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course phy 201
ph1 query 1
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Question: `qExplain in your own words how the standard deviation of a set of numbers is calculated.
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Your solution:
First you get the mean of a set of numbers
second you get the deviations from the mean
third you square those and divide by less than one of the number of them
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You need to add them after squaring them.
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last you take the square root of that number
confidence rating #$&*: 3
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Question: State the given definition of the average rate of change of A with respect to B.
Briefly state what you think velocity is and how you think it is an example of a rate of change.
In reference to the definition of average rate of change, if you were to apply that definition to get an average velocity, what would you use for the A quantity and what would you use for the B quantity?
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Your solution:
something increases by A as B increases
Velocity is the rate of change of speed
A would be a distance feet and B would be time seconds
confidence rating #$&*: 3
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Given Solution:
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. **
NOTE ON NOTATION
Students often quote a formula like v = d / t. It's best to avoid this formula completely.
The average velocity on an interval is defined as the average rate of change of position with respect to clock time. By the definition of average rate, then, the average velocity on the interval is v_ave = (change in position / change in clock time).
One reason we might not want to use v = d / t: The symbol d doesn't look like a change in anything, nor does the symbol t. Also it's very to read 'd' and 'distance' rather than 'displacement'.
Another reason: The symbol v doesn't distinguish between initial velocity, final velocity, average velocity, change in velocity and instantaneous velocity, all of which are important concepts that need to be associated with distinct symbols.
In this course we use `d to stand for the capital Greek symbol Delta, which universally indicates the change in a quantity. If we use d for distance, then the 'change in distance' would be denoted `dd. It's potentially confusing to have two different d's, with two different meanings, in the same expression.
We generally use s or x to stand for position, so `ds or `dx would stand for change in position. Change in clock time would be `dt. Thus
v_Ave = `ds / `dt
(or alternatively, if we use x for position, v_Ave = `dx / `dt).
With this notation we can tell that we are dividing change in position by change in clock time.
For University Physics students (calculus-based note):
If x is the position then velocity is dx/dt, the derivative of position with respect to clock time. This is the limiting value of the rate of change of position with respect to clock time. You need to think in these terms.
v stands for instantaneous velocity. v_Ave stands for the average velocity on an interval.
If you used d for position then you would have the formula v = dd / dt. The dd in the numerator doesn't make a lot of sense; one d indicates the infinitesimal change in the other d.
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Self-critique (if necessary):
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Question: Given average speed and time interval how do you find distance moved?
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Your solution:
Multiply average speed by time
confidence rating #$&*: 3
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Given Solution:
** You multiply average speed * time interval to find distance moved.
For example, 50 miles / hour * 3 hours = 150 miles. **
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Question: Given average speed and distance moved how do you find the corresponding time interval?
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Your solution:
Divide distance by speed
confidence rating #$&*: 3
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Given Solution:
** 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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Question: Given time interval and distance moved how do you get average speed?
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Your solution:
Divide distance by time
confidence rating #$&*: 3
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Given Solution:
** 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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Question: A ball rolls from rest down a book, off that book and onto another book, where it picks up speed before rolling off the end of that book. Consider the interval that begins when the ball first encounters the second book, and ends when it rolls of the end of the book.
For this interval, place in order the quantities initial velocity (which we denote v_0), and final velocity (which we denote v_f), average velocity (which we denote v_Ave).
During this interval, the ball's velocity changes. It is possible for the change in its velocity to exceed the three quantities you just listed? Is it possible for all three of these quantities to exceed the change in the ball's velocity? Explain.
Note that the change in the ball's velocity is denoted `dv.
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Your solution:
Smallest v_0 Middle 2nd v_Ave Largest 3rd v_f
It could exceed The initial
No
Because It will always be greater than the initial since it is an increase of the initial and the change will always be less than the final velocity
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It's not clear what you mean by 'it', but I expect you mean the change in velocity.
However the change in velocity is not necessarily greater than the initial velocity.
The change in velocity is not an increase of the initial velocity (the initial velocity doesn't increase), but it could be described as an increase over the initial velocity.
I understand how you intend to present your argument, and your idea is fine.
Does the following follow your approach? Is this argument airtight?
The initial velocity is positive.
