#$&*
course Phy 231
Give your counts for the four observations made today, in the order you made them:My counts were: For the first slanted direction 13, 12; For the second slanted direction 15, 14.
#$&*
According to your counts, which way did the table slope, to the right or left as it appeared on your screen?
According to my counts, the table sloped to the left because it required less time for the marble to travel in the first slanted direction (for both elevations), than in the second.
#$&*
You will need to know this definition, word for word and symbol for symbol, starting now and for the rest of the course. The definition is about 19 words and a few symbols long and most of the words are single syllables:
Definition of average rate of change: The average rate of change of A with respect to B is (change in A) / (change in B).
You should already recognize this definition as perhaps the most fundamental definition in calculus, though it could be asserted that the most fundamental definition also applies a limiting process to this definition.
You also need the following two definitions:
Average velocity is the average rate of change of position with respect to clock time.
Average acceleration is the average rate of change of velocity with respect to clock time.
According to the definition of average rate of change, then, what is the calculation for velocity?
The calculation for average velocity is change in position with respect to the change in time. This definition is also listed above…..
#$&*
Explain how this calculation is consistent with your experience.
This calculation is consistent with my experience of driving. When sign reads 'speed limit 55 mph,' I realize it means that you are not allowed to travel any more than 55 miles in one hour. 55 miles would be your change in position and 1 hour would be your change in time.
#$&*
Explain how this calculation is consistent with formulas you've learned.
One of the formulas we have learned so far is the formula for calculating the velocity from a s-t graph is derivative of position with respect to time. This is the same as calculating the slope of the line on the v-t graph. The equation for calculating the slope is rise over run (i.e. rise = change in the y-values = change in velocity, and run = change in x-values = change in time).
#$&*
Specifically apply this definition to find the average velocity of the ball in each trial, assuming it traveled 60 cm during each interval of observation.
From the equation for average velocity, the velocities for the first slanted direction are equal to: 4.62 cm/unit of time, 5 cm/unit of time. The velocities for the second slanted direction are equal to: 4 cm/unit of time, 4.29 cm/unit of time.
#$&*
Now this is where things start to get a little tricky. You should answer the following with the best of your common sense, thinking about what the questions mean rather than looking up formulas and explanations. The answers should come from you, not from some other source. And you should do your best to answer the questions without talking to your classmates, though once you have done your own thinking it would be great for you to discuss it with whomever you can.
You know the ball started from rest in each trial, as some of you stated in class today. You've just calculated the average velocities for the four trials.
Knowing that the ball starts from rest and knowing its average velocity, using only common sense and not some formula that might give you the right answer without requiring you to understand anything, explain the most reasonable approach you can think of to finding the final velocity.
To find the final velocity, the most reasonable approach would be to plot a graph of all the data points gathered from a trial and use the data points to calculate an equation for the line connecting those points. From that line you could take the derivative and plug in the final value for time to get an exact answer for the final velocity.
#$&*
Assuming you do know the final velocity and the count, how would you apply the definition of average rate of change and the definition of average acceleration to determine the acceleration of the ball?
To calculate the average acceleration you would take the average rate of velocity with respect to the change in time. This would be like taking, for example, m/s with respect to seconds again - yielding m/(s^2).
#$&*
Using your best estimate of the ball's final velocity for each of the four trials, what is the average acceleration for each? Show in detail how you get the average acceleration for the first trial, then just include the brief details of your calculation for each of the other three trials.
Since we only have two data points for each of the four trials (i.e. (0,0), (unit of time, 60 cm)), the final velocity has to be equal to the average velocity…..since we do not have anything else to show that the velocity either increased or decreased - to give a separate velocity for the final. Therefore the average acceleration for each of the four trials would be the average velocity with respect to the change in time. For the first trial, the average velocity was 4.62 cm/unit of time. To get the average acceleration, we would divide the average velocity by the amount of time it took to achieve that acceleration. For the first trial, the average acceleration would be = .355 cm/(unit of time^2) = 4.62 / 13. The calculations for the following trials would follow the same pattern. The resulting average accelerations are as follows: For the first slant 0.355 cm/(unit of time^2) & 0.417 cm/(unit of time^2), and For the second slant 0.267 cm/(unit of time^2) & 0.306 cm/(unit of time^2).
#$&*
By what percent do you estimate the average frequency of your counts might have varied between trials? Express your answer as the difference between the lowest and highest frequency, as a percent of the average of all the frequencies. Don't go looking up a technical definition of the word ""frequency"", which would probably confuse the whole issue. You have enough intuition about the meaning of that word to come up with a reasonable, if not profoundly accurate, estimate. You also shouldn't have to look up what we mean by the difference between the frequencies as a percent of the average frequency, but that terminology is well-defined, completely applicable and should not be confusing so if you've got to look it up it's OK.
