A time after noticing the garbage can, the driver applies the brakes and slows down at a constantrate before coming to a halt just before the garbage can. What is the magnitude of the car's acceleration after the brakes are applied?

PROBLEM-SOLVING 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 intosymbols.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

,

, and

.

Use or , as appropriate to the problem, rather than the generic .Replace and with numerical subscripts defined in the pictorial representation.Uniform motion with constant velocity has .

t = s s s s s

= 0,5.0,20,30,30 v(0),v(2),v(4),v(6), v(8) m/s

v0d t

= + !tvfs vis as

= + !t + (!tsf si vis12 as )2

= + 2 !sv2fs v2

is as

x y s

i f= 0as



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 these diagrams is the correct pictorialrepresentation of the problem?

ANSWER:

Correct

Part B

As an alternative to the pictorial representation shown above, you could consider using a graphical representation instead. Below are four velocity-versus-time graphs. Which graph correctly represents the situation described in this problem?

ANSWER:

(a)

(b)

(c)

(d)



CorrectNow that you have assessed the information given in the problem, you need to translate it into symbols. You know that when the driver hits thebrakes at time 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. Alignthe x axis with the direction of motion, and let be the moment when the driver sees the garbage can, be the car's position at that moment,and be the car's velocity at that moment. Let , , and be the time, position, and velocity, respectively, when the driver hits the brakes,and let , , and be the time, position, and velocity, respectively, when the car finally comes to rest. Note that it is most convenient to takethe origin of the coordinate axis to be at . At this point, your pictorial representation should look like this:

Now, decide which quantities are known and which need to be found.

Solve

Assess

Problem 2.17

A speed skater moving across frictionless ice at 8.40 hits a 4.80-m-wide patch of rough ice. She slows steadily, then continues on at 5.70 .

Part A

What is her acceleration on the rough ice?

ANSWER:

Correct

Conceptual Question 2.6

Part A

The figure shows the position-versus-time graph for a moving object. At which lettered point or points: Is the object moving the slowest? Is the objectmoving the fastest? Is the object at rest?

A

B

C

D

t

t0 x0

v0x t1 x1 v1x

t2 x2 v2x

x0

m/s m/s

-3.97 m/s2



Drag the appropriate items to their respective bins.

ANSWER:

Correct

Part B

At which lettered point or points is the object moving to the negative direction?

ANSWER:

A

B

C

D

E



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 itsposition as a function of time . A traveling object might move at different speeds and indifferent directions during an interval of time, but if we ask at what constant velocity the objectwould have to travel to achieve the same displacement over the given time interval, that is whatwe call the object's average velocity. We will use the notation to indicate averagevelocity over the time interval from to . For instance, is the average velocity overthe 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.


Average velocity is defined as the constant velocity at which an object would have to travel to achieve a given displacement (difference betweenfinal and initial positions, which can be negative) over a given time interval, from the initial time to the final time . The average velocity istherefore equal to the displacement divided by the given time interval. In symbolic form, average velocity is given by

.

ANSWER:

Correct

Part B

Find the average velocity over the time interval from 1 to 3 seconds.


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

x(t) t

[ , ]vave t1 t2t1 t2 [1,3]vave

t = 1 t = 3

ti tf

[ , ] =vave ti tfx( )"x( )tf ti

"tf ti

= 0 [0,1]vave m/s



ANSWER:


Average velocity is defined as the constant velocity at which an object would have to travel to achieve a given displacement (difference betweenfinal and initial positions, which can be negative) over a given time interval, from the initial time to the final time . The average velocity istherefore equal to the displacement divided by the given time interval. In symbolic form, average velocity is given by

.

ANSWER:

CorrectA note about instantaneous velocity. The instantaneous velocity at a certain moment in time is represented by the slope of the graph at thatmoment. For straight-line graphs, the (instantaneous) velocity remains constant over the interval, so the instantaneous velocity at any time duringan 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 3seconds 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, thetime interval has changed.

ANSWER:

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?


= 40 "xf xi m

ti tf

[ , ] =vave ti tfx( )"x( )tf ti

"tf ti

= 20 [1,3]vave m/s

20 m/s

[0,3]vave

= 13.3 [0,3]vave m/s

[0,3]vave [0,1]vave [1,3]vave[0,1]+ [1,3]vave vave

2



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:

ANSWER:

CorrectNote that though the average velocity is zero for this time interval, the instantaneous velocity (i.e., the slope of the graph) has several differentvalues (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 thetime, 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 2.4

The figure is the position-versus-time graph of a jogger.

= -40 x(6.0) " x(3.0) m

= 3.0 "tf ti s

= -13.3 [3.0,6.0]vave m/s

= 0 x(6.0) " x(0.0) m

= 0 [0.0,6.0]vave m/s



Part A

What is the jogger’s velocity at = 10 ?


ANSWER:

Correct

Part B



ANSWER:

Correct

Part C



ANSWER:

Correct

Problem 2.6

A particle starts from 13 at = 0 and moves with the velocity graph shown in the figure.

t s

= 1.3 v ms

t s

= 0 v ms

t s

= -5.0 v ms

m t0



Part A

Does this particle have a turning point?

