Query 1

You should have included more detail about your solution and your thinking on most problems, and most should have been self-critiqued but weren't. See my notes, in boldface like this one, to be sure you understand what is mean by 'self-critique', and when you need to do it. I only marked the first couple of places where you should have self-critiqued, but most of your responses required it.

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19:11:25 Briefly state what you think velocity is and how you think it is an example of a rate.

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RESPONSE --> IT IS A RATE OF SPEED

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19:11:51 ** A rate is a change in something divided by a change in something else. This question concerns velocity, which is the rate of change of position: change in position divided by change in clock time. **

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RESPONSE --> OK

Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the given solution, and if necessary asking specific questions.

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19:12:41 Given average speed and time interval how do you find distance moved?

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RESPONSE --> NOT SURE

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19:14:31 ** You multiply average speed * time interval to find distance moved. For example, 50 miles / hour * 3 hours = 150 miles. **

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RESPONSE --> OK

This also requires a self-critique.

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19:14:52 Given average speed and distance moved how do you find the corresponding time interval?

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RESPONSE --> DON'T KNOW

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19:16:15 06-24-2005 19:16:15 ** time interval = distance / average speed. For example if we travel 100 miles at 50 mph it takes 2 hours--we divide the distance by the speed. In symbols, if `ds = vAve * `dt then `dt = `ds/vAve. Also note that (cm/s ) / s = cm/s^2, not sec, whereas cm / (cm/s) = cm * s / cm = s, as appropriate in a calculation of `dt. **

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NOTES ------->

Self-critique should be included here.

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19:16:21 ** time interval = distance / average speed. For example if we travel 100 miles at 50 mph it takes 2 hours--we divide the distance by the speed. In symbols, if `ds = vAve * `dt then `dt = `ds/vAve. Also note that (cm/s ) / s = cm/s^2, not sec, whereas cm / (cm/s) = cm * s / cm = s, as appropriate in a calculation of `dt. **

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RESPONSE --> OK

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19:16:39 Given time interval and distance moved how do you get average speed?

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RESPONSE --> HOW?

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19:16:53 06-24-2005 19:16:53 ** Average speed = distance / change in clock time. This is the definition of average speed. For example if we travel 300 miles in 5 hours we have been traveling at an average speed of 300 miles / 5 hours = 60 miles / hour. **

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NOTES ------->

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19:17:52 You should have written up Video Experiment 2 as directed and submitted it. If you have done so within a couple of days of submitting this query then you need not answer any question here that was answered on your writeup. This is a general principle for experiments and can be applied on all queries. 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 --> I WAS THE STUDENT WHO IS WAITING ON CD'S/DVD'S.

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19:18:08 06-24-2005 19:18:08 ** It's very important to express all quantities in terms of the correct units. Slope, being rise / run in units of cm / cm, ends up unitless (the cm 'cancel'). Velocities are expressed in cm / sec. Rates of velocity change are expressed in (cm / sec) / sec or cm / sec^2. Slopes are expressed in units of rise / units of run, in this case giving just cm / sec^2. **

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NOTES ------->

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19:18:40 Explain briefly how you calculated your slopes.

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RESPONSE --> I JUST USED THE SLOPE FORMULA

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19:19:02 ** slopes are caclulated by using the basic rise over run formula. For instance one student reports that when the paper stack was 1.26 cm tall and the desk was 44 cm long the slope was 1.26/44 which is equivalent to 0.0286 **

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RESPONSE --> OK

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19:19:25 Explain briefly how you determined your rates of velocity change.

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RESPONSE --> I DID NOT HAVE REGULAR DVD ACCESS

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19:20:44 06-24-2005 19:20:44 ** rates of velocity change are calculated by dividing change in velocity by change in clock time. From your timing you got average velocity. Since accel is uniform and initial velocity is zero, final velocity must be double average velocity. Change in velocity is obtained by subtracting init vel from final vel. Change in clock time is the time required to accelerate down the incline. Note units: when you divide change in velocity by change in clock time you are dividing cm/sec by sec, giving cm / sec * 1/sec = cm/sec^2. **

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NOTES ------->

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19:22:47 What were the units of the slope of your straight line on the graph of rate of velocity change vs. slope? 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 --> I DID NOT WORRY ABOUT THIS ASSIGNMENT BECAUSE I DID NOT HAVE DVD'S HOWEVER, I DID KEEP NOTES FROM THINGS THAT I DID NOT UNDERSTAND

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19:24:47 06-24-2005 19:24:47 ** Rise is change in the rate of velocity change, in cm/sec^2. Ramp slope is unitless. So graph slope is in units of cm/sec^2. The ideal value would be very close to the acceleration of gravity, which is 980 cm/sec/sec. Typical range when timing with the pendulum is from 600 to 1500 cm/sec/sec. Using very accurate timing (electronic equipment in the lab) with these clips we still end up with an error of about 8% on the high side, which shouldn't happen. The experiment done with accurate timing should be quite accurate. **

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NOTES -------> OK

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