Velocity Relationships

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course Phy 121

6/3 0005 hr

003. Velocity Relationships*********************************************

Question: `q001. Note that there are 13 questions in this assignment.

vAve = `ds / `dt, which is the definition of average velocity and which fits well with our intuition about this concept.

If displacement `ds is measured in meters and the time interval `dt is measured in seconds, in what units will vAve be obtained?

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Your solution:

meters/seconds

confidence rating #$&*:

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Given Solution:

vAve = `ds / `dt.

The units of `ds are m and the units of `dt are sec, so the units of `ds / `dt must be m / sec.

Thus vAve is in m/s.

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Self-critique (if necessary):ok

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Self-critique rating:3

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Question: `q002. If the equation vAve = `ds / `dt is to be solved for `ds we multiply both sides of the equation by `dt to obtain `ds = vAve * `dt.

If vAve is measured in cm / sec and `dt in sec, then in what units must `ds be measured?

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Your solution:

'ds= cm

confidence rating #$&*:

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Given Solution:

Since vAve is in cm/sec and `dt in sec, `ds = vAve * `dt must be in units of cm / sec * sec = cm.

STUDENT QUESTION

I don’t get how sec and sec would cancel each other out

INSTRUCTOR RESPONSE

cm / s * s means

(cm/s) * s, which is the same as

(cm / s) * (s / 1). Multiplying numerators and denominators we have

(cm * s) / (s * 1) or just

(cm * s) / s, which is the same as

cm * (s / s) = cm * 1 = cm.

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Self-critique (if necessary):ok

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Self-critique rating:3

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Question: `q003. Explain the algebra of multiplying the unit cm / sec by the unit sec.

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Your solution:

By multiplying by sec you are actually multiplying cm/sec by sec/1, therefore cancelling out sec leaving cm/1 or cm

confidence rating #$&*:

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Given Solution:

When we multiply cm/sec by sec we are multiplying the fractions cm / sec and sec / 1. When we multiply fractions we will multiply numerators and

denominators. We obtain cm * sec / ( sec * 1). This can be rearranged as (sec / sec) * (cm / 1), which is the same as 1 * cm / 1. Since

multiplication or division by 1 doesn't change a quantity, this is just equal to cm.

STUDENT RESPONSE:

For some reason this question just isn't making sense to me.

INSTRUCTOR RESPONSE:

In a self-critique you need to address the given solution in detail. A general statement such as yours gives me no information on what you

understand. I need this information as a basis for helping you with what you don't understand.

In order to give me the information I need you should be addressing each statement, and each phrase, to show me what you do and do not understand.

The given solution can be broken into individual statements:

1. When we multiply cm/sec by sec we are multiplying the fractions cm / sec and sec / 1.

2. When we multiply fractions we will multiply numerators and denominators.

3. We obtain cm * sec / ( sec * 1).

4. This can be rearranged as (sec / sec) * (cm / 1), which is the same as 1 * cm / 1. Since multiplication or division by 1 doesn't change a

quantity, this is just equal to cm.

Do you understand Statement 1?

If not, have you written out the expressions cm/sec and sec/1 in standard form? (You might want to review the link given at the end of the

Typewriter Notation exercise from Orientation, which should be posted at your access page).

As best you can communicate it, what do you and do you not understand about this statement?

Do you understand Statement 2? If not, what do you and do you not understand about this statement?

Do you understand Statement 3? If not, have you written out the multiplication of cm/sec and sec/1 on paper? The multiplication is (cm / sec) * (sec /

1). Again, if you aren't sure how to write this out, refer to the link at the end of the Typewriter Notation exercise.

Do you understand Statemet 4? If not do you understand that (sec / sec) * (cm / 1) is equal to sec * cm / (sec * 1), which is in turn equal to (cm *

sec) / (sec * 1)? If not, specifically what do you and do you not understand?

If you don't understand anything, then you should start with a review of basic fractions, a topic which is very much neglected in the typical

curriculum in U.S. schools. Then you should return to these questions and give your best answers.

A good link, current as of Sept. 2010:

http://www.themathpage.com/arith/multiply-fractions-divide-fractions.htm

You should submit a copy of question `q003, your solution, the given solution and this note. Insert your answers and/or additional specific questions

and mark with &&&& before and after each insertion, then submit using the Submit Work Form.

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Self-critique (if necessary):ok

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Self-critique rating:3

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Question: `q004. If the definition vAve = `ds / `dt is to be solved for `dt we multiply both sides of the equation by `dt to obtain vAve * `dt =

`ds, then divide both sides by vAve to get `dt = `ds / vAve. If vAve is measured in km / sec and `ds in km, then in what units must `dt be measured?

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Your solution:

km/km/sec= sec

'dt= sec

confidence rating #$&*:

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Given Solution:

Since `dt = `ds / vAve and `ds is in km and vAve in km/sec, `ds / vAve will be in km / (km / sec) = seconds.

STUDENT SOLUTION LACKING DOCUMENTATION

seconds

INSTRUCTOR RESPONSE

You should show the reasoning; we know in advance that `dt will be in seconds, but be sure you understand how to get there from the given units.