The change in velocity is the final velocity minus the initial velocity.
If the final velocity is greater than the initial then the change in velocity is positive and is less than the final velocity.
If the final velocity is less than the initial then the change in velocity is negative and hence less than the initial velocity.
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confidence rating #$&*: 3
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Question: If the velocity at the beginning of an interval is 4 m/s and at the end of the interval it is 10 m/s, then what is the average of these velocities, and what is the change in velocity?
List the four quantities initial velocity, final velocity, average of initial and final velocities, and change in velocity, in order from least to greatest.
Give an example of positive initial and final velocities for which the order of the four quantities would be different.
For positive initial and final velocities, is it possible for the change in velocity to exceed the other three quanities?
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Your solution:
Average 3m/s , Initial 4m/s , Change 6 , Final 10m/s
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The velocity is never less than 4 m/s, so the average velocity could not be 3 m/s.
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It can exeed the average and the initial
confidence rating #$&*:
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Question: If the position of an object changes by 5.2 meters, with an uncertainty of +-4%, during a time interval of 1.3 seconds, with an uncertainty of +-2%, then
What is the uncertainty in the change in position in meters>
What is the uncertainty in the time interval in seconds?
What is the average velocity of the object, and what do you think ia the uncertainty in the average velocity?
(this last question is required of University Physics students only, but other are welcome to answer): What is the percent uncertainty in the average velocity of the object, and what is the uncertainty as given in units of velocity?
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Your solution:
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Solution?
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confidence rating #$&*:
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Question: A ball rolls from rest down a book, off that book and onto another book, where it picks up speed before rolling off the end of that book. Consider the interval that begins when the ball first encounters the second book, and ends when it rolls of the end of the book.
For this interval, place in order the quantities initial velocity (which we denote v_0), and final velocity (which we denote v_f), average velocity (which we denote v_Ave).
During this interval, the ball's velocity changes. It is possible for the change in its velocity to exceed the three quantities you just listed? Is it possible for all three of these quantities to exceed the change in the ball's velocity? Explain.
Note that the change in the ball's velocity is denoted `dv.
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Your solution:
Smallest v_0 Middle 2nd v_Ave Largest 3rd v_f
It could exceed The initial
No
Because It will always be greater than the initial since it is an increase of the initial and the change will always be less than the final velocity
@&
It's not clear what you mean by 'it', but I expect you mean the change in velocity.
However the change in velocity is not necessarily greater than the initial velocity.
The change in velocity is not an increase of the initial velocity (the initial velocity doesn't increase), but it could be described as an increase over the initial velocity.
I understand how you intend to present your argument, and your idea is fine.
Does the following follow your approach? Is this argument airtight?
The initial velocity is positive.
The change in velocity is the final velocity minus the initial velocity.
If the final velocity is greater than the initial then the change in velocity is positive and is less than the final velocity.
If the final velocity is less than the initial then the change in velocity is negative and hence less than the initial velocity.
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confidence rating #$&*: 3
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Question: If the velocity at the beginning of an interval is 4 m/s and at the end of the interval it is 10 m/s, then what is the average of these velocities, and what is the change in velocity?
List the four quantities initial velocity, final velocity, average of initial and final velocities, and change in velocity, in order from least to greatest.
Give an example of positive initial and final velocities for which the order of the four quantities would be different.
For positive initial and final velocities, is it possible for the change in velocity to exceed the other three quanities?
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Your solution:
Average 3m/s , Initial 4m/s , Change 6 , Final 10m/s
@&
The velocity is never less than 4 m/s, so the average velocity could not be 3 m/s.
*@
It can exeed the average and the initial
confidence rating #$&*:
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Question: If the position of an object changes by 5.2 meters, with an uncertainty of +-4%, during a time interval of 1.3 seconds, with an uncertainty of +-2%, then
What is the uncertainty in the change in position in meters>
What is the uncertainty in the time interval in seconds?
What is the average velocity of the object, and what do you think ia the uncertainty in the average velocity?
(this last question is required of University Physics students only, but other are welcome to answer): What is the percent uncertainty in the average velocity of the object, and what is the uncertainty as given in units of velocity?
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Your solution:
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Solution?
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confidence rating #$&*:
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