Assignment 1 query

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course PHY 201

2/15 about 3:00 pm

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:

Find the average of the set of numbers. Subtract the average from each of the set of numbers. Square each of the numbers and add them together. Divide this sum of squared numbers by the number of original numbers, minus 1. So if you had four numbers to start with subtract one to get three. After dividing, take the square root of the result and this is the standard deviation.

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:

average rate of change of A with respect to B = (‘delta A) / (‘delta B)

Velocity is distance over time. To find the average velocity, you look at the distance traveled during a time interval. So you have a change in distance, for example from 0 cm at start to 50 cm at the end of the interval. You then can see that it took, for example 10 seconds to go from the beginning at 0 seconds to the end at 10 seconds. Your change in distance is 50 cm and your change in time is 10 seconds. You then take 50 cm / 10 s and get 5 cm/s for an average velocity. To know what to use for A and B in this example, you have to know the units associated with the rate of change you are looking for. I knew that velocity is a distance over a time so I knew that distance was A and time was B.

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):I understand in your solution why not to use v=d/t.

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Self-critique rating:3

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

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Your solution:

V_ave. = ‘ds/’dt -We want to find distance moved so we solve this for ‘ds

‘ds= v_ave * ‘dt

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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Self-critique rating: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:

V_ave. = ‘ds/’dt -We want to find time so we solve this for ‘dt

‘dt = ‘ds/v_ave

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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Self-critique (if necessary):OK

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Self-critique rating:OK

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

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Your solution:

V_ave=’ds/’dt -- take the change in the distance and divide by the change in 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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Self-critique (if necessary):OK

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Self-critique rating: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:

v_0

v_ave

v_f

I am really unsure of the second part to this question. The only way I can see it possible for ‘dv to be greater than v_0, v_ave, and v_f would be if v_0 was negative. ‘dv could definitely be less than all three if say ‘dv = 1, v_0 was 4, v_ave. is 4.5 and v_f is 5.0.

confidence rating #$&*: 2

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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 I guess would be (4+10)/2 = 7 m/s. ‘dv = 6 m/s

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Terminology note: this is the average of the initial and final velocities, which might or might not be the average velocity.

If the velocity changes at a uniform rate, then the average of initial and final velocities will be the average velocity. However in many situations the velocity does not change at a uniform rate.

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v_0, ‘dv, v_ave, v_f

v_0 = 4; v_f = 5; v_ave. is therefore 4.5 and ‘dv = 1.0 so the order would be ‘dv, v_0, v_ave, v_f for this example.

I do not think it would be possible for ‘dv to exceed the other three quantities unless v_0 is negative.

confidence rating #$&*:2

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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:

According to the formula, you would multiply the value (A) by the decimal value of the percent to get the variance.

+/- 0.208 = +/- 0.21

+/- 0.026

I think you would take the extremes (upper and lower) using the variances to get the velocities variance. Using significant figures I got 5.4 and 5.0 for the position changes and 1.3 and 1.3 for the time changes. This gives me 4.2 m/s and 3.9 m/s. The average velocity would be 4.05 m/s or 4.1 m/s. The uncertainty, I believe, would be 0.15/4.05 * 100 = +/- 3.7 percent.

confidence rating #$&*: 2

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If the position change is 4% higher than the calculated value, and the time interval is 2% less, then you can verify that the resulting average velocity will be about 6% greater than the calculated value.

This, and a similar argument for when position change is 4% less and time interval 2% greater than calculated values, indicate that the uncertainty would be +- 6%.

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