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Self-critique (if necessary):ok

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Self-critique rating:3

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Question: `q005. Explain the algebra of dividing the unit km / sec into the unit km.

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Your solution:

km/1 divided by km/sec, means you flip the fraction in denominator and multiply the top fraction. By doing so you cancel out the km, leaving the

seconds.

confidence rating #$&*:

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Given Solution:

The division is km / (km / sec). Since division by a fraction is multiplication by the reciprocal of the fraction, we have km * (sec / km). This is

equivalent to multiplication of fractions (km / 1) * (sec / km). Multiplying numerators and denominators we get (km * sec) / (1 * km), which can be

rearranged to give us (km / km) * (sec / 1), or 1 * sec / 1, or just sec.

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Self-critique (if necessary):ok

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Self-critique rating:3

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Question: `q006. If an object moves from position s = 4 meters to position s = 10 meters between clock times t = 2 seconds and t = 5 seconds, then

at what average rate is the position of the object changing (i.e., what is the average velocity of the object) during this time interval? What is the

change `ds in position, what is the change `dt in clock time, and how do we combine these quantities to obtain the average velocity?

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Your solution:

Vave= 'ds/'dt

=(10-4)/(5-2)

=2 m/s

confidence rating #$&*:

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Given Solution:

We see that the changes in position and clock time our `ds = 10 meters - 4 meters = 6 meters and `dt = 5 seconds - 2 seconds = 3 seconds. We see also

that the average velocity is vAve = `ds / `dt = 6 meters / (3 seconds) = 2 meters / second. Comment on any discrepancy between this reasoning and

your reasoning.

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Self-critique (if necessary):ok

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Self-critique rating:3

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Question: `q007. Symbolize this process: If an object moves from position s = s1 to position s = s2 between clock times t = t1 and t = t2, then

what expression represents the change `ds in position and what expression represents the change `dt in the clock time?

What expression therefore symbolizes the average velocity between the two clock times.

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Your solution:

vave=(s2-s1)/(t2-t1)

confidence rating #$&*:

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Given Solution:

The change in position is `ds = s2 - s1, obtained as usual by subtracting the first position from the second. Similarly the change in clock time is

`dt = t2 - t1.

The symbolic expression for the average velocity is therefore

vAve = `ds / `dt = (s2 - s1) / (t2 - t1).

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Self-critique (if necessary):ok

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Self-critique rating:3

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Question: `q008. On a graph of position s vs. clock time t we see that the first position s = 4 meters occurs at clock time t = 2 seconds, which

corresponds to the point (2 sec, 4 meters) on the graph, while the second position s = 10 meters occurs at clock time t = 5 seconds and therefore

corresponds to the point (5 sec, 10 meters). If a right triangle is drawn between these points on the graph, with the sides of the triangle parallel

to the s and t axes, the rise of the triangle is the quantity represented by its vertical side and the run is the quantity represented by its

horizontal side. This slope of the triangle is defined as the ratio rise / run. What is the rise of the triangle (i.e., the length of the vertical

side) and what quantity does the rise represent? What is the run of the triangle and what does it represent?

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Your solution:

rise is 10-4=6 meters

run is 5-2= 3 seconds

confidence rating #$&*:

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Given Solution:

The rise of the triangle represents the change in the position coordinate, which from the first point to the second is 10 m - 4 m = 6 m. The run of

the triangle represents the change in the clock time coordinate, which is 5 s - 2 s = 3 s.

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Self-critique (if necessary):ok

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Self-critique rating:3

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Question: `q009. What is the slope of this triangle and what does it represent?

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Your solution:

slope equals rise over run

=6/3

slope=2 m/s

confidence rating #$&*:

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Given Solution:

The slope of this graph is 6 meters / 3 seconds = 2 meters / second.

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Self-critique (if necessary):ok

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Self-critique rating:3

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Question: `q010. In what sense does the slope of any graph of position vs. clock time represent the velocity of the object? For example, why does a

greater slope imply greater velocity?

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Your solution:

the graph s vs t; the slope shows 'ds vs 'dt, which is what is used in velocity. vave='ds/'dt which can be read from graph and is the slope rise over

run. A greater velocity means a higher rise which means higher 'ds.

confidence rating #$&*:

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Given Solution:

Since the rise between two points on a graph of velocity vs. clock time represents the change `ds in position, and since the run represents the change

`dt clock time, the slope represents rise / run, or change in position / change in clock time, which is `ds / `dt. This is equal to the average rate

of change of position with respect to clock time, which is the definition of average velocity.

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Self-critique (if necessary): I think I understand the concept, but when I explain what I am saying I forget to use respect to time.

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Self-critique rating:3

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Question: `q011. As a car rolls from rest down a hill, its velocity increases. Describe a graph of the position of the car vs. clock time.

If you have not already done so, tell whether the graph is increasing at an increasing rate, increasing at a decreasing rate, decreasing at an

increasing rate, decreasing at a decreasing rate, increasing at a constant rate or decreasing at a constant rate.

