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course Phy201+L

004. `Query 4NOTE PRELIMINARY TO QUERY:

** Questions about velocity, average velocity, acceleration, etc. can be very confusing because so many of the concepts have similar definitions. People have trouble distinguishing things like average velocity, which for uniform acceleration can be obtained in a process that adds two velocities, from average acceleration, which involves subtracting two velocities; one of these processes involves dividing by 2 and the other dividing by the time interval `dt.

It is essential to keep the definitions and the meanings of the terms very clear and to work everything from definitions. It is equally important to have a good common-sense understanding of every definition so you can develop the intuition to make sense of everything you do.

That inevitably takes people a little time. But in the process you develop the habits you will need to succeed in the course. **

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Question: How is acceleration an example of a rate of change?

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Your solution:

Acceleration is a rate of change of velocity with respect to clock time.

confidence rating #$&*: 3

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Given Solution:

** Velocity is the rate of change of position with respect to clock time. Acceleration is rate of change of velocity with respect to clock time.

To find the acceleration from a v vs. t graph you take the rise, which represents the change in the average velocity, and divide by the run, which represents the change in clock time.

note that the term 'average rate of change of velocity with respect to clock time' means the same thing as 'acceleration' **

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Question: If you know average acceleration and time interval what can you find?

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Your solution:

If you know acceleration and you know the time interval, you can find velocity from the equation, velocity= acceleration * time interval.

confidence rating #$&*:3

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Given Solution:

** Accel = change in vel / change in clock time, so if you know accel and time interval (i.e., change in clock time) you can find change in vel = accel * change in clock time.

In this case you don't know anything about how fast the object is traveling. You can only find the change in its velocity.

COMMON ERROR (and response): Average acceleration is the average velocity divided by the time (for the change in the average velocity)so you would be able to find the average velocity by multiplying the average acceleration by the change in time.

INSTRUCTOR RESPONSE: Acceleration is rate of change of velocity--change in velocity divided by the change in clock time. It is not average velocity / change in clock time.

COUNTEREXAMPLE TO COMMON ERROR: Moving at a constant 60 mph for 3 hours, there is no change in velocity so acceleration = rate of change of velocity is zero. However average velocity / change in clock time = 60 mph / (3 hr) = 20 mile / hr^2, which is not zero. This shows that acceleration is not ave vel / change in clock time.

COMMON ERROR and response: You can find displacement

INSTRUCTOR RESPONSE: From average velocity and time interval you can find displacement.

However from average acceleration and time interval you can find only change in velocity. Acceleration is the rate at which velocity changes so average acceleration is change in velocity/change in clock time. From this it follows that change in velocity = acceleration*change in clock time. **

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Question: Can you find velocity from average acceleration and time interval?

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Your solution:

If we have average acceleration and time interval, we cannot find velocity, only change in velocity, as this is what is needed to determine acceleration.

confidence rating #$&*: 3

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Given Solution:

** Ave accel = change in vel / change in clock time. If acceleration is constant, then this relationship becomes acceleration = change in velocity/change in clock time.

Change in clock time is the time interval, so if we know time interval and acceleration we can find change in velocity = acceleration * change in clock time = acceleration * change in clock time.

We cannot find velocity, only change in velocity. We would need additional information (e.g., initial velocity, average velocity or final velocity) to find an actual velocity.

For example if we know that the velocity of a car is changing at 2 (mi/hr) / sec then we know that in 5 seconds the speed will change by 2 (mi/hr)/s * 5 s = 10 mi/hr. But we don't know how fast the car is going in the first place, so we have no information about its actual velocity.

If this car had originally been going 20 mi/hr, it would have ended up at 30 miles/hr. If it had originally been going 70 mi/hr, it would have ended up at 80 miles/hr.

Similarly if an object is accelerating at 30 m/s^2 (i.e., 30 (m/s) / s) for eight seconds, its velocity will change by 30 meters/second^2 * 8 seconds = 240 m/s. Again we don't know what the actual velocity will be because we don't know what velocity the object was originally moving.

ANOTHER SOLUTION:

The answer is 'No'.

You can divide `ds (change in position) by `dt (change in clock time) to get vAve = `ds / `dt.

Or you can divide `dv (change in vel) by `dt to get aAve.

So from aAve and `dt you can get `dv, the change in v.

But you can't get v itself.

EXAMPLE: You can find the change in a quantity from a rate and a time interval, but you can't find the actual value of the quantity. For example, accelerating for 2 sec at 3 mph / sec, your velocity changes by 6 mph, but that's all you know. You don't know how fast you were going in the first place. Could be from 5 mph to 12 mph, or 200 mph to 206 mph (hopefully not down the Interstate).

COMMON ERROR: Yes. Final velocity is average velocity multiplied by 2.

INSTRUCTOR RESPONSE: We aren't given ave velocity and time interval, we're give ave accel and time interval, so this answer is not valid.

Note also that final velocity is average velocity multiplied by 2 ONLY when init vel is zero. Be sure you always state it this way.

