phy -201
Your 'cq_1_01.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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The problem:
Here is the definition of rate of change of one quantity with respect to another:
The average rate of change of A with respect to B on an interval is
average rate of change of A with respect to B = (change in A) / (change in B)
Apply the above definition of average rate of change of A with respect to B to each of the following. Be sure to identify the quantity A, the quantity B and the requested average rate.
If the position of a ball rolling along a track changes from 10 cm to 20 cm while the clock time changes from 4 seconds to 9 seconds, what is the average rate of change of its position with respect to clock time during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line): The difference between changes A 10cm and the difference in the change in B is 5 secs, applying the foruma of rate of change 10cm/5s= 2cm/s. I guess since the ball had moved 10cm in 4 secs and only move another 10cm in 5 secs kinda shows that the ball is losing speed
Good answer to the question that was posed.
Be careful about assuming extraneous information. There is no reason to suppose that the ball moved 10 cm in 4 sec. You don't know where it was at t = 0 sec, so you have no information on how far it might have moved between t = 0 and t = 4 sec. All the information you have is for the interval between t = 0 and t = 4 sec.
#$&*
If the velocity of a ball rolling along a track changes from 10 cm / second to 40 cm / second during an interval during which the clock time changes by 3 seconds, then what is the average rate of change of its velocity with respect to clock time during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line): The difference between 10cm/sec and 40cm/sec is 30cm/s and if you divide that by the change in 3 seconds then the rate of change would be 10cm/s.
#$&*
If the average rate at which position changes with respect to clock time is 5 cm / second, and if the clock time changes by 10 seconds, by how much does the position change?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line): The rate change is 50 cm. If you take the formula given above and you know that that 10 sec is the rate of change for b and 5cm/second is the average rate if you plug everything into the formula it comes out that 50 cm /10 secs equals 5 cm/s.
#$&*
You will be expected hereafter to know and apply, in a variety of contexts, the definition given in this question. You need to know this definition word for word. If you try to apply the definition without using all the words it is going to cost you time and it will very likely diminish your performance. Briefly explain how you will ensure that you remember this definition.
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line): It easy for me to remember that the change of A is divided by the change in B with respect to one another.
#$&*
You are asked in this exercise to apply the definition, and given a general procedure for doing so. Briefly outline the procedure for applying this definition, and briefly explain how you will remember to apply this procedure.
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):(change in A/ the change b), It easy for me to remember that the change of A is divided by the change in B with respect to one another.
You're in good shape with the 'seed question'.
Did you submit the information below?
1.29
.922
.996
.887
.996
1.24
1.01
1.17
.953
1.15
This data was collected from a ball (golf ball) rolling down a ramp, the first five are from left to right, the second five are from the ball rolling right to left.
** **
686
.859
.688
.687
.734
.766
.672
.875
.641
.824
** **
.605
.641
.687
.703
.714
.589
.636
.500
.390
.531
** **
1.02, .159
1.11, .119
.731, .075
.756, .099
.670, .046
.529, .094
** **
27.5
25.2
38.3
37.0
41.8
52.9
Vave='ds/'dt. I divided the distance of 28 cm of the ball by the mean of the time it for the ball to travel down the ramp in my Ball down a ramp experiment. So for 1 domino vave= 28cm/1.02s = 27.5 for right to left, the same for left to right, and the same calculation for each left to right, right to left times for 2 and 3 dominoes. The vave is recorded in cm per second.
** **
26.9
22.7
52.4
48.9
62.4
100.1
The average rate of acceleration is 'dv/'dt (change in velocity divided by change in time interval)I used the same velocity as calculated for the vave becuase the initial velocity started each time from zero, so the change in velocity would be the same as the vave. So for the first domino I had a velocity of 27.5cm/s / 1.02s (the mean clocktime) which gave me the acceleration 26.9cm/s^2. I did this same calculation for two and three dominoes right to left and then left to right. This is recorded in cm/s^2
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26.4cm/s
37.7cm/s
47.3cm/s
The above number are the average of the the velocities obtained by adding the two velocity found for each trial left and right and dividing those by 2. So for the first trial with one dominoe the velocities were found to be (27.5 + 25.2)/2=26.4. I did this for each trial of both two and three dominoes.