I would estimate that my frequencies weren't off any more than 1, either direction. The estimation of the average frequency of my counts would be the difference in the lowest and highest frequencies with respect to the average of all the frequencies. An estimation of being off no more than +/-1 frequency gives a difference of 2 (for the difference between the lowest and highest frequency) and the average frequency was 13.5 = ((12+13+14+15)/4). Therefore, the resulting percent is 14.81% = 2 / ((12+13+14+15)/4) = 2 / 13.5.
#$&*
If the frequency for a trial was off by 2%, by what percent would the resulting calculation of velocity be off?
If the frequency for a trial was off by 2%, the resulting calculation for velocity would be off 2% since the only calculation wouldn't involve the variable in question more than once. That is, the only calculation would be taking the distance covered by the time…….time is only involved once and would alter the data 2%.
#$&*
If the frequency for a trial was off by 2%, by what percent would the resulting calculation of acceleration be off?
If the frequency for a trial was off by 2%, the resulting calculation for acceleration would be off by 4%. I arrived at this decision because the calculation for acceleration involves the variable time twice…..therefore it would be 2 * 2% = 4%. For example, the calculation for acceleration is: velocity (distance/time) / time…..involving the questionable variable twice! Twice 2% is 4%.
#$&*
"
`gr41
#$&*
course Phy 231
Give your counts for the four observations made today, in the order you made them:My counts were: For the first slanted direction 13, 12; For the second slanted direction 15, 14.
#$&*
According to your counts, which way did the table slope, to the right or left as it appeared on your screen?
According to my counts, the table sloped to the left because it required less time for the marble to travel in the first slanted direction (for both elevations), than in the second.
#$&*
You will need to know this definition, word for word and symbol for symbol, starting now and for the rest of the course. The definition is about 19 words and a few symbols long and most of the words are single syllables:
Definition of average rate of change: The average rate of change of A with respect to B is (change in A) / (change in B).
You should already recognize this definition as perhaps the most fundamental definition in calculus, though it could be asserted that the most fundamental definition also applies a limiting process to this definition.
You also need the following two definitions:
Average velocity is the average rate of change of position with respect to clock time.
Average acceleration is the average rate of change of velocity with respect to clock time.
According to the definition of average rate of change, then, what is the calculation for velocity?
The calculation for average velocity is change in position with respect to the change in time. This definition is also listed above…..
#$&*
Explain how this calculation is consistent with your experience.
This calculation is consistent with my experience of driving. When sign reads 'speed limit 55 mph,' I realize it means that you are not allowed to travel any more than 55 miles in one hour. 55 miles would be your change in position and 1 hour would be your change in time.
#$&*
Explain how this calculation is consistent with formulas you've learned.
One of the formulas we have learned so far is the formula for calculating the velocity from a s-t graph is derivative of position with respect to time. This is the same as calculating the slope of the line on the v-t graph. The equation for calculating the slope is rise over run (i.e. rise = change in the y-values = change in velocity, and run = change in x-values = change in time).
#$&*
Specifically apply this definition to find the average velocity of the ball in each trial, assuming it traveled 60 cm during each interval of observation.
From the equation for average velocity, the velocities for the first slanted direction are equal to: 4.62 cm/unit of time, 5 cm/unit of time. The velocities for the second slanted direction are equal to: 4 cm/unit of time, 4.29 cm/unit of time.
#$&*
Now this is where things start to get a little tricky. You should answer the following with the best of your common sense, thinking about what the questions mean rather than looking up formulas and explanations. The answers should come from you, not from some other source. And you should do your best to answer the questions without talking to your classmates, though once you have done your own thinking it would be great for you to discuss it with whomever you can.
You know the ball started from rest in each trial, as some of you stated in class today. You've just calculated the average velocities for the four trials.
Knowing that the ball starts from rest and knowing its average velocity, using only common sense and not some formula that might give you the right answer without requiring you to understand anything, explain the most reasonable approach you can think of to finding the final velocity.
To find the final velocity, the most reasonable approach would be to plot a graph of all the data points gathered from a trial and use the data points to calculate an equation for the line connecting those points. From that line you could take the derivative and plug in the final value for time to get an exact answer for the final velocity.
#$&*
Assuming you do know the final velocity and the count, how would you apply the definition of average rate of change and the definition of average acceleration to determine the acceleration of the ball?
