course PHY: 201
1. State the definition of rate of change.vvvv
&&&& the average rate of change of A with respect to B is average rate = change in A/ change in B &&&&
2. State the definition of velocity.
&&&& velocity is the distance travelled divided by the time interval &&&&
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3. State the definition of acceleration.
&&&& acceleration is the average rate of change of velocity with respect to clock time &&&&
4. A ball rolls along a path, moving from position 20 cm to position 50 cm as its velocity increases from 5 cm/s to 15 cm/s.
• What is its change in velocity and how do you obtain it from the given information? &&&& 3 cm2/s, I took the distance travelled, or 50 – 20 = 30cm and divided it by the time interval, 15 – 5 = 10 cm/s, and came up with 3 as my answer. Then for the units I took cm/s and cm/1 and multiplied them and got cm2/s &&&&
5 cm/s and 15 cm/s are not clock times, and their difference is not a time interval.
Their difference is 10 cm/s, but this isn't a time interval. It doesn't have units of time.
• What is its change in position and how do you obtain it from the given information? &&&& 30 cm, you take the final position, 50 cm and subtract the initial position, 20 cm and get 30 cm &&&&
5. A ball accelerates from velocity 30 cm/s to velocity 80 cm/s during a time interval lasting 10 seconds.
Explain in detail how to use the definitions you gave above to reason out
• the average velocity of the ball during this interval, &&&& the average velocity is the change in A, or position of the ball, with respect to change in B, or the time interval. So you would take the change in the position and divide it by the change in clock time. In this case you would take 80 cm/s and subtract 30 cm/s coming to an answer of 50 cm/s. This is how you get the change in A or ball position.
Excellent reasoning. However 80 cm/s and 30 cm/s do not have units of position, and are not positions.
Then the change in B would be the 10 seconds. It’s not stated in the problem, so we assume the initial clock time is 0 seconds. So we would take 10 seconds and subtract 0 seconds getting 10 seconds for the change in B.
It is valid to assume a clock that starts at 0 seconds, but it's not necessary. You already know the change in clock time is 10 sec. You could equally well have assumted that the initial clock time is, say, 873 seconds so that the final clock time is 883 seconds. Either way you would have ended up with what you know, that clock time changes by 10 seconds.
Then you divide ball position, 50 seconds by the clock time, 10 seconds and get 5 cm/ s2.
The change in ball position isn't 50 seconds, and the result of dividing seconds by seconds is not cm/s^2.
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• its acceleration during this interval. &&&& Acceleration is the change in velocity with respect to clock time. In this case velocity would be the change in A and the clock time would be the change in B. So it would be like the problem above. You would take 80 cm/s and subtract 30 cm/s coming to an answer of 50 cm/s. This is how you get the change in A or velocity. Then the change in B would be the 10 seconds. It’s not stated in the problem, so we assume the initial clock time is 0 seconds. So we would take 10 seconds and subtract 0 seconds getting 10 seconds for the change in B. Then you divide ball position, 50 seconds by the clock time, 10 seconds and get 5 cm/ s2.
Very good. Your reasoning is good, as before, and on this problem you correctly identified the quantities and got all the details right.
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Remember, the main goal is to use a detailed reasoning process which connects the given information to the two requested results. You should use units with every quantity that has units, units should be included at every step of the calculation, and the algebraic details of the units calculations should be explained.
6. A ‘graph trapezoid’ has ‘graph altitudes’ of 40 cm/s and 10 cm/s, and its base is 6 seconds. Explain in detail how to find each of the following:
• The rise of the graph trapezoid. &&&& the rise of the graph is the vertical distance between the two points. In the instance we would take 40 and subtract 10 from it to get a rise of 30 cm/s. &&&&
• The run of the graph trapezoid. &&&& the run of the graph is the horizontal distance between the two points. The base is 6 seconds so we would assume that the initial is 0 so the run would be 6 seconds. &&&&
• The slope associated with the trapezoid. &&&& the slope of a graph is rise over run. In this problem it would be 30/6 which would give us 5. Then to get the units, we take cm/s and multiply it by seconds. the two quantities were divided; so you would divide the units, not multiply them
The seconds will cancel out and we are left with cm. So the slope is 5 centimeters. &&&&
• The dimensions of the equal-area rectangle associated with the trapezoid. &&&& the dimensions would be 25 cm/s and 6 seconds. &&&&
• The area of the trapezoid. &&&& The area of the trapezoid would be 150 centimeters. &&&&
This is correct. You could have included more steps in your reasoning, but I believe you know them.
Each calculation should include the units at every step, and the algebraic details of the units calculations should be explained.
