ph1_query_1

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course Phy 121

Thu 04 Nov 2010 09:50:44 PM EDT

Question: `qExplain in your own words how the standard deviation of a set of numbers is calculated.

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Your solution:

1. Average the set of numbers.

2. The deviation is the distance of an individual number from the mean of its set (n is a member of N, distance between average of N and an n).

3. The standard deviation is when you take the square root of the average (using length of N - 1 when the length of N is less than 30 -- not sure about why yet) of all the deviations of each member along the set of numbers.

I think.

confidence rating #$&*: 2

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

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Your solution:

Velocity is a rate of speed, which is an average distance compared to an average time. A rate is a ratio between two measurements.

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 change in position / change in clock time.

The symbol d doesn't look like a change in anything, nor does the symbol t.

And 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. Very 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):

Well if I need a delta, then: Δd / Δt. I think the purpose of assigning variables is to depict something meaningful to the author, and eventually so it can be translated to others. If the context of an assignment calls for an initial velocity, couldn't I use vi = di / ti (they would all be sub-text if possible). I guess too many years of programming has dulled my sensitivity to mathematics. I suppose in mathematics, we try to say something very absolutely in definition. Is that right? That would be contrary to my existing system.

Quinitial velocityestion: Given average speed and time interval how do you find distance moved?

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Your solution:

Average speed * Time interval = Distance interval

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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Self-critique (if necessary): OK

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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:

Distance interval / Average speed = Time interval

confidence rating #$&*:

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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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Self-critique (if necessary): OK

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Question: Given time interval and distance moved how do you get average speed?

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Your solution:

Average velocity = Distance interval / Time interval

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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Self-critique (if necessary): OK

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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:

1. initial velocity (smallest), v_0

2. average velocity, v_Ave

3. final velocity (largest), v_f

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:

The average velocity throughout the interval is: (10 m/s + 4 m/s) / 2 = 7 m/s

The chance in velocity is 10 m/s - 4 m/s = 6 m/s.

initial velocity = 4 m/s

change in velocity = 6 m/s

average velocity = 7 m/s

final velocity = 10 m/s

confidence rating #$&*: 3

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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 is 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:

What is the uncertainty in the change in position in meters?

As low as (1-0.04)*5.2=4.992 meters and as high as (1+0.04)*5.2=5.408 meters.

What is the uncertainty in the time interval in seconds?

As low as (1-0.02)*1.3=1.274 seconds and as high as (1+0.02)*1.3=1.326 seconds.

What is the average velocity of the object, and what do you think is the uncertainty in the average velocity?

v_Ave = Δd / Δt

v_Ave = 5.2 / 1.3

v_Ave = 4

Uncertainty... just guessing here: average(0.04, 0.02) = 0.03. Could just as certainly to me be sum(0.04, 0.02) = 0.06.

Divide the lowest possible velocity by the highest possible time interval and see by what percent that differs from 4 m/s.

That will confirm your alternative result .06.

confidence rating #$&*: 2

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&#Your work looks good. See my notes. Let me know if you have any questions. &#

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