ANOTHER EXAMPLE:

You can't find velocity from ave accel and time interval--you can only find change in velocity from this information. For example a velocity change of 10 mph would result from ave accel 2 m/s^2 for 2 seconds; this change could be between 10 and 20 mph or between 180 and 190 mph, and if all we know is ave accel and time interval we couldn't tell the difference. ONE MORE RESPONSE:

You can find the change in velocity. The actual velocity cannot be found from ave accel and time interval. For example you would get the same result for acceleration if a car went from 10 mph to 20 mph in 5 sec as you would if it went from 200 mph to 220 mph in 10 sec. **

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Question: `qCan you find change in velocity from average acceleration and time interval?

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Your solution:

You can find change in velocity from average acceleration and time interval.

confidence rating #$&*: 3

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Given Solution:

Average acceleration is ave rate of change of velocity with respect to clock time, which is `dv / `dt.

Given average acceleration and time interval you therefore know aAve = `dv / `dt, and you know `dt.

The obvious use of these quantities is to multiply them:

• aAve * `dt = `dv / `dt * `dt = `dv

So with the given information aAve and `dt, we can find `dv, which is the change in velocity.

From this information we can find nothing at all about the average velocity vAve, which is a quantity which is completely unrelated to `dv .

`a**Good student response:

Yes, the answer that I provided previously is wrong, I didn't consider the 'change in velocity' I only considered the velocity as being the same as the change in velocity and that was not correct.

Change in velocity is average accel * `dt.

CALCULUS-RELATED ANSWER WITH INSTRUCTOR NOTE(relevant mostly to University Physics students)

Yes, you take the integral with respect to time

INSTRUCTOR NOTE:

That's essentially what you're doing if you multiply average acceleration by time interval.

In calculus terms the reason you can't get actual velocity from acceleration information alone is that when you integrate acceleration you get an arbitrary integration constant. You don't have any information in those questions to evaluate c. **

IMPORTANT INSTRUCTOR NOTE: Always modify the term 'velocity' or the symbol 'v'.

I do not use v or the unmodified term 'velocity' for anything at this stage of the course, and despite the fact that your textbook does, you should at this stage consider avoiding it as well.  At this point in the course the word 'velocity' should always be modified by an adjective.

Motion on any interval involves the following quantities, among others:

• initial velocity v0, the velocity at the beginning of the interval

• final velocity vf, the velocity at the end of the interval

• average velocity vAve, defined as average rate of change of position with respect to clock time, `ds / `dt

• change in velocity `dv, which is the difference between initial and final velocities

• (midpoint velocity vMid), which is the same as vAve provided the v vs. t graph is linear (i.e., provided acceleration is constant); since most motion problems will involve uniform acceleration this quantity will be seen than the others

If you aren't specific about which velocity you mean, you will tend to confuse one or more of these quantities.

The symbol v, and the unmodified term 'velocity', have more complex and ambiguous meanings than the specific terms outlined above:

• The symbol v stands for 'instantaneous velocity', a concept that is challenging to understand well without a calculus background (which isn't expected or required for your the General College Physics or Principles of Physics courses). Your text (along with most others) uses v to stand for the instantaneous velocity at clock time t, but sometimes it uses the symbol v for the average velocity.

• To denote an instantanous velocity I consider it more appropriate to use the functional notation v(t)., which clearly denotes the velocity at a specific instant.

The ambiguous use of the word 'velocity' and the symbol 'v' are the source of almost universal confusion among students in non-calculus-based physics courses. (Students in calculus-based courses are expected to have the background to understand these distinctions, though most such students can also profit from the specific terminology outlined here.

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Question: `qCan you find average velocity from average acceleration and time interval?

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Your solution:

You cannot find average velocity from average acceleration and time interval.

confidence rating #$&*: 3

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Given Solution:

`a** CORRECT STATEMENT BUT NOT AN ANSWER TO THIS QUESTION:

The average acceleration would be multiplied by the time interval to find the change in the velocity

INSTRUCTOR RESPONSE:

Your statement is correct, but as you say you can find change in vel, which is not the same thing as ave vel.

You cannot find ave vel. from just accel and time interval. There is for example nothing in accel and time interval that tells you how fast the object was going initially. The same acceleration and time interval could apply as well to an object starting from rest as to an object starting at 100 m/s; the average velocity would not be the same in both cases. So accel and time interval cannot determine average velocity.

CALCULUS-RELATED ERRONEOUS ANSWER AND INSTRUCTOR CLAIRIFICATION(relevant mostly to University Physics students:

Yes, you take the integral and the limits of integration at the time intervals

CLARIFICATION BY INSTRUCTOR:

A definite integral of acceleration with respect to t gives you only the change in v, not v itself. You need an initial condition to evaluate the integration constant in the indefinite integral.

To find the average velocity you would have to integrate velocity (definite integral over the time interval) and divide by the time interval. **

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Question: `qYou can find only change in velocity from average acceleration and time interval. To find actual velocity you have to know at what velocity you started. Why can't you find average velocity from acceleration and time interval?

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Your solution:

To find average velocity you need the beginning velocity and the end velocity, so you cannot average velocity from acceleration and time interval alone.

confidence rating #$&*: 3

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Given Solution:

** Average velocity is change in position/change in clock time. Average velocity has no direct relationship with acceleration.

CALCULUS-RELATED ANSWER you dont know the inital velocity or the final velocity

INSTRUCTOR COMMENT:

. . . i.e., you can't evaluate the integration constant. **

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Question: Give at least three possible units for velocity, and at least three possible units for clock time.

