course Phy 121
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15:37:16 NOTE PRELIMINARY TO QUERY:
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15:37:41 ** Questions about velocity, average velocity, acceleration, etc. are 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.
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15:37:53 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.
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15:37:58 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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15:38:23 Query You should write up Video Experiment 2 as directed and submit it. What were your final results? Give your slopes and your rates of velocity change as rate vs. slope ordered pairs according to the y vs. x convention (slope first, rate second), and specify the slope of the straight line you got at the end.
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RESPONSE --> not assigned
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15:38:44 ** Common Error: A common error is to divide ave velolcity by the time required, which does not give acceleration. If you did these calculations correctly but didn't justify them on the experiment, justify your calculations now (just show one calculation in detail). If you did the calculations incorrectly then revise your calculations and give the same sort of justification. Note that only a Summary is required in the Query: This query just calls for a few summary items; if the experiment has been submitted and returned at OK there's no need to go into great detail in the Query. **
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15:38:55 Explain briefly how you determined your accelerations.
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15:39:15 **To determine acceleration we first find the average velocity by dividing displacement by elapsed time: vAve = `ds / `dt. Then since acceleration is uniform and init velocity is zero we can double the average velocity to get the final velocity. Finally we find acceleration, which is rate of velocity change. The rate of velocity change is (change in velocity) / (change in clock time). Change in velocity is the change from initial to final velocity: `dv = vf - v0. Change in clock time is the time required to accelerate down the ramp, which is what you observed. **
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RESPONSE --> ok
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15:39:54 What were the units of the slope of your straight line? Hint: The slope of a ramp has no units; what were the units of the rise between two points? (you won't be timed on this one, so don't worry about taking a few minutes if you need it).
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RESPONSE --> The rise is cm/s
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15:40:26 ** The rise measures change in acceleration in cm/s while the run measures change in clock time in s, so the units of the graph of acceleration vs. clock time are (cm/s) / s = cm/s^2. **
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RESPONSE --> Oops this is acceleration, cm/s^2.
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15:40:55 How is acceleration an example of a rate of change?
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RESPONSE --> You are dividing a change in velocity by a change in time.
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15:41:22 ** Velocity is the rate of change of position. Acceleration is rate of change of velocity--change in velocity divided by the time period. 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 time. The average rate of change of velocity with respect to clock time is the same as the acceleration **
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RESPONSE --> Got it.
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15:46:52 If you know average acceleration and time interval what can you find?
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RESPONSE --> You can find the velocity by multiplying average acceleration by the change in time.
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15:48:22 ** 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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RESPONSE --> we can only find the `dv not the vAve.
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15:49:08 Can you find velocity from average acceleration and time interval?
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RESPONSE --> No, unless we are given other information, we can only find the change in velocity.
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15:49:33 ** 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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RESPONSE --> OK
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15:50:15 Can you find change in velocity from average acceleration and time interval?
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RESPONSE --> Yes. aAve * `dt = `dv
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15:50:42 **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. **
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RESPONSE --> ok
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15:51:03 Can you find average velocity from average acceleration and time interval?
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RESPONSE --> No
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15:51:13 ** 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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RESPONSE --> ok
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15:51:49 You 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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RESPONSE --> because we don't know v0 or vf
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15:52:04 ** 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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RESPONSE --> ok
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15:52:14 Introductory Force Experiment Give your data for the number of 1/4 cups of water vs. the length of the rubber band in ordered-pair form (i.e., (13,1), (14,2), (14.8,3), ... would indicate lengths of 13 cm, 14 cm and 14.8 cm for 1, 2 and 3 1/4-cups).
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15:52:18 **COMMON ERROR: (0, 8.5) (1, 9) (2, 9.4) (3, 9.8) (4, 10.4) (5, 10.8) INSTRUCTOR RESPONSE: You have your coordinates backwards. The numbers of quarter-cups are 0, 1, 2, 3, 4. The rubber band lengths obtained for your specific rubber band, as you report them, are 8.5 cm, 8.8 cm, etc. Different students might get different results. Remember, the order is y vs. x. The y coordinate comes before 'vs.', the x coordinate after. When listing ordered pairs the x coordinate is the first and the y coordinate the second. So a graph of quarter-cups vs. length in cm. might contain the points (8.5, 0), (8.8, 1), etc. , etc.; but it wouldn't contain the points (0, 8.5), (1, 8.8) etc. **
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15:52:31 Give your results for weight vs. length in ordered-pair form. [ Note that the requested order of the variables seems to violate our convention for independent vs. dependent variable. However, we will later exert forces by stretching the rubber band, and will regard the stretch as the variable we control with force being dependent on stretch. So in this experiment we switch what seems to be the logical order of the variables. ]
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15:52:37 ** Each quarter-cup weighs about .55 Newtons, as given in the experiment, so that 0, 1, 2, . . . quarter-cups weigh 0, .55 N, 1.1 N, etc.. If the graph points for the preceding question include (0, 8.5), (1, 8.8), (2, 9.4) etc. then the graph of weight (in Newtons) vs. length (in cm) will include the points(8.5, 0), (8.8, .55), (9.4, 1.1), etc.. Your results will likely differ a bit but this example should show you whether you did the graph correctly. **
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15:52:44 What length you would expect for .8 Newtons of weight and for 2 Newtons of weight?
