#$&* course Phy 231 6/21/11 11:52PM Question: How is acceleration an example of a rate of change?
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Given Solution: ** Velocity is the rate of change of position with respect to clock time. Acceleration is rate of change of velocity with respect to clock time. To find the acceleration from a v vs. t graph you take the rise, which represents the change in the average velocity, and divide by the run, which represents the change in clock time. note that the term 'average rate of change of velocity with respect to clock time' means the same thing as 'acceleration' ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: If you know average acceleration and time interval what can you find? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: With an average accelearation and a time frame, you can find the average velocity. confidence rating #$&*: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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I made the common mistake of thinking I could find the average velocity with the average acceleration and time. Instead, I can only find the change in velocity. ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: Can you find velocity from average acceleration and time interval? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: No, you can only find the change in velocity. 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qCan you find change in velocity from average acceleration and time interval? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: yes, you can through multiplication. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qCan you find average velocity from average acceleration and time interval? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: No, you can only find the change in velocity. 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Average velocity is found by dividing the change in distance by the change in time. Average acceleration is found by dividing the change in velocity (not the average velocity) by the change in time. Since the average acceleration is found with the change in velocity, you can't find average velocity form average acceleration and time. 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: velocity: m/s, mi/h, cm/min time: ms, s, min, h slope: m/s^2, ft/min^2, km/h^2 The slope of a graph of velocity vs. clock time is acceleration. I just provided units for acceleration. 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'). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (5+13)/2= 9m/s=average velocity (8m/s)/4s= 2m/s^2=average acceleration confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: Explain how to solve the relationship aAve = `dv / `dt for `dt. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Just divide both sides by `dv to isolate `dt. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok? ------------------------------------------------ Self-critique Rating:ok Question for University Physics Students: What is the instantaneous rate of change of v with respect to t at t = 2, given that v(t) = 2 t^2 - t + 3? Explain how you obtained this result. What is the expression for the instantaneous rate of change of v with respect to t at general clock time t, given the same velocity function? Explain how you determined this. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4t-1 8-1=7 instantaneous acceleration = 7 (units not determined) I took the derivative of v(t) to find a(t), then I plugged 2 into t to find the instantaneous acceleration. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I took the derivative of v(t) and solved for a(2) instead of solving for v(2). I arrived at the same answer. ------------------------------------------------ Self-critique Rating:ok ********************************************* 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? Since we know the clock time at all three significant points (t1=beginnging of the first book, t2=beginning of the second book, t3=the end of both books) we can calculate the change in time for both books. We also know the velocity at three significant points(v1=beginning of the first book=0 since the ball starts at rest, v2=velocity at the beginning of the second book, v3=velocity at the end of the second book) so we can calculate the change in velocityfor both books. we calculate the average velocity on the first book with (v2-v1)/2 The second book has a similar velocity calculation (v3-v2)/2 We can calculate acceleration on book 1 with (v2-v1)/(t2-t1) We can calculate acceleration on book 2 with (v3-v2)/(t3-t2) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `qQuery Add any surprises or insights you experienced as a result of this assignment. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I didn't expect the calculus portion yet. It's ok, though. I was able to solve it without much issue. confidence rating #$&*:ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating:
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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: ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: