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course Phy 121

C?}????????p?assignment #003003. Velocity Relationships

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Physics I

06-06-2007

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17:56:32

`q001. Note that there are 11 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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RESPONSE -->

vAve will be in meters/second.

confidence assessment: 3

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17:57:08

vAve = `ds / `dt. The units of `ds are cm and the units of `dt are sec, so the units of `ds / `dt must be cm / sec. Thus vAve is in cm/s.

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RESPONSE -->

I thought it said the units of `ds were in meters. I must have read the problem wrong.

self critique assessment: 2

you read it correctly

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17:58:05

`q002. If the definition 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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RESPONSE -->

`ds must be in cm.

confidence assessment: 3

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17:58:16

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

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RESPONSE -->

ok.

self critique assessment: 3

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17:58:56

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

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RESPONSE -->

When you multiply cm/sec by sec, you are left with cm because the two sec's cancel eachother out.

confidence assessment:

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17:59:37

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.

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RESPONSE -->

I was thinking of it as more of a cross multiplying situation, but this makes sense also.

self critique assessment: 2

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18:00:06

`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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RESPONSE -->

`dt must be measured in seconds.

confidence assessment: 3

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18:00:15

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

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RESPONSE -->

ok.

self critique assessment: 3

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18:01:29

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

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RESPONSE -->

When you divide km/sec into km, you have to flip the km/sec fraction over and multiply. When you do that you are cross multiply km*km and they cancel leaving seconds.

confidence assessment: 3

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18:01:51

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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RESPONSE -->

ok.

self critique assessment: 3

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18:03:38

`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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RESPONSE -->

The average velocity is 2m/sec. The change in `ds is 10m-4m = 6m. The change in `dt is 5sec-2sec = 3sec. Average velocity is going to be 6m/3sec = 2 m/sec.

confidence assessment: 3

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18:03:49

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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RESPONSE -->

ok.

self critique assessment: 3

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18:04:29

`q007. Symbolize this process: If an object moves from position s = s1 to position s = s2 between clock times t = t1 and t = t2, when what expression represents the change `ds in position and what expression represents the change `dt in the clock time?

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RESPONSE -->

`ds = s2-s1

`dt = t2-t1.

confidence assessment: 3

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18:05:10

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

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RESPONSE -->

The avg velocity would be (s2-s1) / (t2-t1).

self critique assessment: 2

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18:07:11

`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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RESPONSE -->

The rise is `ds which is 10-4 = 6. The run is `dt which is 5-2 = 3.

confidence assessment: 3

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18:07:43

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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RESPONSE -->

ok.

self critique assessment: 3

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18:08:26

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

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RESPONSE -->

The slope is 2 and represents the average velocity.

confidence assessment: 3

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18:08:32

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

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RESPONSE -->

ok.

self critique assessment: 3

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18:09:30

`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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RESPONSE -->

A greater slope would imply a greater velocity because that means the graph rose more steeply which means that the object got to the next point more quickly.

confidence assessment: 3

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18:09:48

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

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RESPONSE -->

ok.

self critique assessment: 3

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18:11:50

`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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RESPONSE -->

The graph would look like a u shape because it starts of slower and then goes faster and faster which will make it steeper. It would be increasing at an increasing rate. The slope of the graph would also be increasing. The slope is going to tell that the velocity is increaseing because it is positive. If the slope were decreasing then the velocity would be negative.

confidence assessment: 3

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18:12:04

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.

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RESPONSE -->

ok.

self critique assessment: 3

&#I believe you submitted this as part of a previous submission. Let me know if I'm wrong about that; if I'm right, then be sure to avoid this sort of redundancy. &#

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?????z????n???

assignment #004

004. Acceleration

Physics I

06-06-2007

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22:58:03

`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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RESPONSE -->

5 m/sec/sec

confidence assessment: 2

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23:00:51

06-06-2007 23:00:51

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.

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NOTES -------> I got this problem correct. However, I get confused with the wording rate of change of speed and velocity and acceleration and what to apply to each of these. Is there any way to sort them out?

&#You might find the following helpful at this point in the course.

Remember that average velocity is an average rate of change of position with respect to clock time and

average acceleration is an average rate of change of velocity with respect to clock time.

The example below defines the term 'average rate', connects the definition to the slope of a graph, and applies that definition to a situation involving temperature and clock time. &#

&#

The definition of an average rate of change of a quantity A with respect to another quantity B is

ave rate = (change in A) / (change in B).

From this it follows that

change in A = ave rate * change in B

and also that

change in B = change in A / ave rate

where 'ave rate' here is an abbreviation for the specific term 'average rate of change of A with respect to B.