• Give at least three possible units for the slope between two points of a graph of velocity vs. clock time.

• Explain how you reasoned out the answer to this question.

• Explain the meaning of the slope of this graph.

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Your solution:

M/s, feet/sec, km/hr. Second, minute, hour.

m/s/s, ft/s/s, km/hr/hr.

confidence rating #$&*: 3

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Given Solution:

(Remember that it is essential for most students to write out the more complicated expressions on paper, in standard notation. There are two reasons for this. In the first place, this will make them be easier to read and comprehend. In the second place, this writing things on paper reinforces the process better than viewing it on a screen. Some students can read and understand these expressions from the 'typewriter notation' form given here, and with practice everyone quickly gets better at reading this notation, but at this stage most will need to write at least some of these expressions out.)

Possible units for velocity might include millimeters / hour, kilometers / second, or meters / minute. The standard unit is meters / second.

Possible units for clock time might include microseconds, minutes, years. The standard unit is the second.

The slope between two points of a graph is the rise of the graph divided by its run.

• The rise between two points of a graph of velocity vs. clock time represents the change in the velocity between these points. So the rise might have units of, say, millimeters / hour or meters / minute. The standard unit would be meters / second.

• The run between two points of a graph of velocity vs. clock time represents the change in the clock time between these points. So the run might have units of, say, microseconds or years. The standard unit is the second.

• The units of slope are units of rise divided by units of run. So the units of the slope might be any of the following:

• (millimeters / hour) / microsecond, which by the rules for multiplying and dividing fractions would simplify to (millimeters / hour) * (1 / microsecond) = millimeters / (hour * microsecond).

• (meters / minute) / year, which by the rules for multiplying and dividing fractions would simplify to (meters / minute) * (1 / year) = meters / (minute * year)

• or the standard unit, (meters / second) / second, which by the rules for multiplying and dividing fractions would simplify to to (meters / second) * (1 / second) = meters / (second * second) = meters / second^2.

Note on expected levels of understanding for various courses:

Principles of Physics students should understand the following from the preceding:

• meters / second, millimeters / hour, kilometers / second, or meters / minute are all possible units for velocity, and microseconds, minutes, years are all possible units for clock time, and therefore

• (millimeters / hour) / microsecond, (meters / minute) / year, (meters / second) / second are all possible units of the slope of a velocity vs. clock time graph

and will hopefully understand the simplifications. If the simplifications are not clear, or if the units are not understood, it is very important to follow the usual instructions and give a detailed and focused self-critique demonstrating what is understood and what is not.

Some Principles of Physics students (usually those with more extensive mathematical backgrounds) typically understand the subsequent process for conversion of units, but most students who haven't had mathematics courses beyond Algebra II will have some difficulty with the complexity of the expressions.

General College Physics students, who have had a precalculus or high school analysis background, are expected to fully understand this solution, but nevertheless might still have some difficulty in places, and should give focused self-critiques if this is the case.

University Physics students, with their calculus prerequisite, should have the mathematical background and experience to understand everything in this solution very easily; if not they should be very sure to include detailed and specific self-critiques, and immediately take steps to address this issue.

Note that a unit like millimeters / (hour * microsecond) could be converted to standard units.

• Since 1000 millimeters = 1 meter we can use conversion factor (1000 millimeters) / (1 meter) or (1 meter) / (1000 millimeters)

• Since 1 hour = 3600 seconds we can use conversion factor (1 hour) / (3600 seconds), or (3600 seconds) / (1 hour)

• Since 10^6 microseconds = 1 second we have conversion factors (10^6 microseconds) / (1 second) and (1 second) / (10^6 microseconds).

• If we understand the rules for fractions, we can easily apply these conversion factors to get the following:

millimeters / (hour * microsecond) =

mm / (hr microsec) * 1 m / (1000 mm) * 1 hr / (3600 sec) * 10^6 microsec / (1 sec) =

mm * hr * 10^6 microsec m/ (hr microsec mm * 3600 sec * sec) =

(10^6 / 3600) * (mm hr microsec m) / (hr microsec mm sec sec) =

(10^6 / 3600) * m / sec^2

In the second step

we use 1 m / (1000 mm) to convert the original mm in the numerator to meters (the mm in the denominator of the conversion factor 'matches up' with the mm in the numerator of our original expression so the two will later 'divide out' and leave us with m in our numerator)

we use 1 hr / (3600 sec) to convert the original hr in the denominator to second (the hr in the numerator of the conversion factor 'matches up' with the hr in the denominator of our original expression so the two will later 'divide out' and leave us with sec in our denominator)

* we use 10^6 microsec / (1 sec) to convert the original microsec in the denominator to second (the microsed in the numerator of the conversion factor 'matches up' with the microsed in the denominator of our original expression so the two will later 'divide out' and leave us with sec in our denominator)

In the simplification the third step simply multiplies all the numerators from the second step, and all the denominators, to get the numerator and denominator of the third step.

The fourth step breaks the fraction into the product of two fractions, the first being 10^6 / 3600 to represent all the numbers in the fraction, the second being (mm hr microsec m) / (hr microsec mm sec sec) to represent all the units.