Is the slope of your graph increasing or decreasing?

How does the behavior of the slope of your graph indicate the condition of the problem, namely that the velocity is increasing?

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Your solution:

The graph is increasing at an increasing rate, because velocity is increasing.

confidence rating #$&*:

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Given Solution:

The graph should have been increasing, since the position of the car increases with time (the car gets further and further from its starting point).

The slope of the graph should have been increasing, since it is the slope of the graph that indicates velocity. An increasing graph within increasing

slope is said to be increasing at an increasing rate (an alternative description would be that the graph is increasing and concave up).

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Self-critique (if necessary):ok

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Self-critique rating:3

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Question: `q012. If at clock time t = t_1 the position of an object is x = x_1, while at clock time t = t_2 its position is x - x_2, then what is

its average velocity during the corresponding interval?

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Your solution:

vave= (x_2-x_1) / (t_2-t_1)

confidence rating #$&*:

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Question: `q013. On a graph of position x vs. clock time t, what is the average slope between the point (t_1, x_1) and (t_2, x_2)? What is the

meaning of this average slope, and why?

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Your solution:

average slope= (x_2 - X_1)/ (t_2-t_1)

average slope means slope, because slope is an average of a distance divided by a time interval.

confidence rating #$&*:

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Self-critique rating:3

You should submit the above questions, along with your answers and self-critiques. You may then begin work on the Questions, Problem and Exercises,

as instructed below.

Questions, Problems and Exercises

You should answer the questions and work the problems as given below, as appropriate to your course. Your work should normally be handwritten, should

include diagrams and sketches where appropriate, and should go into your Physics notebook (not into your lab notebook).

If the course is not specified for a problem, then students in all physics courses should do that problem.

Principles of Physics students need not do the questions or problems that are specified for General College Physics or University Physics.

General College Physics students need not do questions or problems specified for University Physics.

University Physics students should do all questions and problems.

Principles of Physics students may if they wish do some of the questions and problems specified for General College Physics, but this is neither

expected nor required. These problems are accessible to Principles of Physics students, but are generally more challenging that what is expected of

students in this course. (Some University Physics problems will also be accessible to Principles of Physics students, though some will not.)

General College Physics students who wish to do so are welcome to work some or all of the University Physics questions and problems, though this is

neither expected nor required. Many of the University Physics questions and problems are more challenging than those expected of General College

Physics students, and a number of the problems require the use of calculus, which is not expected of General College Physics students.

You are not expected to submit these questions and problems. It would take too much of your time to key in all the answers and solutions. You should

solve these problems and answer these questions in your notebook, in a form you can later reference and, if you later desire, revise. The Query at

the end of the assignment will ask you selected questions, which you will at that time be expected to answer based on the work you have done in your

notebook.

Remember that you are always welcome to ask questions at any time. Any question about a problem should include a copy of the problem and a summary of

what you do and do not understand about it.

Questions related to q_a_

1. If we were to multiply the unit cm / sec by the unit sec / hr what unit would we get?

2. If we were to divide the unit cm by the unit cm / sec what unit would we get?

3. In this qa we began with the definition of average velocity. We have seen that this definition is consistent with our idea of speed as distance

traveled divided by time required. However our definition of velocity can produce a negative result, which was not the case for speed. The negative

result will, for example, occur when the change in position is negative while the change in clock time is positive.

How is it that the change in position can be negative?

4. Having defined average velocity, we used symbols to stand for the initial and final positions and clock times. What expression did this give us

for the average velocity?

5. Having defined average velocity, we then represented a pair of initial and final positions and clock times on a graph of position vs. clock time.

What did we learn about the graph?

6. What are some possible units for position? What are some possible units for clock time? What therefore are some possible units for rate of

change of position with respect to clock time?

Questions related to text

1. What is 3.2 km + 340 meters, to the correct number of significant figures?

2. Find 1.80 m + 142.5 cm + 5.34 * 10^5 micro m, to the correct number of significant figures. You should know that 'micro' indicates multiplication

by 10^-6, so a micro m is 10^-6 meter.

3. Find 3.84 kilograms - 3842 grams, to the appropriate number of significant figures.

Questions related to key systems

1. Pendulums of lengths 20 cm and 25 cm are counted for one minute. The counts are respectively 69 and 61. To how many significant figures do we

know the difference between these counts?

Questions/problems for University Physics Students

1. Find the x and y components of two displacements, one of 4.0 meters at 60 degrees and the other of 3.0 meters at 135 degrees. Both angles are

measured in the counterclockwise direction from the positive x axis.

2. If you drive 2.6 km north, then 4 km east before turning 45 deg to your left and traveling an addition 3.1 km, then what is the magnitude and

direction of your displacement from the original position to the final position? Sketch a figure depicting this motion, describe the figure and

describe how it appears to support your result.

3. What do you think is the uncertainty in the magnitude of a vector whose components are given as 0.7 cm in the x direction and 5.932 cm in the y

direction? What do you think is the uncertainty in its direction?"

Self-critique (if necessary):

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