To calculate the average acceleration you would take the average rate of velocity with respect to the change in time. This would be like taking, for example, m/s with respect to seconds again - yielding m/(s^2).
#$&*
Using your best estimate of the ball's final velocity for each of the four trials, what is the average acceleration for each? Show in detail how you get the average acceleration for the first trial, then just include the brief details of your calculation for each of the other three trials.
Since we only have two data points for each of the four trials (i.e. (0,0), (unit of time, 60 cm)), the final velocity has to be equal to the average velocity…..since we do not have anything else to show that the velocity either increased or decreased - to give a separate velocity for the final. Therefore the average acceleration for each of the four trials would be the average velocity with respect to the change in time. For the first trial, the average velocity was 4.62 cm/unit of time. To get the average acceleration, we would divide the average velocity by the amount of time it took to achieve that acceleration. For the first trial, the average acceleration would be = .355 cm/(unit of time^2) = 4.62 / 13. The calculations for the following trials would follow the same pattern. The resulting average accelerations are as follows: For the first slant 0.355 cm/(unit of time^2) & 0.417 cm/(unit of time^2), and For the second slant 0.267 cm/(unit of time^2) & 0.306 cm/(unit of time^2).
#$&*
By what percent do you estimate the average frequency of your counts might have varied between trials? Express your answer as the difference between the lowest and highest frequency, as a percent of the average of all the frequencies. Don't go looking up a technical definition of the word ""frequency"", which would probably confuse the whole issue. You have enough intuition about the meaning of that word to come up with a reasonable, if not profoundly accurate, estimate. You also shouldn't have to look up what we mean by the difference between the frequencies as a percent of the average frequency, but that terminology is well-defined, completely applicable and should not be confusing so if you've got to look it up it's OK.
I would estimate that my frequencies weren't off any more than 1, either direction. The estimation of the average frequency of my counts would be the difference in the lowest and highest frequencies with respect to the average of all the frequencies. An estimation of being off no more than +/-1 frequency gives a difference of 2 (for the difference between the lowest and highest frequency) and the average frequency was 13.5 = ((12+13+14+15)/4). Therefore, the resulting percent is 14.81% = 2 / ((12+13+14+15)/4) = 2 / 13.5.
#$&*
If the frequency for a trial was off by 2%, by what percent would the resulting calculation of velocity be off?
If the frequency for a trial was off by 2%, the resulting calculation for velocity would be off 2% since the only calculation wouldn't involve the variable in question more than once. That is, the only calculation would be taking the distance covered by the time…….time is only involved once and would alter the data 2%.
#$&*
If the frequency for a trial was off by 2%, by what percent would the resulting calculation of acceleration be off?
If the frequency for a trial was off by 2%, the resulting calculation for acceleration would be off by 4%. I arrived at this decision because the calculation for acceleration involves the variable time twice…..therefore it would be 2 * 2% = 4%. For example, the calculation for acceleration is: velocity (distance/time) / time…..involving the questionable variable twice! Twice 2% is 4%.
#$&*
"
`gr41
#$&*
course Phy 231
Give your counts for the four observations made today, in the order you made them:My counts were: For the first slanted direction 13, 12; For the second slanted direction 15, 14.
#$&*
According to your counts, which way did the table slope, to the right or left as it appeared on your screen?
According to my counts, the table sloped to the left because it required less time for the marble to travel in the first slanted direction (for both elevations), than in the second.
#$&*
You will need to know this definition, word for word and symbol for symbol, starting now and for the rest of the course. The definition is about 19 words and a few symbols long and most of the words are single syllables:
Definition of average rate of change: The average rate of change of A with respect to B is (change in A) / (change in B).
You should already recognize this definition as perhaps the most fundamental definition in calculus, though it could be asserted that the most fundamental definition also applies a limiting process to this definition.
You also need the following two definitions:
Average velocity is the average rate of change of position with respect to clock time.
Average acceleration is the average rate of change of velocity with respect to clock time.
According to the definition of average rate of change, then, what is the calculation for velocity?
The calculation for average velocity is change in position with respect to the change in time. This definition is also listed above…..
#$&*
Explain how this calculation is consistent with your experience.
This calculation is consistent with my experience of driving. When sign reads 'speed limit 55 mph,' I realize it means that you are not allowed to travel any more than 55 miles in one hour. 55 miles would be your change in position and 1 hour would be your change in time.
#$&*
Explain how this calculation is consistent with formulas you've learned.