In the last step all we do is divide out units of the numerator which are matched by units of the denominator (a process you have seen referred to as 'cancellation').

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Self-critique (if necessary):

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Question: If the velocity of an object changes at a uniform rate from 5 m/s to 13 m/s between clock times t = 7 s and t = 11 s then on this interval what is its average velocity, and what is the average rate at which its velocity changes with respect to clock time?

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Your solution:

(13 m/s + 5 m/s)/2= 9 m/s (average velocity)

13 m/s - 5 m/s= 8 m/s

8 m/s / 4 s= 2 m/s/s

confidence rating #$&*: 3

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Question:

Explain how to solve the relationship

• aAve = `dv / `dt

for `dt.

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Your solution:

aAve = ‘dv / ‘dt. Multiply both sides by ‘dt to get the equation ‘dt * aAve= ‘dv. Then divide both sides to get the final equation of

‘dt = ‘dv/ aAve.

confidence rating #$&*: 3

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Given Solution:

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Question for University Physics Students: What is the instantaneous rate of change of v with respect to t at t = 2, given that v(t) = 2 t^2 - t + 3? Explain how you obtained this result.

What is the expression for the instantaneous rate of change of v with respect to t at general clock time t, given the same velocity function? Explain how you determined this.

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Your solution:

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Given Solution:

v(2) = 2 * 2^2 - 2 + 3 = 9.

v(2.1) = 2 * (2.1)^2 - 2.1 + 3 = 9.72

So the average rate of change of v with respect to t for the interval from t = 2 to t = 2.1 is

• ave rate = change in v / change in t = (9.72 - 9) / (2.1 - 2) = 7.2

v(2.01) = 2.0702, and v(2.001) = 2.007002. Using these values along with v(2) = 9 we find that

• on interval t = 2 to 2 = 2.01 the average rate is 7.02

• on interval t = 2 to 2 = 2.001 the average rate is 7.002

It is therefore reasonable to conjecture that the instantaneous rate at t = 2 is exactly 7.

In fact the instantaneous rate of change function is the derivative function

• v ' (t) = dv / dt = 4 t - 1.

This function gives the instantaneous acceleration at clock time t:

• a(t) = v ' (t) = 4 t - 1.

Evaluating this function at t = 2 we obtain

a(2) = 4 * 2 - 1 = 8 - 1 = 7,

which confirms the conjecture we make based on the series of intervals above.

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Question:

A ball rolls from rest down a book, off that book and onto another book, where it picks up additional speed before rolling off the end of that book.

Suppose you know all the following information:

• How far the ball rolled along each book.

• The clock time at which the ball is released, the clock time at which it first reaches the end of the first book, and the clock time at which it first reaches the end of the second book.

• How fast the ball is moving at each end of each book.

How would you use your information to calculate the ball's average speed on each book?

How would you use your information to calculate how quickly the ball's speed was changing on each book?

confidence rating #$&*:

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Question: `qQuery Add any surprises or insights you experienced as a result of this assignment.

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Your solution:

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Given Solution:

`a** Student Response: I think I really confused what information stood for what in the Force and Pendulum Experiment. However, I enjoy doing the flow diagrams. They make you think in a different way than you are used to. INSTRUCTOR NOTE: These diagrams are valuable for most people. Not all--it depends on learning style--but most. **

Comments:

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Question:

A ball rolls from rest down a book, off that book and onto another book, where it picks up additional speed before rolling off the end of that book.

Suppose you know all the following information:

• How far the ball rolled along each book.

• The clock time at which the ball is released, the clock time at which it first reaches the end of the first book, and the clock time at which it first reaches the end of the second book.

• How fast the ball is moving at each end of each book.

How would you use your information to calculate the ball's average speed on each book?

How would you use your information to calculate how quickly the ball's speed was changing on each book?

confidence rating #$&*:

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Question: `qQuery Add any surprises or insights you experienced as a result of this assignment.

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Your solution:

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Given Solution:

Comments:

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course Phy201+L

004. Acceleration

Goals:

• Definition of average rate of change of velocity with respect to clock time.

• Apply the definition of average rate of change to define acceleration.

• Interpret the slope of a velocity vs. clock time graph

• (understand the defining characteristic of the v vs. t graph for constant acceleration: not well emphasized at this point)

• Describe the v vs t graph of an object with nonuniform acceleration

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Question: `q001 Note that there are 10 questions in this assignment.

At a certain instant the speedometer of a car reads 5 meters / second (of course cars in this country generally read speeds in miles per hour and km per hour, not meters / sec; but they could easily have their faces re-painted to read in meters/second, and we assume that this speedometer has been similarly altered). Exactly 4 seconds later the speedometer reads 25 meters/second (that, incidentally, indicates very good acceleration, as you will understand later). At what average rate is the speed of the car changing with respect to clock time?

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Your solution:

25 m/sec - 5 m/sec=20 m/sec

20m/sec / 4 sec= 5 m/sec/sec.

confidence rating #$&*:32; 3

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Given Solution:

The rate of change of the speed with respect clock time is equal to the change in the speed divided by the change in the clock time. So we must ask, what is the change in the speed, what is the change in the clock time and what therefore is the rate at which the speed is changing with respect to clock time?