** **
1.07s
.744s
.599s
24.7cm/s^2
50.7cm/s^2
78.9cm/s^2
To find the average rate of change with respect to clocktime I averagesd the two time interval for each trial of the One, two, and three domino heights. I divided the average of the velocities of the one, two and three dominoes by the average of the time, so for one domino the average velocity was 26.4cm/s / 1.07 (average of the two times) = 24.7cm/s^2
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I would expect that the average rate of change in velocit with respect to clocktime would be the same for each slope.
If I average the two acceleration rate found, ie for one dominoe the acceleration rate were found to be 26.9 and 22.7cm/s^2, the average of those is 24.8, for two the average acceleration is 50.6cm/s^2, and the average for the acceration for the three dominoes is 81.2cm/s^2. So, the average of the acceleration are the same or nearly the same as the average of the velocities divided by the average of the time.
I would expect this because the average velociies are calculated from a distance divided by a clocktime, and each clocktime was obtained by the same incline in height by the same distance of the same ramp, the only difference was one was right to left the other was left to right. When you obtain the average of two number you are finding the midpoint of those numbers, therefore, the number will lay within the same interval whether it is clocktime or velocity. This is why I would expect both methods to be the same.
** **
.03, 24.7
.06, 50.7
.09, 78.9
These are the slope,average rate of change with respect to clocktime, for 1, 2, and 3 dominoes. The avergae rate of change with respect to clocktime is reported in cm/s^2, and the ramp slope are unitless.
I obtained these results from the calculation of average rate of change with respect to clocktime, and the slopes of the ramps with 1, 2 and 3 dominoes.
** **
.4
35
The units of the horizontal intercept are not specified.
The unitls of the vertical intercept are in cm.s^2
The graph of acceleration vs ramp slope will have the acceleration on the vertical axis and the ramp slope on the horizontal axis. As ramp slope increases the acceleration increases, so the graph is increasing at an increasing rate. The horizontal intercept appears to be at .4, which is between the first and second points, The vertical intercept appears to be at about 35cm/s^2 vertically, between the first and second points.
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.10, 85
From the graph I created, the point that would correspond to a ramp slope of .10 would be vertically asbout 85cm/s^2. I think this is close to accurate since when the ramp slope increases so does the velocity. In our original experiment the ramp slope increased by apprximately .3, ie they went from .3 to .6 to .9, nd as the incllined tripled in height the velocities seemed to rise by double the first and then triple the first. If we plot a point from .9 to .10 that is only a one point increase and if at .9 the acceleration is almost 80cm/s^2 it would make sense that the increase for the slope at .10 in between 80cm/s^2 and 90cm/s^2.
** **
.6
50
83.3
The first number is the number between the horizontal intercept of the best fit line ony my graph and the point on that line that corresponds to ramp slope .10 horizontally. The second number is the point on the graph that corresponds to the number that lies between the vertical intercept on my graph and .10 vertically. I calculated the rise/run by 50/.6 to be 83.3, which is close to the acceleration that I estimated to be at .10. This tells us the the slope is the same as the accelertaion rate for the graph.
** **
1.02, 1.11, 1.06
.159, .119, .139
The number on the first line are the mean of the time interval from right to left, left to right and the average. On the second line are the standard deviation from right to left, left to right and their average. The signifigance of these numbers are that for i dominoe the average time was about 1.1 seconds, and the average standard deviation was about .14, this could be compared to the uncertainty, of about .1, which is the same as the average of the standard deviation.
** **
.921, 1.19
So the bounderies of this interval are bounded on the left at .921 and on the right at 1.19. I obtained these number by 1.06 - .139 =.921
and 1.06 + .139 = 1.19.
