course Phy 121 004. `Query 4*********************************************
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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 #$&* OK ********************************************* Question: If you know average acceleration and time interval what can you find? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The change in velocity. confidence rating #$&* 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: Can you find velocity from average acceleration and time interval? Your solution: No. Only the change in velocity. confidence rating #$&* 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qCan you find change in velocity from average acceleration and time interval? Your solution: Yes confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: `qCan you find average velocity from average acceleration and time interval? Your solution: NO confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. 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 #$&* OK ********************************************* Question: `qYou can find only change in velocity from average acceleration and time interval. To find actual velocity you have to know at what velocity you started. Why can't you find average velocity from acceleration and time interval? Your solution: The average velocity is deltax/deltat. It has no direct connection with acceleration. confidence rating #$&*: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: Give at least three possible units for velocity, and at least three possible units for clock time. Give at least three possible units for the slope between two points of a graph of velocity vs. clock time. Explain how you reasoned out the answer to this question. Explain the meaning of the slope of this graph. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: mph, m/s, ft/min. Hr, sec, min. mph^2, m/s^2, ft/min^2. Its simply about confidence rating #$&* 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK. I could have given an example of how I solved for the units though. ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: If the velocity of an object changes at a uniform rate from 5 m/s to 13 m/s between clock times t = 7 s and t = 11 s then on this interval what is its average velocity, and what is the average rate at which its velocity changes with respect to clock time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 13m/s-5m/s=8m/s/2=4m/s 8m/s/4s=2m/s^2 confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: Explain how to solve the relationship aAve = `dv / `dt for `dt. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: multiply both sides by dt then divide dv by aAve confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: A ball rolls from rest down a book, off that book and onto another book, where it picks up additional speed before rolling off the end of that book. Suppose you know all the following information: How far the ball rolled along each book. The clock time at which the ball is released, the clock time at which it first reaches the end of the first book, and the clock time at which it first reaches the end of the second book. How fast the ball is moving at each end of each book. How would you use your information to calculate the ball's average speed on each book? How would you use your information to calculate how quickly the ball's speed was changing on each book? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: I would determine the time interval and the position interval to determine the average speed. Then I would break up the data for each book to determine the average speed on each book. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `qQuery Add any surprises or insights you experienced as a result of this assignment. Your solution: I learned about the difference between average velocity and change in velocity. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** "