The change in speed from 5 meters/second to 25 meters/second is 20 meters/second. This occurs in a time interval lasting 4 seconds. The average rate of change of the speed is therefore (20 meters/second)/(4 seconds) = 5 meters / second / second. This means that on the average, per second, the speed changes by 5 meters/second.

STUDENT QUESTION

Would we not have s^2 since we are multiplying s by s? INSTRUCTOR RESPONSE

That is correct. However in this question I've chosen not to confuse the issue by simplifying the complex fraction m/s/s, which we address separately.

To clarify, m / s / s means, by the order of operations, (m/s) / s, which is (m/s) * (1/s) = m/s^2.

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Question: `q002. Explain in commonsense terms of the significance for an automobile of the rate at which its velocity changes. Do you think that a car with a more powerful engine would be capable of a greater rate of velocity change?

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Your solution:

The rate at which an automobile’s velocity changes is significant because it is the automobile’s acceleration. The faster an automobile’s acceleration the more demand it will have on the market. A car with a more powerful engine would most defiantly have a greater rate of velocity change, being that two car’s weights stay relatively equal.

confidence rating #$&*:32; 3

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Given Solution:

A car whose velocity changes more rapidly will attain a given speed in a shorter time, and will be able to 'pull away from' a car which is capable of only a lesser rate of change in velocity. A more powerful engine, all other factors (e.g., weight and gearing) being equal, would be capable of a greater change in velocity in a given time interval.

STUDENT COMMENT:

The significance for an automobile of the rate at which its velocity changes is the amount of speed it takes to travel to a place in a certain amount of time. If one car is traveling along side another, and they are going to the same location the velocity will be how long it take this car to get to this location going at a speed other than the other car. If a car with a more powerful engine were to travel the same distance its velocity would be capable of a greater rate if increased speed occurred. INSTRUCTOR RESPONSE:

It's necessary here to distinguish between velocity, which is a pretty intuitive concept, and rate of change of velocity, which is much less intuitive and less familiar. An object can change velocity at a constant rate, from rest to a very high velocity. All the while the rate of change of velocity with respect to clock time can be unchanging. So the rate of change of velocity with respect to clock time has nothing to do with how fast the object is moving, but rather with how quickly the velocity is changing. Moving from one location to another, the displacement is the change in position. If the displacement is divided by the time required we get the average rate of change of position with respect to clock time, or average velocity.

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Question: `q003. Explain how we obtain the units meters / second / second in our calculation of the rate of change of the car's speed.

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Your solution:

We obtain the units meters/sec/sec by finding the difference between the two velocities over a given time period. Once we have the difference in the two velocities, we then divide this number by the given time period, which gives us the units of meters/sec/sec.

confidence rating #$&*:32; 3

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Given Solution:

When we divide the change in velocity, expressed in meters/second, by the duration of the time interval in seconds, we get units of (meters / second) / second, often written meters / second / second.

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Question: `q004. The unit (meters / second) / second is actually a complex fraction, having a numerator which is itself a fraction. Such a fraction can be simplified by multiplying the numerator by the reciprocal of the denominator. We thus get (meters / second) * (1/ second). What do we get when we multiply these two fractions?

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Your solution:

When these two fractions are multiplied, we get the units of meters/sec2.

confidence rating #$&*:32; 3

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Given Solution:

Multiplying the numerators we get meters * 1; multiplying the denominators we get second * second, which is second^2. Our result is therefore meters * 1 / second^2, or just meters / second^2. If appropriate you may at this point comment on your understanding of the units of the rate of change of velocity.

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Question: `q004. If the velocity of an object changes from 10 m/s to -5 m/s during a time interval of 5 seconds, then at what average rate is the velocity changing with respect to clock time?

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Your solution:

10 m/s - (-5 m/sec) = -15 m/sec

-15 m/s / 5 sec =-3 m/sec/sec

confidence rating #$&*:32; 3

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Given Solution:

We see that the velocity changes from 10 meters/second to -5 meters/second, a change of -15 meters / second, during a five-second time interval. A change of -15 m/s during a 5 s time interval implies an average rate of -15 m/s / (5 s) = -3 (m/s)/ s = -3 m/s^2. This is the same as (-3 m/s) / s, as we saw above. So the velocity is changing by -3 m/s every second.

STUDENT QUESTION

Do you have to do the step -3 m/s /s. Because I get the same answer not doing that.

INSTRUCTOR RESPONSE

Your solution read ' -5 m/s - 10 m/s = -15 m/s / 5 seconds = -3 m/s '.

Everything was right except the units on your answer. So the answer to you question is 'Yes. It is very important to do that step.' The final answer in the given solution is '-3 m/s every second', which is not at all the same as saying just '-3 m/s'. -15 m/s / (5 s) = -3 m/s^2, which means -3 m/s per s or -3 cm/s every second or -3 m/s/s.

-3m/s is a velocity. The question didn't ask for a velocity, but for an average rate of change of velocity.

-3 m/s per second, or -3 m/s every second, or -3 m/s/s, or -3 m/s^2 (all the same) is a rate of change of velocity with respect to clock time.