** **
30.4, 33.0
23.5, 19.8
19.8, 33.0
These number represent the average velocity and the acceleration for the left boundary on the first line and on the second the average velocity and accleration on the right hand boundary. The third line are the minimum and maximum possible values of acceleration. The acceleration would be between these two results because the change in velocity will not exceed or fall below this interval for the given distance of 28 cm and time interval obtained, since acceleration is the change in velocity with respect to clocktime.
** **
.744, .087
.657, .831
42.6, 33.6
40.4, 64.8
The first line is the average of the means from left to right and right to left and the average of the stnard deviation from left to right and right to left. The second line is the boundary of the interval on the left hand side and the boundary of the interval on the right hand side. The third line are the maximum and minimum velocitie obtained by the division of the distance of 28cm by the maximum or minimum time intervals. And the fourth line is the minimum ans maximum acceleration for the 2 domino data.
** **
.599, .07
.529, .669
52.9, 41.9
62.6, 100
The first line is the average of the means for the 3 dominoe data and the average of the standard deviations. The second line is the minimum and maximum boundaries for the timed interval from left to right. The third line is the minimum and maximum velocities, and the fourth line is the minimum and maximum accelerations.
** **
.03, 19.8
.06, 40.4
.09, 62.6
.03, 33.0
.06, 64.8
.09, 100
** **
Yes, my graph fits this description. Yes my best fit line passes through the three short vertical segments. The best fit straight line runs through the points of the acceleration vs ramp slope, which indicates that the ramp slope and the acceleration are closely related.
** **
93.8
75,.8
The rise of my line is between 85 and 10 vertically and .10 and .2 horizontally, so 75/.8= 93.8 is my slope
** **
87.5
70,.8
The least steep passes through vertically between 80 and 10 at 70, and horizontally between .10 and .2 at .8 70/.8= 87.5 slope.
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3 hours
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`gr52
phy -201
Your 'cq_1_01.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
** **
** **
The problem:
Here is the definition of rate of change of one quantity with respect to another:
The average rate of change of A with respect to B on an interval is
average rate of change of A with respect to B = (change in A) / (change in B)
Apply the above definition of average rate of change of A with respect to B to each of the following. Be sure to identify the quantity A, the quantity B and the requested average rate.
If the position of a ball rolling along a track changes from 10 cm to 20 cm while the clock time changes from 4 seconds to 9 seconds, what is the average rate of change of its position with respect to clock time during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line): The difference between changes A 10cm and the difference in the change in B is 5 secs, applying the foruma of rate of change 10cm/5s= 2cm/s. I guess since the ball had moved 10cm in 4 secs and only move another 10cm in 5 secs kinda shows that the ball is losing speed
Good answer to the question that was posed.
Be careful about assuming extraneous information. There is no reason to suppose that the ball moved 10 cm in 4 sec. You don't know where it was at t = 0 sec, so you have no information on how far it might have moved between t = 0 and t = 4 sec. All the information you have is for the interval between t = 0 and t = 4 sec.
#$&*
If the velocity of a ball rolling along a track changes from 10 cm / second to 40 cm / second during an interval during which the clock time changes by 3 seconds, then what is the average rate of change of its velocity with respect to clock time during this interval?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line): The difference between 10cm/sec and 40cm/sec is 30cm/s and if you divide that by the change in 3 seconds then the rate of change would be 10cm/s.
#$&*
If the average rate at which position changes with respect to clock time is 5 cm / second, and if the clock time changes by 10 seconds, by how much does the position change?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line): The rate change is 50 cm. If you take the formula given above and you know that that 10 sec is the rate of change for b and 5cm/second is the average rate if you plug everything into the formula it comes out that 50 cm /10 secs equals 5 cm/s.
#$&*
You will be expected hereafter to know and apply, in a variety of contexts, the definition given in this question. You need to know this definition word for word. If you try to apply the definition without using all the words it is going to cost you time and it will very likely diminish your performance. Briefly explain how you will ensure that you remember this definition.
