Query 1

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course Phy 241

1/16 at 8:45

ph1 query 1delim #$&*

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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, determine the mean time of the numbers by averaging them. Next, take the absolute value of the difference between each number and the mean, this is the deviation for each data point. Next, square each of the deviations and add them. Finally, take the square root of this number; this gives you the standard deviation of the set of numbers.

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:

Rate of change, is the change in something (say distance traveled) divided by some other measurement (say the time taken to travel that distance.

Velocity is a measurement of change in distance over time, for example miles per hour.

The B quantity would be a distance measurement and A would be a quantity of time. So it might be B mi / A hour

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): OK

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Self-critique rating: OK

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

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Your solution: If you have an average speed in mi/hr and a time in hr then you multiply the two (mi/hr) * hr = mi to get the value in miles.

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:

Given an average speed mi/hr and a distance mi, you would divide distance by speed mi / (mi/hr) = hr to get a time in hours.

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:

Given time in hr and distance moved in mi you would divide distance by time mi / hr = mi/hr to get an average rate of speed

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

It is possible for change in velocity to exceed v_0 and v_ave, but not v_f since the change in velocite is always v_f - v_0 and the ball is accelerating.

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With positive velocities this would be the case. With negative initial velocity, though, it would be possible for change in velocity to exceed final velocity.

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If v_0=2 and v_f=10 then v_Ave=6 and `dv=8

All three of these quantities can exceed `dv.

If v_0=2.1 and v_f=2.2 then v_Ave=2.15 and `dv=0.1

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:

If v_0=4m/s and v_f=10m/s then v_Ave=7m/s and `dv=6m/s

v_0, `dv, v_Ave, v_f

If v_0=2m/s and v_f=10m/s then v_Ave=6m/s and `dv=8m/s

v_0, v_Ave, `dv, v_f

The change in velocity can exceed everything except the final velocity.

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 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:

It is +-0.208m or 5.408m to 4.992m.

It is +-0.026s or 1.326s to 1.274s.

v_Ave = 5.2m / 1.3s = 4m/s I believe the uncertainty is the addition of the two % uncertainties or +-6%

Since v_Ave = 4m/s +-6% or +-0.24m/s

confidence rating #$&*: 2

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@&

Right.

Don't spend a lot of time puzzling over this, but here's an enrichment note (connecting error anslysis to the product and quotient rules of calculus) that you might find worth thinking about:

The reason we add the percent uncertainties in this calculation is because (f / g) ' = (f ' g - g ' f) / g^2, from which it follows by simple algebra that

(f / g) ' / (f / g) = f ' / f - g ' / g

so that in differential form

d (f / g) / (f / g) = df / f - dg / g.

Here f is regarded as the observed displacement and df as the uncertainty in that observation. Similarly g is the observed time interval, and dg the uncertainty. f / g would be the average velocity and d(f / g) the uncertainty in that average velocity.

So df / f is the uncertainty in the displacement as a proportion of the observed displacement, dg / g the uncertainty in time interval as proportion of the observed time interval, and d (f / g) / (f / g) the resulting uncertainty in the calculated velocity as a proportion of the actual velocity.

Since df and dg can be of opposite signs it is possible for | df / f - dg / g | to be equal to | df / f | + | dg / g |, so the proportional uncertainty in f / g is regarded as the sum, not the difference, of the proportional uncertainties in f and g.

Note also that if we multiply all the proportional uncertainties by 100 we get the percent uncertainties in these quantities, so the same statement can be made in terms of percent uncertainties.

So when we divide two observed quantities, the percent uncertainty in the result is the sum of the percent uncertainties.

If we multiply two quantities, we come to the same conclusion. This is because (f g) ' = f ' g + g ' f, so that d ( f g ) = df / f + dg / g.

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