course PHY 121 _zM\ʦݞassignment #002
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16:56:02 How is acceleration an example of a rate of change?
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RESPONSE --> because when you accelerate you are changing either your speed or direction. confidence assessment:
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16:56:19 ** 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 --> ok self critique assessment:
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16:57:25 If you know average acceleration and time interval what can you find?
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RESPONSE --> change in velocity? confidence assessment: 1
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16:57:56 ** 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 --> ok self critique assessment:
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16:59:07 Can you find velocity from average acceleration and time interval?
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RESPONSE --> yes. ave velocity = change in position/ change in clock time confidence assessment:
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16:59:47 ** 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. so we can only find change in velocity not the actual velocity. self critique assessment:
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17:00:55 Can you find change in velocity from average acceleration and time interval?
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RESPONSE --> yes. change in position / change in clock time = ave velocity confidence assessment:
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17:01:24 **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. self critique assessment:
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17:01:49 Can you find average velocity from average acceleration and time interval?
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RESPONSE --> yes. using the same equation as before confidence assessment:
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17:02:11 ** 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. self critique assessment:
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17:02:47 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 you would need to know the initial velocity at which you started confidence assessment:
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17:03:28 ** 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. average velocity has no direct relationship with acceleration. self critique assessment:
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17:03:48 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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RESPONSE --> confidence assessment:
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17:03:53 ** 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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RESPONSE --> self critique assessment:
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17:05:11 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> i am still a little confused about rate of change of velocity and acceleration. i think i got ahead of myself thinking that it is as basic as i teach my current students however it is more indepth. i need to review these topics further. confidence assessment:
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17:06:04 ** 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 --> great. will do. self critique assessment:
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17:12:36 Principles of Physics Students and General College Physics Students: Problem 14. What is your own height in meters and what is your own mass in kg, and how did you determine these?
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RESPONSE --> 5ft 3 in. 5 ft * 30.48 cm = 152.4 cm 3 in * 2.54 cm = 7.62 cm = 160.02 cm or 1.6 m when you move the decimal to places to the right. 119 lbs. 2.2 lbs = 1kg when g = 9.8 m/s^2 119/2.2 = 54.09 kg confidence assessment:
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17:13:07 Presumably you know your height in feet and inches, and your weight in pounds. Presumably also, you can convert your height in feet and inches to inches. To get your height in meters, you would first convert your height in inches to cm, using the fact that 1 inch = 2.54 cm. Dividing both sides of 1 in = 2.54 cm by either 1 in or 2.54 cm tells us that 1 = 1 in / 2.54 cm or that 1 = 2.54 cm / 1 in, so any quantity can be multiplied by 1 in / (2.54 cm) or by 2.54 cm / (1 in) without changing its value. Thus if you multiply your height in inches by 2.54 cm / (1 in), you will get your height in cm. For example if your height is 69 in, your height in cm will be 69 in * 2.54 cm / (1 in) = 175 in * cm / in. in * cm / in = (in / in) * cm = 1 * cm = cm, so our calculation comes out 175 cm.
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RESPONSE --> ok. think i got it right. may not have been that way but... self critique assessment:
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19:27:29 Query Principles of Physics and General College Physics: Summarize your solution to Problem 1.19 (1.80 m + 142.5 cm + 5.34 `micro m to appropriate # of significant figures)
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RESPONSE --> 1.8 m 142.5 cm = 1.425 m 5.34 micro m = 0 m therefore 3.225 m confidence assessment: 0
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19:28:18 ** 1.80 m has three significant figures (leading zeros don't count, neither to trailing zeros unless there is a decimal point; however zeros which are listed after the decimal point are significant; that's the only way we have of distinguishing, say, 1.80 meter (read to the nearest .01 m, i.e., nearest cm) and 1.000 meter (read to the nearest millimeter). Therefore nothing smaller than .01 m can be distinguished. 142.5 cm is 1.425 m, good to within .001 m. 5.34 * `micro m means 5.34 * 10^-6 m, or .00000534 m, accurate to within .00000001 m. When these are added you get 3.22500534 m; however the 1.80 m is not resolved beyond .01 m so the result is 3.23 m. Remaining figures are meaningless, since the 1.80 m itself could be off by as much as .01 m. **
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RESPONSE --> ok. not in that detail but... self critique assessment:
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19:28:37 University Physics #34: Summarize your solution to Problem 1.34 (4 km on line then 3.1 km after 45 deg turn by components, verify by scaled sketch).
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RESPONSE --> confidence assessment:
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19:28:46 ** THE FOLLOWING CORRECT SOLUTION WAS GIVEN BY A STUDENT: The components of vectors A (2.6km in the y direction) and B (4.0km in the x direction) are known. We find the components of vector C(of length 3.1km) by using the sin and cos functions. }Cx was 3.1 km * cos(45 deg) = 2.19. Adding the x component of the second vector, 4.0, we get 6.19km. Cy was 2.19 and i added the 2.6 km y displacement of the first vector to get 4.79. So Rx = 6.19 km and Ry = 4.79 km. To get vector R, i used the pythagorean theorem to get the magnitude of vector R, which was sqrt( (6.29 km)^2 + (4.79 km)^2 ) = 7.3 km. The angle is theta = arctan(Ry / Rx) = arctan(4.79 / 6.19) = 37.7 degrees. **
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RESPONSE --> self critique assessment:
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