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Question: `q005. You should have noted that velocity can be positive or negative, as can the change in velocity or the rate at which velocity changes. The average rate at which a quantity changes with respect to time over a given time interval is equal to the change in the quantity divided by the duration of the time interval. In this case we are calculating the average rate at which the velocity changes. If v represents velocity then we we use `dv to represent the change in velocity and `dt to represent the duration of the time interval. What expression do we therefore use to express the average rate at which the velocity changes?

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Your solution:

‘dv/’dt represents the average rate at which velocity changes.

confidence rating #$&*:32; 3

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Given Solution:

The average rate would be expressed by [ave rate of velocity change with respect to clock time] = `dv / `dt. The expression [ave rate of velocity change with respect to clock time] is pretty cumbersome so we give it a name. The name we give it is 'average acceleration', abbreviated by the letter aAve. Using a to represent acceleration, write down the definition of average acceleration. The definition of average acceleration is aAve = `dv / `dt. Please make any comments you feel appropriate about your understanding of the process so far.

STUDENT COMMENT:

It’s average velocity so it would be aAve.

INSTRUCTOR RESPONSE:

Good, but note: It’s average acceleration (not average velocity) so it would be aAve.  In most of your course acceleration is constant, so initial accel = final accel = aAve.

• In this case we can just use 'a' for the acceleration.

STUDENT QUESTION

If I understand this correctly, the average rate in which velocity changes is acceleration???? Where did average acceleration fit into the problem, the problem asked for the average rate that velocity changed? INSTRUCTOR RESPONSE

Acceleration is rate of change of velocity with respect to clock time. So the terms 'average acceleration' and 'average rate of change of velocity with respect to clock time' are identical. The term 'average rate of change of velocity' actually leaves off the 'with respect to clock time', but in the context of uniformly accelerated motion 'average rate of change of velocity' is understood to mean 'average rate of change of velocity with respect to clock time' .

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Question: `q006. If a runner is moving at 6 meters / sec at clock time t = 1.5 sec after starting a race, and at 9 meters / sec at clock time t = 3.5 sec after starting, then what is the average acceleration of the runner between these two clock times?

If you can, answer the question as posed. If not, first consider the two questions below:

• What is the change `dv in the velocity of the runner during the time interval, and what is the change `dt in clock time during this interval?

• What therefore is the average rate at which the velocity is changing with respect to clock time during this time interval?

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Your solution:

9 m/s - 6 m/sec = 3 m/s

3.5 sec- 1.5 sec = 2 sec

3 m/s / 2 sec= 1.5 m/s/s

confidence rating #$&*:32; 3

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Given Solution:

We see that the runner's velocity changes from 6 meters/second to 9 meters/second, a change of `dv = 9 m/s - 6 m/s = 3 m/s, during a time interval lasting from t = 1.5 sec to t = 3.5 sec.

The duration of the interval is `dt = 3.5 s - 1.5 s = 2.0 s.

The rate at which the velocity changes is therefore 3 m/s / (2.0 s) = 1.5 m/s^2.

STUDENT QUESTION

I'm not understanding why you have the power of 2 for. INSTRUCTOR RESPONSE

When you divide m/s by s you do the algebra of the fractions and get m/s^2. You don't get m/s.

The distinction is essential:

• m/s^2 is a unit of acceleration.

• m/s is a unit of velocity.

Velocity and acceleration are two completely different aspects of motion.

The algebra of dividing m/s by s was given in a previous question in this document. In a nutshell, (m/s) / s = (m/s) * (1/s) = m/s^2.

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Question: `q007. On a graph of velocity vs. clock time, we can represent the two events of this problem by two points on the graph. The first point will be (1.5 sec, 6 meters/second) and the second point will be (3.5 sec, 9 meters / sec). What is the run between these points and what does it represent?

What is the rise between these points what does it represent?

What does the slope between these points what does it represent?

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Your solution:

The rise is a value of 2 seconds, which gives us our clock value over a given period of time. The rise value is a value of 3 m/sec, which represents our velocity. The slope between these two points represents the acceleration of the object over the given time.

confidence rating #$&*:32; 3

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Given Solution:

The rise from the first point to the second is from 6 meters/second to 9 meters/second, or 3 m/s. This represents the change `dv in velocity. The run is from 1.5 seconds to 3.5 seconds, or 2 seconds, and represents the change `dt in clock time. The slope, being the rise divided by the run, is 3 m/s / (2 sec) = 1.5 m/s^2. This slope represents `dv / `dt, which is the average acceleration during the time interval. You may if you wish comment on your understanding to this point.

STUDENT QUESTION

Are we going to use the terms acceleration and average acceleration interchangeably in this course? I just want to make sure I understand. INSTRUCTOR RESPONSE

Good question.

The term 'acceleration' refers to instantaneous acceleration, the acceleration at a given instant.

The term 'average acceleration' refers to the average acceleration during an interval, calculated by subtracting initial from final velocity and dividing by the change in clock time.

If acceleration is uniform, it's always the same. If acceleration is uniform, then, it is unchanging. In that case the instantaneous acceleration at any instant is equal to the average acceleration over any interval.

So when acceleration is uniform, 'acceleration' and 'average acceleration' are the same and can be used interchangeably.

Acceleration isn't always uniform, so before using the terms interchangeably you should be sure you are in a situation where acceleration is expected to be uniform.