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line): It easy for me to remember that the change of A is divided by the change in B with respect to one another.
#$&*
You are asked in this exercise to apply the definition, and given a general procedure for doing so. Briefly outline the procedure for applying this definition, and briefly explain how you will remember to apply this procedure.
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):(change in A/ the change b), It easy for me to remember that the change of A is divided by the change in B with respect to one another.
You're in good shape with the 'seed question'.
Did you submit the information below?
1.29
.922
.996
.887
.996
1.24
1.01
1.17
.953
1.15
This data was collected from a ball (golf ball) rolling down a ramp, the first five are from left to right, the second five are from the ball rolling right to left.
** **
686
.859
.688
.687
.734
.766
.672
.875
.641
.824
** **
.605
.641
.687
.703
.714
.589
.636
.500
.390
.531
** **
1.02, .159
1.11, .119
.731, .075
.756, .099
.670, .046
.529, .094
** **
27.5
25.2
38.3
37.0
41.8
52.9
Vave='ds/'dt. I divided the distance of 28 cm of the ball by the mean of the time it for the ball to travel down the ramp in my Ball down a ramp experiment. So for 1 domino vave= 28cm/1.02s = 27.5 for right to left, the same for left to right, and the same calculation for each left to right, right to left times for 2 and 3 dominoes. The vave is recorded in cm per second.
** **
26.9
22.7
52.4
48.9
62.4
100.1
The average rate of acceleration is 'dv/'dt (change in velocity divided by change in time interval)I used the same velocity as calculated for the vave becuase the initial velocity started each time from zero, so the change in velocity would be the same as the vave. So for the first domino I had a velocity of 27.5cm/s / 1.02s (the mean clocktime) which gave me the acceleration 26.9cm/s^2. I did this same calculation for two and three dominoes right to left and then left to right. This is recorded in cm/s^2
** **
26.4cm/s
37.7cm/s
47.3cm/s
The above number are the average of the the velocities obtained by adding the two velocity found for each trial left and right and dividing those by 2. So for the first trial with one dominoe the velocities were found to be (27.5 + 25.2)/2=26.4. I did this for each trial of both two and three dominoes.
** **
1.07s
.744s
.599s
24.7cm/s^2
50.7cm/s^2
78.9cm/s^2
To find the average rate of change with respect to clocktime I averagesd the two time interval for each trial of the One, two, and three domino heights. I divided the average of the velocities of the one, two and three dominoes by the average of the time, so for one domino the average velocity was 26.4cm/s / 1.07 (average of the two times) = 24.7cm/s^2
** **
I would expect that the average rate of change in velocit with respect to clocktime would be the same for each slope.
If I average the two acceleration rate found, ie for one dominoe the acceleration rate were found to be 26.9 and 22.7cm/s^2, the average of those is 24.8, for two the average acceleration is 50.6cm/s^2, and the average for the acceration for the three dominoes is 81.2cm/s^2. So, the average of the acceleration are the same or nearly the same as the average of the velocities divided by the average of the time.
I would expect this because the average velociies are calculated from a distance divided by a clocktime, and each clocktime was obtained by the same incline in height by the same distance of the same ramp, the only difference was one was right to left the other was left to right. When you obtain the average of two number you are finding the midpoint of those numbers, therefore, the number will lay within the same interval whether it is clocktime or velocity. This is why I would expect both methods to be the same.
** **
.03, 24.7
.06, 50.7
.09, 78.9
These are the slope,average rate of change with respect to clocktime, for 1, 2, and 3 dominoes. The avergae rate of change with respect to clocktime is reported in cm/s^2, and the ramp slope are unitless.
I obtained these results from the calculation of average rate of change with respect to clocktime, and the slopes of the ramps with 1, 2 and 3 dominoes.
** **
.4
35
The units of the horizontal intercept are not specified.