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15:52:50 ** You should look at your graph of weight vs. length. .8 Newtons will occur as the 'y' coordinate. Look straight across from that coordinate on the y axis to find the point on the graph where y = .8 Newtons. Then look straight down to find the corresponding 'x' coordinate, which will represent the stretch. If the weight vs. stretch points include (8.5, 0), (8.8, .55), (9.4, 1.1) then the .8 Newtons weight would be between .55 N and 1.1 N, so that the length would likely lie between 8.8 cm and 9.4 cm. The precise estimate of the length would depend on exactly how the graph is drawn. **
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15:52:52 What weight would you expect to stretch the rubber band by 1, 2, 3, 4, and 5 cm?
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15:52:56 ** You should look at your graph of weight vs. length. This graph can be relabeled to give you weight vs. stretch: Simply subtract the 0-weight length from each of the given lengths and write this number beneath the length on the x axis. Each stretch will occur as a newly labeled 'x' coordinate. Locate on the x axis the points corresponding to stretches of 1, 2, 3, 4, and 5 cm. Look straight up from each coordinate on the x axis to find the corresponding point on the graph. Then look straight across to the 'y' axis to find the corresponding 'y' coordinate, which will represent the weight necessary for that stretch. **
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15:52:59 Describe your force vs. stretch graph (i.e.: Do the points seem to lie on a straight line or a curve? If it is a curve for what stretches is the curve steepest, and for what stretches is it least steep? Does the direction of the curvature change and if so how?).
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15:53:07 **The graph will generally increase but at a decreasing rate. If you stretched the rubber band further than you should have, the graph might eventually straighten out for an instant then begin increasing at an increasing rate. **
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15:53:12 What do you think makes the rubber band stretch?
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15:53:16 ** The gravitational force between Earth and the water, which is what we mean by the 'weight' of the water, pulls the water down. The further the rubber band is stretched the harder it pulls back up, until the pull of the rubber band is equal and opposite to the gravitational force on the water. **
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15:53:18 Is there a force on the bag and water? If so what is the source of the force? What is the direction of this force?
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15:53:21 ** The source of the force is the gravitational attraction between water and Earth **
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15:53:25 Is there more than one force on the bag and water? If so what are the sources of these forces? What are the directions of these forces?
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15:53:27 ** gravity pulls down, the hook pulls up, and the two are in balance so they must be equal and opposite GOOD (but slightly flawed) STUDENT ANSWER: There is more than one force on the bag and the water. the rubber band would be exerting an upward force on the bag and the water although the force of the water in the bag pulling down on the rubber band is greater than the force of the rubber band pulling upward on the bag and water. INSTRUCTOR RESPONSE: Good, but since the bag isn't accelerating in either the upward or downward direction the two forces must be equal and opposite. If one was greater the bag would accelerate in the direction of that force. The rubber band stretches until it exerts exactly enough force to create this equilibrium with the gravitational force. **
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15:53:32 General College Physics only: Problem #10 Summarize your solution Problem 1.10 (approx. uncertainty in area of circle given radius 2.8 * 10^4 cm).
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15:53:37 ** Radius 2.8 * 10^4 cm means that the radius is between 2.75 * 10^4 cm and 2.85 * 10^4 cm. This means that the area is between pi * (2.75 * 10^4 cm)^2 = 2.376 * 10^9 cm^2 and pi * (2.85 * 10^4 cm)^2 = 2.552 * 10^9 cm^2. The difference is .176 * 10^9 cm^2 = 1.76 * 10^8 cm^2, which is the uncertainty in the area. Note that the .1 * 10^4 cm uncertainty in radius is about 4% of the radius, which the .176 * 10^9 cm uncertainty in area is about 8% of the area. This is because the area is proportional to the squared radius. A small percent uncertainty in the radius gives very nearly double the percent uncertainty in the squared radius. **
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15:55:42 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> Neither of the experiments discussed in this query were on the assignment page.
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15:57:21 ** COMMON STUDENT COMMENT AND INSTRUCTOR RESPONSE: I am really confused about velocity, acceleration changes in velocity and acceleration, etc. I guess I am the type that works with a formula and plugs in a number. I have went back to the class notes and the problem sets to summarize formulas. Any suggestions? RESPONSE: I note that you are expressing most of your answers in the form of formulas. Ability to use formulas and plug in numbers is useful, but it doesn't involve understanding the concepts, and without an understanding of the concepts we tend to plug our numbers into equations that don't apply. So we deal first with concepts. However formulas do come along fairly soon. The concepts of velocity, acceleration, etc. are very fundamental, but they are tricky and they take awhile to master. You are doing OK at this point. You'll see plenty more over the next few assignments. If you look at the Linked Outline (on the main Physics 1 page--the one where you click on the Assts button--click on the Overviews button, then on the Linked Outline. You will see a table with a bunch of formulas and links to explanations. You might find this page very useful. Also the Introductory Problem Sets give you formulas in the Generalized Solutions. **
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RESPONSE --> I think I am starting to grasp the material. "