This can be visualized in terms of graphs:

The instantaneous acceleration can be represented by the slope of the line tangent to the graph of v vs. t, at the point corresponding to the specified instant.

The average acceleration can be represented by the average slope between two points on a graph of v vs. t.

If acceleration is uniform then the slope of the v vs. t graph is constant--i.e., the v vs. t graph is a straight line, and between any two points of the straight line the slope is the same. In this case the tangent line at a point on the graph is just the straight-line graph itself.

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Question: `q008. In what sense does the slope of any graph of velocity vs. clock time represent the acceleration of the object? For example, why does a greater slope imply greater acceleration?

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Your solution:

Sense the slope represents the formula rise/run, the graph of velocity vs. clock time’s slope represents the formula velocity/time, or acceleration. A greater slope implies a greater acceleration because this would show that the velocity is higher over a given time period, meaning it has a greater acceleration.

confidence rating #$&*:32; 3

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Given Solution:

Since the rise between two points on a graph of velocity vs. clock time represents the change in `dv velocity, and since the run represents the change `dt clock time, the slope represents rise / run, or change in velocity /change in clock time, or `dv / `dt. This is the definition of average acceleration.

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Question: `q009. This is the same situation as in the preceding problem: An automobile coasts down a hill with a constant slope. At first its velocity increases at a very nearly constant rate. After it attains a certain velocity, air resistance becomes significant and the rate at which velocity changes decreases, though the velocity continues to increase. Describe a graph of velocity vs. clock time for this automobile (e.g., neither increasing nor decreasing; increasing at an increasing rate, a constant rate, a decreasing rate; decreasing at an increasing, constant or decreasing rate; the description could be different for different parts of the graph).

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Your solution:

A graph of this situation would be an exponential graph, as the graph will increase dramatically, then level off as the velocity reaches a constant rate that cannot increase.

confidence rating #$&*:32; 3

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Given Solution:

Your graph should have velocity as the vertical axis and clock time as the horizontal axis. The graph should be increasing since the velocity starts at zero and increases. At first the graph should be increasing at a constant rate, because the velocity is increasing at a constant rate. The graph should continue increasing by after a time it should begin increasing at a decreasing rate, since the rate at which the velocity changes begins decreasing due to air resistance. However the graph should never decrease, although as air resistance gets greater and greater the graph might come closer and closer to leveling off. Critique your solution by describing or insights you had or insights you had and by explaining how you now know how to avoid those errors.

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Question: `q010. An automobile coasts down a hill with a constant slope. At first its velocity increases at a very nearly constant rate. After it attains a certain velocity, air resistance becomes significant and the rate at which velocity changes decreases, though the velocity continues to increase. Describe a graph of acceleration vs. clock time for this automobile (e.g., neither increasing nor decreasing; increasing at an increasing rate, a constant rate, a decreasing rate; decreasing at an increasing, constant or decreasing rate; the description could be different for different parts of the graph).

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Your solution:

A graph of acceleration vs. clock time would show a concave down structure, where the graph increases to a point then begins to decrease.

confidence rating #$&*:32; 3

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Given Solution:

Your graph should have acceleration as the vertical axis and clock time as the horizontal axis. At first the graph should be neither increasing nor decreasing, since it first the acceleration is constant. Then after a time the graph should begin decreasing, which indicates the decreasing rate at which velocity changes as air resistance begins having an effect. An accurate description of whether the graph decreases at a constant, increasing or decreasing rate is not required at this point, because the reasoning is somewhat complex and requires knowledge you are not expected to possess at this point. However it is noted that the graph will at first decrease more and more rapidly, and then less and less rapidly as it approaches the t axis.

STUDENT QUESTION: Can you clarify some more the differences in acceleration and velocity?

INSTRUCTOR RESPONSE: ** Velocity is the rate at which position changes and the standard units are cm/sec or m/sec.

Acceleration is the rate at which velocity changes and its standard units are cm/s^2 or m/s^2.

Velocity is the slope of a position vs. clock time graph.

Acceleration is the slope of a velocity vs. clock time graph. **

STUDENT QUESTION:

In the problem it states that velocity continues to increase even though the rate at which velocity changes decreases. I don’t understand your the slope will decrease if this is true. I can understand a diminish in velocity and time, but not a down turn of the slope, which is what your solution leans to. INSTRUCTOR RESPONSE

Your thinking is good, but you need carefully identify what it is you're describing. The question here concerns the acceleration vs. clock time graph, whereas most of your comments apply to the velocity vs. clock time graph. Under these conditions the slope of the velocity vs. clock time graph will decrease, this will occur as long as the acceleration vs. clock time graph decreases, regardless of whether that decrease is at a constant, an increasing or a decreasing rate.

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Self-critique (if necessary):

My acceleration actually does not increase it only decreases, so the graph would be more like a reverse exponential graph.

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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. 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.

Questions related to q_a_

1. If we divide the unit sec by the unit cm/sec, what unit do we get? Does this unit correspond to any quantity we have defined? Answer the same questions for the following calculations:

• dividing the unit cm / sec by the unit cm

• dividing the unit cm by the unit cm/sec

• dividing the unit cm/sec by the unit sec

(Note: It is not uncommon that a number of students have difficulty with the algebra of units. This is largely a result of a lack of practice with fractions, which is perhaps the result of a tendency to permit overuse of calculators in secondary school.