On a graph of A vs. B, the quantity A is represented relative to the vertical axis and the quantity B relative to the horizontal. Between two graph points, therefore, the change in A is the 'rise' from one point to the other, and the change in B is the 'run' from the same first point to the second. The average rate of change of A with respect to B is then

ave rate = (change in A) / (change in B) = rise / run.

Since rise / run between two points is the slope of the straight line segment between those points, we can identify an average rate of change of A with respect to B as the slope between two points on the graph of A vs. B.

Since slope = rise / run, we see that

rise = slope * run and

run = rise / slope

Interpreting slope as ave rate, rise as change in A and run as change in B, this again tells us that change in A = ave rate * change in B, and change in B = change in A / ave rate. &#

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23:00:56

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.

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RESPONSE -->

self critique assessment: 2

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23:02:47

`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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RESPONSE -->

This is telling the car how fast it is going usually per hour in the US. This is helpful with maintaining speed limits. I think that a more powerful engine will definately be capable of a greater rate of velocity change.

confidence assessment: 2

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23:03:27

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.

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RESPONSE -->

ok.

self critique assessment: 3

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23:04:43

`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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RESPONSE -->

Meters per second per second comes from dividing seconds into meters per second. When you do this you really are multiplying m/sec * sec/1 to get m/sec/sec.

confidence assessment: 3

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23:04:56

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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RESPONSE -->

ok.

self critique assessment: 3

&#

Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the given solution, and if necessary asking specific questions.

&#

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23:09:44

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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RESPONSE -->

We would get meters/sec^2. I get the rate of change of velocity units. Is rate of change of velocity also referred to as acceleration?

acceleration is defined to be rate of change of velocity with respect to clock time

self critique assessment: 2

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23:15:08

`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?

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RESPONSE -->

It would change at a rate of -3 m/sec/sec.

confidence assessment: 3

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23:15:37

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.

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RESPONSE -->

ok.

self critique assessment: 2

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23:18:27

`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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RESPONSE -->

vAve = `ds/`dt

confidence assessment: 3

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23:22:10

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.

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RESPONSE -->

I understand that the rate of change of velocity is average acceleration and I get the formula too.

self critique assessment: 2

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23:24:10

`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?

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RESPONSE -->

The average acceleration is 3/2 m/sec/sec.

confidence assessment: 3

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23:25:03

`q006a. 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?

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RESPONSE -->

change in velocity is 3 m/sec. the change in clock time is 3 seconds.

confidence assessment: 2

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23:25:38

`q006b. What therefore is the average rate at which the velocity is changing during this time interval?

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RESPONSE -->

The average rate is 1.5m/sec/sec.

confidence assessment: 3

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23:26:05

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 their runs from t = 1.5 sec to t = 3.5 sec so that 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. Please comment if you wish on your understanding of the problem at this point.

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RESPONSE -->

I understand the average acceleration problems.

self critique assessment: 3

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23:27:46

`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 is the slope between these points what does it represent?

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RESPONSE -->

The run is the distance between the seconds which is 2. The rise is the change in velocities which is 3. The slope is the acceleration.

confidence assessment: 3

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23:28:03

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.

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RESPONSE -->

ok.

self critique assessment: 3

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23:28:50

`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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RESPONSE -->

A greater slope would imply greater acceleration because that would mean that the rise is increasing at a faster pace than the run and the graph would be increasing at an even more increasing rate.

confidence assessment: 3

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23:29:02

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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RESPONSE -->

ok.

self critique assessment: 3

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23:30:43

`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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RESPONSE -->

The graph would start out increasing at an increasing rate because the velocity and acceleration are increasing. The graph would then start to become more level and then probably completely level as the acceleration stopped, but velocity maintained.

confidence assessment: 3

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23:32:01

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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RESPONSE -->

I said the graph was increasing at an increasing rate, but it is increasing at a constant rate. I'm not really sure why this is because isn't the care getting faster? However, I was right about the graph only leveling off and not decreasing at all.

self critique assessment: 2

It can get faster at an increasing, a decreasing or a constant rate.

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23:33:42

`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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RESPONSE -->

The beginning of the graph is increasing at a constant rate. After that I'm not really sure about how velocity is increasing but acceleration is decreasting. Don't you have to accelerate to increase velocity?

confidence assessment: 2

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23:34:50

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.

In answer to the following question posed at this point by a student: Can you clarify some more the differences in acceleration and velocity?

** 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. **

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RESPONSE -->

This is difficult, but I think I have the hang of this problem.

self critique assessment: 3

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"

You are asking good questions and doing a good job of sorting out these ideas. See my notes and let me know if you have additional questions.