assign 3 

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14:41:05 Given a graph of velocity vs. time how do we determine the rate which velocity increases?

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RESPONSE --> You divide the average velocity by the time.

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14:42:16 ** To find the average rate of change of velocity we use the slope of the graph between two points, which indicates rise / run = change in velocity/change in clock time. This is equivalent to using the slope of the graph, which indicates rise / run = change in velocity/change in clock time. COMMON ERROR: rate = Velocity/time INSTRUCTOR COMMENT: If you divide v by t you do not, except in certain special cases, get the average rate at which velocity increases. You have to divide change in velocity by change in clock time. For example if you're traveling at a constant 60 mph for 1 hour and divide 60 mph by 1 hour you get a nonzero acceleration of 60 mph / hr while in fact you are traveling at a constant rate and have acceleration 0. **

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RESPONSE --> That is what i was trying to describe but didn't come out quite right. So you divide change in velocity by change in clock time.

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14:42:49 If we know the velocities at two given clock times how do we calculate the average rate at which the velocity is changing during that time interval?

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RESPONSE --> You divide the average velocity by the time interval

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14:42:56 ** We divide the difference `dv in velocities by the duration `dt of the time interval. **

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RESPONSE --> OK

Be careful not to confuse change in velocity with average velocity.

ave accel = change in velocity / change in clock time ave vel = change in position / change in clock time.

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14:43:15 Does the rate at which a speedometer needle moves tell us how fast the vehicle is moving?

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RESPONSE --> Yes

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14:43:53 ** The rate at which the speedometer needle moves does not tell us how fast the vehicle is moving. The fact that the speedometer needle moves tells us that the car is either speeding up or slowing down, not at what rate the car is moving. The reading on the speedometer tells us how fast we are moving. A speedometer that simply sits there on 60 mph doesn't move at all, but presumably the vehicle is moving pretty fast. If the speedometer moves, that indicates that velocity is changing. If the speedometer moves quickly, the velocity is changing quickly. If the speedometer moves slowly, velocity is changing slowly. A quickly moving speedometer needle doesn't imply a quickly moving vehicle. The speedometer might go past the 5 mph mark very fast. But at the instant it passes the mark the car isn't moving very quickly. The needle can move quickly from 1 mph to 2 mph, or from 100 mph to 101 mph. The speed of the needle has little to do with the speed of the vehicle. Of course if the needle is moving quickly for an extended period of time, this implies a large change in the velocity of the vehicle. However this information still does not tell you what that velocity actually is. **

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RESPONSE --> Ok i understand what it is saying now

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14:44:01 Does a stationary speedometer needle implies stationary vehicle?

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RESPONSE --> No

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14:44:08 ** A stationary speedometer needle does not imply a stationary vehicle. The speedometer could be constant at the 60 mph mark while the car is moving 60 mph. **

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RESPONSE --> OK

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14:44:20 Does a quickly moving speedometer needle imply a quickly moving vehicle?

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RESPONSE --> no it implies quicker velocity

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14:44:29 ** A quickly moving speedometer needle does not imply a quickly moving vehicle. It merely implies that the velocity of car is changing quickly. **

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RESPONSE --> OK

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14:45:08 What does it feel like inside a car when the speedometer needle is moving fast?

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RESPONSE --> You dont really feel anything, you feel the same whether the speedometer needle is moving fast or slow

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14:45:18 ** When the speedometer is moving fast, as for example when you are first starting out in a hurry, in a powerful vehicle, you feel yourself pushed back in your seat. The speedometer can also move quickly when you press hard on the brakes. In this case you tend to feel pressed forward toward the steering wheel. **

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RESPONSE --> OK

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14:45:34 What does the speed of the speedometer needle tell us?

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RESPONSE --> It tell us how fast we are going

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14:45:43 ** The position on the speedometer tells us how fast we are moving--e.g., 35 mph, 50 mph. However the speed of the needle tells us at what rate we are speeding up or slowing down--specifically the speed of the speedometer needle gives us the rate at which velocity changes. **

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RESPONSE --> OK

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�Y�v�t̜w�����ݒ���Ȣ����� Student Name: assignment #003

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13:59:57 `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 --> m/s

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14:00:19 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 --> OK

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14:03:08 `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 --> cm

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14:03:14 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

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14:03:50 `q003. Explain the algebra of multiplying the unit cm / sec by the unit sec.

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RESPONSE --> You put cm/sec x sec/1

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14:04:05 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 --> OK

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14:04:29 `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 --> seconds

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14:04:34 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

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14:04:51 `q005. Explain the algebra of dividing the unit km / sec into the unit km.

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RESPONSE --> Km/sec x km/1

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14:05:23 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, i didn't put the sec over the km.

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14:08:40 `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 2. The change in `ds position is 6 meters. The change in clock time is 3 seconds. You divide the change in `ds by the change in clock time.

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14:09:06 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 --> My reasoning was the same.

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14:10:41 `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 --> The change in ds is s2-s1 The change in clock time is t2-t1

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14:11:21 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 --> s2-s1 / t2-t1

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14:17:48 `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 --> Rise=3 Run = 6

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14:18:44 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

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14:20:12 The slope of this graph is 6 meters / 3 seconds = 2 meters / second.

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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of prob I dont understand how you are getting this. I thought it was rise/run which would be 3/6 not the other way around. Please explain.

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14:21:26 `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 that it would take longer to have a greater velocity.

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14:21:40 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

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14:23:25 `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 car is rolling down the hill at an increasing at an increasing rate. The slope of the graph is decreasing.

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14:23:37 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

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�򠠥��V����Ӵ���Ř���_���� Student Name: assignment #003 003. Misc: Surface Area, Pythagorean Theorem, Density "

You are giving good answers to many questions, which leads me to believe you're doing OK.

However you aren't giving enough detail in answers or self-critiques for me to be sure, or to spot and help you address potential errors in your approach.