7. If the altitudes of a ‘graph trapezoid’ represent the initial and final positions of a ball rolling down an incline, in meters, and the based of the trapezoid represents the time interval between these positions in seconds, then
• What is the rise of the graph trapezoid and what are its units? &&&& the rise of the graph would be the final ball position in meters minus the initial ball position in meters. The units will be meters. &&&&
• What is the run of the graph trapezoid and what are its unit? &&&& the run of the graph would be the time interval in seconds. The units will be seconds. &&&&
• What is the slope of the trapezoid and what are its units? &&&& the slope of the trapezoid would be the ball position in meters divided by the time interval in seconds. The final units would be meters/seconds. &&&&
• What is the area of the trapezoid and what are its units? &&&& the area of the trapezoid would be the average ball position in meters times the time interval in seconds. The units would be meters per seconds.
meters per second indicates division of meters by seconds, i.e., meters / second.
Your calculation does not involve dividing a quantity in meters by a quantity in seconds.
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• What, if anything, does the slope represent? &&&& the slope represents the average position of the ball. &&&&
The average position of the ball might be the position midway between the initial and final positions, but the slope does not represent this quantity.
• What is the altitude of the equal-area rectangle and what are its units? &&&& the area of the equal area rectangle would be [(final ball position(meters) + initial ball position(meters)) / 2] times time interval(seconds). The units would be meters per seconds. &&&&
You have multiplied meters by seconds, not divided.
• What is the base of the equal-area rectangle and what are its units? &&&& the base of the equal-area rectangle is 6 seconds and its units are seconds. &&&&
• What, if anything, does the area represent? &&&& the area represents position of the ball &&&&
the area doesn't necessarily have a meaningful interpretation, but its units are not units of position, so it doesn't represent the position of the ball
Each answer should include a complete explanation, reasoned out from the geometry of the trapezoid and the definitions you gave at the beginning.
8. If the altitudes of a ‘graph trapezoid’ represent the initial and final velocities of a ball rolling down an incline, in meters / second, and the base of the trapezoid represents the time interval between these velocities in seconds, then
• What is the slope of the trapezoid and what are its units? &&&& the slope of the graph is the [final velocity (meter / seconds) - the initial velocity in (meters / seconds)] / time intervals (seconds); the units would be in meters. &&&&
You are dividing a quantity with units of meters / second by a quantity with units of seconds. You don't get a quantity with units of meters.
• What is the area of the trapezoid and what are its units? &&&& the area of the graph is the [(final velocity (meter / seconds) + the initial velocity in (meters / seconds)) /2] * time interval in seconds; the units would be in meters.
Correct. meters / second * seconds = meters.
• What, if anything, does the slope represent? &&&& The slope represents the velocity of the ball &&&&
the altitudes represent velocities; the slope doesn't
• What, if anything, does the area represent? &&&& The area represents the position of the ball &&&&
you've left out a couple of key words; the ball has a position at the beginning and end of the interval, so it doesn't have 'a position'.
Each answer should include a complete explanation, reasoned out from the geometry of the trapezoid and the definitions you gave at the beginning.
9. A ball rolls along a path, moving from position 20 cm to position 50 cm as its velocity increases from 5 cm/s to 15 cm/s.
• If its acceleration is uniform, then how long does this take, and what is the ball’s acceleration? &&&& it takes 15 seconds; the acceleration is 3 cm2/s &&&&
If it took 15 seconds then the change in velocity would be 10 cm/s, the change in clock time 15 seconds, and the acceleration would be 10 cm/s / (15 s) = .667 cm/s^2.
If it took 15 seconds then since the change in position is 30 cm, its average velocity would be 30 cm / (15 s) = 2 cm/s, which is less than either its initial or final velocity.
Can you detail your reasoning on this question?
I'm assessing your work on several criteria, each assessed on a 5-point scale. This isn't a grade and won't be used as a grade. It's for your information.
On the 5-point scale, 1 represents something you seldom did well and 5 something you almost always did well. On the average, most of the class at this point appears to be around the middle of the scale.
Here are the criteria, listed more or less in the order of importance at this stage of the course:
Overall level of reasoning process (i.e., are you basing your answers on the definitions and explaining your details).
My assessment on this criterion: 4.
accuracy of definitions: 3 (should have been a 5 but see my note on # 2)
accuracy of wording: 4+
consistent use of units: 4+
correct identification of quantities (still a weak area for the class as a whole): 3
correct units calculations (still a weak area for the class as a whole): 3
I recommend that you submit a revision of your work, according to my notes. You doing quite well at this stage, so try to 'lock in' all of these ideas as soon as possible.