The unitls of the vertical intercept are in cm.s^2
The graph of acceleration vs ramp slope will have the acceleration on the vertical axis and the ramp slope on the horizontal axis. As ramp slope increases the acceleration increases, so the graph is increasing at an increasing rate. The horizontal intercept appears to be at .4, which is between the first and second points, The vertical intercept appears to be at about 35cm/s^2 vertically, between the first and second points.
** **
.10, 85
From the graph I created, the point that would correspond to a ramp slope of .10 would be vertically asbout 85cm/s^2. I think this is close to accurate since when the ramp slope increases so does the velocity. In our original experiment the ramp slope increased by apprximately .3, ie they went from .3 to .6 to .9, nd as the incllined tripled in height the velocities seemed to rise by double the first and then triple the first. If we plot a point from .9 to .10 that is only a one point increase and if at .9 the acceleration is almost 80cm/s^2 it would make sense that the increase for the slope at .10 in between 80cm/s^2 and 90cm/s^2.
** **
.6
50
83.3
The first number is the number between the horizontal intercept of the best fit line ony my graph and the point on that line that corresponds to ramp slope .10 horizontally. The second number is the point on the graph that corresponds to the number that lies between the vertical intercept on my graph and .10 vertically. I calculated the rise/run by 50/.6 to be 83.3, which is close to the acceleration that I estimated to be at .10. This tells us the the slope is the same as the accelertaion rate for the graph.
** **
1.02, 1.11, 1.06
.159, .119, .139
The number on the first line are the mean of the time interval from right to left, left to right and the average. On the second line are the standard deviation from right to left, left to right and their average. The signifigance of these numbers are that for i dominoe the average time was about 1.1 seconds, and the average standard deviation was about .14, this could be compared to the uncertainty, of about .1, which is the same as the average of the standard deviation.
** **
.921, 1.19
So the bounderies of this interval are bounded on the left at .921 and on the right at 1.19. I obtained these number by 1.06 - .139 =.921
and 1.06 + .139 = 1.19.
** **
30.4, 33.0
23.5, 19.8
19.8, 33.0
These number represent the average velocity and the acceleration for the left boundary on the first line and on the second the average velocity and accleration on the right hand boundary. The third line are the minimum and maximum possible values of acceleration. The acceleration would be between these two results because the change in velocity will not exceed or fall below this interval for the given distance of 28 cm and time interval obtained, since acceleration is the change in velocity with respect to clocktime.
** **
.744, .087
.657, .831
42.6, 33.6
40.4, 64.8
The first line is the average of the means from left to right and right to left and the average of the stnard deviation from left to right and right to left. The second line is the boundary of the interval on the left hand side and the boundary of the interval on the right hand side. The third line are the maximum and minimum velocitie obtained by the division of the distance of 28cm by the maximum or minimum time intervals. And the fourth line is the minimum ans maximum acceleration for the 2 domino data.
** **
.599, .07
.529, .669
52.9, 41.9
62.6, 100
The first line is the average of the means for the 3 dominoe data and the average of the standard deviations. The second line is the minimum and maximum boundaries for the timed interval from left to right. The third line is the minimum and maximum velocities, and the fourth line is the minimum and maximum accelerations.
** **
.03, 19.8
.06, 40.4
.09, 62.6
.03, 33.0
.06, 64.8
.09, 100
** **
Yes, my graph fits this description. Yes my best fit line passes through the three short vertical segments. The best fit straight line runs through the points of the acceleration vs ramp slope, which indicates that the ramp slope and the acceleration are closely related.
** **
93.8
75,.8
The rise of my line is between 85 and 10 vertically and .10 and .2 horizontally, so 75/.8= 93.8 is my slope
** **
87.5
70,.8
The least steep passes through vertically between 80 and 10 at 70, and horizontally between .10 and .2 at .8 70/.8= 87.5 slope.
** **
3 hours
** **
&#Your work looks good. See my notes. Let me know if you have any questions. &#