One of the formulas we have learned so far is the formula for calculating the velocity from a s-t graph is derivative of position with respect to time. This is the same as calculating the slope of the line on the v-t graph. The equation for calculating the slope is rise over run (i.e. rise = change in the y-values = change in velocity, and run = change in x-values = change in time).
#$&*
Specifically apply this definition to find the average velocity of the ball in each trial, assuming it traveled 60 cm during each interval of observation.
From the equation for average velocity, the velocities for the first slanted direction are equal to: 4.62 cm/unit of time, 5 cm/unit of time. The velocities for the second slanted direction are equal to: 4 cm/unit of time, 4.29 cm/unit of time.
#$&*
Now this is where things start to get a little tricky. You should answer the following with the best of your common sense, thinking about what the questions mean rather than looking up formulas and explanations. The answers should come from you, not from some other source. And you should do your best to answer the questions without talking to your classmates, though once you have done your own thinking it would be great for you to discuss it with whomever you can.
You know the ball started from rest in each trial, as some of you stated in class today. You've just calculated the average velocities for the four trials.
Knowing that the ball starts from rest and knowing its average velocity, using only common sense and not some formula that might give you the right answer without requiring you to understand anything, explain the most reasonable approach you can think of to finding the final velocity.
To find the final velocity, the most reasonable approach would be to plot a graph of all the data points gathered from a trial and use the data points to calculate an equation for the line connecting those points. From that line you could take the derivative and plug in the final value for time to get an exact answer for the final velocity.
#$&*
Assuming you do know the final velocity and the count, how would you apply the definition of average rate of change and the definition of average acceleration to determine the acceleration of the ball?
To calculate the average acceleration you would take the average rate of velocity with respect to the change in time. This would be like taking, for example, m/s with respect to seconds again - yielding m/(s^2).
#$&*
Using your best estimate of the ball's final velocity for each of the four trials, what is the average acceleration for each? Show in detail how you get the average acceleration for the first trial, then just include the brief details of your calculation for each of the other three trials.
Since we only have two data points for each of the four trials (i.e. (0,0), (unit of time, 60 cm)), the final velocity has to be equal to the average velocity…..since we do not have anything else to show that the velocity either increased or decreased - to give a separate velocity for the final. Therefore the average acceleration for each of the four trials would be the average velocity with respect to the change in time. For the first trial, the average velocity was 4.62 cm/unit of time. To get the average acceleration, we would divide the average velocity by the amount of time it took to achieve that acceleration. For the first trial, the average acceleration would be = .355 cm/(unit of time^2) = 4.62 / 13. The calculations for the following trials would follow the same pattern. The resulting average accelerations are as follows: For the first slant 0.355 cm/(unit of time^2) & 0.417 cm/(unit of time^2), and For the second slant 0.267 cm/(unit of time^2) & 0.306 cm/(unit of time^2).
#$&*
By what percent do you estimate the average frequency of your counts might have varied between trials? Express your answer as the difference between the lowest and highest frequency, as a percent of the average of all the frequencies. Don't go looking up a technical definition of the word ""frequency"", which would probably confuse the whole issue. You have enough intuition about the meaning of that word to come up with a reasonable, if not profoundly accurate, estimate. You also shouldn't have to look up what we mean by the difference between the frequencies as a percent of the average frequency, but that terminology is well-defined, completely applicable and should not be confusing so if you've got to look it up it's OK.
I would estimate that my frequencies weren't off any more than 1, either direction. The estimation of the average frequency of my counts would be the difference in the lowest and highest frequencies with respect to the average of all the frequencies. An estimation of being off no more than +/-1 frequency gives a difference of 2 (for the difference between the lowest and highest frequency) and the average frequency was 13.5 = ((12+13+14+15)/4). Therefore, the resulting percent is 14.81% = 2 / ((12+13+14+15)/4) = 2 / 13.5.
#$&*
If the frequency for a trial was off by 2%, by what percent would the resulting calculation of velocity be off?
If the frequency for a trial was off by 2%, the resulting calculation for velocity would be off 2% since the only calculation wouldn't involve the variable in question more than once. That is, the only calculation would be taking the distance covered by the time…….time is only involved once and would alter the data 2%.
#$&*
If the frequency for a trial was off by 2%, by what percent would the resulting calculation of acceleration be off?
If the frequency for a trial was off by 2%, the resulting calculation for acceleration would be off by 4%. I arrived at this decision because the calculation for acceleration involves the variable time twice…..therefore it would be 2 * 2% = 4%. For example, the calculation for acceleration is: velocity (distance/time) / time…..involving the questionable variable twice! Twice 2% is 4%.
#$&*
"
&#Your work looks very good. Let me know if you have any questions. &#