Fortunately the fractions required to do units are not difficult. Multiplication and division of fractions is an elementary-school topic, and a little practice is all that's required. The most difficult operations with fractions are addition and subtraction of fractional quantities, and this skill is not required to do unit calculations. The simplest operations with fractions are multiplication and division, which are easily refreshed with a little practice.

Click for a synopsis of fractions as related to unit calculations.

You may also access numerous Web resources, easily accessed by searching under the headings 'Multiplication of Fractions', 'Division of Fractions', 'Powers of Fractions', etc.)

2. Give at least 3 possible units for position, and 3 possible units for clock time.

In terms of these units:

(note: it would take way too long to write out:detailed answers to all these questions in your notebook; however you can easily answer these questions in the form of sketches, diagrams and abbreviations that could then be expanded into detailed answers if you were later asked for the details)

• give at least 3 possible units for rate of change of position with respect to clock time, explaining specifically how each given unit follows from the definition of average rate of change

• give at least 3 possible units for rate of change of velocity with respect to clock time, explaining specifically how each given unit follows from the definition of average rate of change

• explain what the rise between two points of a position vs. clock time graph represents, and give at least three possible units for the rise of this graph

• explain what the rise between two points of a velocity vs. clock time graph represents, and give at least three possible units for the rise of this graph

• explain what the run between two points of a position vs. clock time graph represents and give three possible units for the run

• explain what the run between two points of a velocity vs. clock time graph represents and give three possible units for the run

• explain what the slope between two points of a position vs. clock time graph represents and give three possible units for the slope

• explain what the slope between two points of a velocity vs. clock time graph represents and give three possible units for the slope

3. If the slope of a graph is constant (i.e., the slope never changes), then what can be said about the shape of the graph?

• If the slope of a position vs. clock time graph for the motion of some object is constant, then what quantity is constant for the motion?

• If the slope of a velocity vs. clock time graph for the motion of some object is constant, then what quantity is constant for the motion?

4. What can be said about the motion of an object if the slope of its position vs. clock time graph is increasing?

What can be said about the motion of an object if the slope of its velocity vs. clock time graph is decreasing?

Questions related to text

(no questions on this assignment)

Questions related to Introductory Problem Sets

1. Give the rate-of-change definition of acceleration, and explain how this definition leads to the equivalent definition

• aAve = `dv / `dt.

2. Explain how to solve the relationship

• aAve = `dv / `dt

for `dv.

3. Explain how to solve the relationship

• aAve = `dv / `dt

for `dt.

4. The figure below is from Introductory Problem Set 2, Problem # 2.

Explain how this figure depicts the definition of aAve in terms of `dv and `dt.

Questions related to Class Notes

1. If an object accelerates uniformly from rest, moving 80 cm in 10 seconds, then what are its average velocity, change in velocity and acceleration?

Average velocity= 8 cm/s

Change in velocity= 8 cm/s

Acceleration=

2. If the velocity of an object changes at a uniform rate from 5 m/s to 13 m/s between clock times t = 7 s and t = 11 s then on this interval what is its average velocity, and what is the average rate at which its velocity changes with respect to clock time?

5 m/s + 13 m/s = 9 m/s (average velocity)

13 m/s - 5 m/s = 8 m/s

8 m/s / 4 s= 2 m/s/s ( acceleration)

3. The water level in a container has values 50 cm, 30 cm, 15 cm and 7 cm at respective clock times 3 second, 7 seconds, 10 seconds and 12 seconds. What is the average velocity during each of the three corresponding intervals? What is the midpoint clock time for each interval?

Sketch a graph of average velocity vs. midpoint clock time and justify your answer to the question:

• Is this information consistent with the hypothesis that the acceleration of the water surface is uniform?

50cm - 30cm = 20 cm

7sec- 3sec= 4sec

20 cm/4 sec= 5cm/sec (average velocity)

(7 sec + 3 sec)/2= 5 sec (midpoint clock time)

30 cm- 15 cm=15 cm

10sec-7sec=3 sec

15cm/3sec= 5 cm/sec (average velocity)

(10sec+7sec)/2= 8.5 sec (midpoint clock time)

15cm -7cm= 8 cm

12 sec- 10 sec= 2 sec

8cm/ 2 sec = 4 cm/sec (average velocity)

(12 sec +10 sec)/2= 11 sec (midpoint clock time)

Questions/problems for University Physics Students

1. The instantaneous rate of change of A with respect to B is defined to be the limiting value, as the B interval approaches zero, of the average rate of change of A with respect to B. This definition isn't specific enough to be truly rigorous, but it will do for the moment.

What therefore would be the definition of the instantaneous rate of change of position with respect to clock time?

What would be the definition of the instantaneous rate of change of velocity with respect to clock time?

2. If the velocity of an object is given by the function v(t) = 2 t^2 - t + 3, then find the average rate of change of v with respect to t for each of the following intervals:

• the interval from t = 2 to t = 2.1

• the interval from t = 2 to t = 2.01

• the interval from t = 2 to t = 2.001

What do you think is the instantaneous rate of change of velocity with respect to clock time at the instant when t = 2?

3. What is the instantaneous rate of change of v with respect to t at clock time t?