ANSWER:

Correct

Part B

If so, at what time?

Express your answer using two significant figures and include the appropriate units.

ANSWER:

Correct

Part C

What is the object's position at = 2, 3, 4 ?

Express your answers using two significant figures separated by commas.

ANSWER:

Correct

Problem 2.13

A jet plane is cruising at 330 when suddenly the pilot turns the engines up to full throttle. After traveling 4.1 , the jet is moving with a speed of 400

0

Yes

No

= 1.0 t s

t s

= 13,19,29 , ,x2 x3 x4 m

m/s km

m/s



.

Part A

What is the jet's acceleration, assuming it to be a constant acceleration?


ANSWER:

Correct

Problem 2.16

A Porsche challenges a Honda to a 100- race. Because the Porsche's acceleration of is larger than the Honda's , the Honda gets a1.0 head start.

Part A

Who wins?

ANSWER:

Correct

Enhanced EOC: Problem 2.23

A particle moving along the x-axis has its position described by the function 4.00 4.00 4.00 , where is in s. At = 4.00, what are theparticle's (a) position, (b) velocity, and (c) acceleration?

You may want to review ( pages 38 - 42) .

For help with math skills, you may want to review:

Differentiation of Polynomial Functions

Part A



Evaluate the position at time = 4.00 .

ANSWER:

m/s

= 6.2 a ms2

m 3.5 m/s2 3.0 m/s2

s

Honda

Porsche

x = ( "t3 t + ) m t t

t s

244 m

http://session.masteringphysics.com/myct/eBookLaunch?pageID=38&bookCode=knight3

http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3179124#



Correct

Part B



How do you determine the velocity as a function of time, , from the position, ? What calculus operation do you have to perform?

Once you have , how do you determine at a particular time?

ANSWER:

Correct

Part C



How do you determine the acceleration as a function of time, , from the velocity, ? What calculus operation do you have to perform?

Once you have , how do you determine the acceleration at a particular time?

ANSWER:

Correct

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 severalfactors, including the weight of the aircraft and the wind velocity.

Part A

A plane accelerates from rest at a constant rate of 5.00 along a runway that is 1800 long. Assume that the plane reaches the required takeoffvelocity at the end of the runway. What is the time needed to take off?

Express your answer in seconds using three significant figures.


As the plane travels along the runway, it has constant acceleration. To solve the problem, you'll need to use the kinematics equations for suchmotion. In particular, you need to use the equation relating the distance traveled and time.

v(t) x(t)

v(t) v

188 ms

a(t) v(t)

a(t)

96.0 ms2

m/s2 mtTO



Hint 2. Find the equation for the distance traveled by the plane

Which expression best describes the distance traveled by the plane during a certain interval of time ? Let and be, respectively, theinitial position and speed of the plane, and use for the acceleration of the plane. Remember that the plane accelerates from rest.

ANSWER:

ANSWER:

CorrectIf using this answer in subsequent parts, please retain full precision and only round as a final step before submitting your answer.

Part B

What is the speed of the plane as it takes off?

Express your answer numerically in meters per second.


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 ofthe 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 of the plane after a certain interval of time ? Let be the initial velocity of the plane, and use for the acceleration of the plane. Remember that the plane starts from rest.

ANSWER:

ANSWER:

d t s0 v0

a

= d

+ av012 t2

+ t + as0 v012 t2

a12 t2

+ + as0 v012 t2

+ as012 t2

= 26.8 tTO s

vTO

v t v0 a

= v

tv0

+ atv0

t + av012 t2

at

= 134 vTO m/s




Part C

What is the distance traveled by the plane in the first second of its run?

Express your answer numerically in meters.


Apply the same equation that you used to solve Part A.

ANSWER:


Part D

What is the distance traveled by the plane in the last second before taking off?

Express your answer numerically in meters.


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 secondbefore 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:

CorrectSince 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 firstsecond of the run. Not surprisingly, so is the distance traveled.If using this answer in subsequent parts, please retain full precision and 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.


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 thesame method that you used to determine the takeoff velocity.

ANSWER:

dfirst

= 2.50 dfirst m

dlast

= 132 dlast m



CorrectThis 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 candetermine whether the plane will successfully take off before the end of the runway by verifying that the plane has gained 70% of the takeoffvelocity 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 toreach the needed takeoff velocity before the plane reaches the end of the runaway. At that point, applying the brakes and aborting the takeoff isthe safest course of action.

Problem 2.58

A toy train is pushed forward and released at with a speed of 2.0 . It rolls at a steady speed for 2.0 , then one wheel begins to stick. Thetrain comes to a stop 6.0 from the point at which it was released.

Part A

What is the magnitude of the train’s acceleration after its wheel begins to stick? Assume acceleration is constant after wheel begins to stick.


ANSWER:

Correct

Score Summary:Your score on this assignment is 99.5%.You received 41.78 out of a possible total of 42 points.

70 %

= 2.0 mx0 m/s sm

= 1.0 a ms2

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Formal Homework Assignment 2