course Phy 201
Class 090902&$&$ means 'insert your answer starting on the next line
Your instructor has been having trouble locating your answers in the work you have been submitting; actually not so much having trouble as having to spend a lot of time hunting for them. This is not your fault. Your instructor should have picked up on this much more quickly.
The symbol string '&$&$' indicates that an answer is expected. Your answer should follow the string, and should begin on a new line. This should ensure that your answers are easily located, and that the instructor doesn't miss anything important.
For the same reason, if you insert a question or an answer at any point not marked with &$&$, then if it's an answer please use the string &&&& to 'flag' it for the instructor, and if it's a question use ????.
Actually you can use just about anything that will catch the instructor's eye, except maybe that picture of my dear old Aunt Martha.
Thanks.
Delta notation for 'change in'
`d is the symbol for the Greek capital letter Delta, the triangle thing you saw me write on the board. It has other meanings, but in any context involving rate of change, Delta pretty much universally stands for 'change in'.
• If we were to write `dA, it would be read 'delta A' of 'change in A'.
• If we were to write `dA / `dB, it would read 'change in A divided by change in B'. Thus the average rate of change of A with respect to B is
ave rate = `dA / `dB.
The position vs. clock time graph
We extensively discussed the position vs. clock time graph in class.
In class I saw a number of carefully constructed graphs, many with sufficient detail to give good results. Some of the graphs did not have particular smooth curves, and/or were too small to yield good results.
If necessary you should sketch another graph, of sufficient size and accuracy to give you reasonably good results. Refer to the instructions given for the preceding class.
Using your graph estimate the following. You may use cycles of your pendulum, or half-cycles, as your time unit, or you can convert these to seconds.
• What was the length of your pendulum (you can give this in centimeters, inches, miles, textbook widths, lines on your notebook paper or whatever units are convenient, as long as these units can later be measured in centimeters)? &$&$
20 cm
• What is the change in position corresponding to the first half of the time interval corresponding to motion down the incline (we will use `dt_total to refer to this time interval)? &$&$
`ds= 1 in
• Answer the same for the second half of the interval. &$&$
`ds= 2in
• Darken the part of the graph which corresponds to motion down the fourth ramp. For this interval estimate the change in position and the change in clock time. &$&$
`ds= 4in, `dt= 3 half-cycles
• Mark the point of the graph that corresponds to the ball's first contact with the seventh ramp. Give the coordinates of that point. &$&$
`ds= 10 in, `dt= 6 half-cycles, (10, 6)
• Do the same for the ball's last contact with the seventh ramp. &$&$
• What is q_rise between these points (recall that q_rise stands for 'the quantity represented by the rise')? &$&$
• What is q_run between these points? &$&$
• What therefore is q_slope between these points?
• Mark on your graph the points corresponding to the transitions from one ramp to the next (i.e., the ball leaves one ramp and first encounters the other at the same instant; mark each on the graph at which this occurs).
• Sketch a series of short straight line segments connecting these points.
• Find q_rise, q_run and q_slope for each of these line segments. Report q_rise, q_run and q_slope, in that order and separated by commas, starting in the line below. Report three numbers in each line, so that each line represents the quantities represented by the rise, run and slope of one of your segments.&$&$
The v0, vf, `dt trapezoid
The altitudes of a certain graph trapezoid are symbolically represented by v0 and vf, indicating initial and final velocity.
The base is represented by `dt, the change in clock time t. The base therefore represents the time interval `dt.
Sketch a graph trapezoid. Label its altitudes v0 and vf and its base `dt.
Now answer the following questions:
If v0 = 5 meters / second and vf = 13 meters / second, with `dt = 4 seconds, then
• What is the rise of the trapezoid and what does it represent? &$&$
v0= 5 m/s and vf= 13 m/s change of the altitudes
• What is the run of the trapezoid and what does it represent? &$&$
the base of the trapezoid which is `dt = 4s
• What is the slope of the trapezoid and what does it represent? &$&$
The slope is 2 m/s^2, which represents the increase acceleration between the two points
• What are the dimensions of the equal-area rectangle and what do they represent? &$&$
6.5 x 4; the average velocity times time
• What therefore is the area of the trapezoid and what does it represent? &$&$
The area equals 26 m
The altitude will have the same units as the 'graph altitudes' of the trapezoid. The 'graph altitudes' are expressed in meters/second, so the altitude of the equal-area rectangle will be meters/second.
The altitude of the equal-area rectangle would be halfway between the altitudes 5 meters/second and 13 meters/second. Halfway between 5 and 13 is 9 (most easily calculated by adding 5 and 13 and dividing by 2). The altitude of the equal-area rectangle is therefore 9 meters/second.
The width is not 4 but 4 seconds.
So the area is 9 meters/second * 4 seconds = 36 meters.
In terms of just the symbols v0, vf and `dt:
• What expression represents the rise? &$&$
The change in velocity (vf – v0)
• What expression represents the run? &$&$
`dt
• What expression therefore represents the slope? &$&$
The change of velocity over the change in time; (vf - v0) / `dt
• What expression represents the width of the equal-area rectangle? &$&$
The base of the rectangle, which is `dt
• What expression represents the altitude of the equal-area rectangle? &$&$
The average velocity; (vf + v0) / 2
• What expression therefore represents the area of the trapezoid? &$&$
(vf + v0 ) / 2 * `dt
• What is the meaning of the slope? &$&$
The slope represents the acceleration
• What is the meaning of the area? &$&$
Represents the distance at which the object traveled
If the ball on the ramp changes its velocity from v0 to vf during time interval `dt, then
• If you have numbers for v0, vf and `dt how would you use them to find the following:
• the change in velocity on this interval &$&$
(vf – v0)
• the change in clock time on this interval &$&$
(tf - t0) = `dt
• the average velocity on this interval, assuming a straight-line v vs. t graph &$&$
(vf + v0) / 2
• the average acceleration on this interval &$&$
(vf – v0) / `dt
• the change in position on this interval &$&$
(vf + v0) / 2 * `dt
• In terms of the symbols for v0, vf and `dt, what are the symbolic expressions for each of the following:
• the change in velocity on this interval &$&$
(vf – v0)
• the change in clock time on this interval &$&$
(tf - t0) = `dt
• the average velocity on this interval, assuming a straight-line v vs. t graph &$&$
(vf – v0) / `dt
• the average acceleration on this interval &$&$
(vf – v0) / `dt
• the change in position on this interval &$&$
(vf + v0) / 2 * `dt
• How are your answers to the above questions related to the v0, vf, `dt trapezoid? &$&$
Not much difference, they are the same
If v0 = 50 cm / sec and vf = 20 cm / sec, and the area of the trapezoid is 140 cm, then
• What is the rise of the trapezoid and what does it represent? &$&$
(vf – v0) = (20 – 50 cm / sec) = -30 cm/sec
• What is the altitude of the equal-area rectangle? &$&$
Ave vel. = (vf + v0) / 2 = (80 cm/s) / 2 = 40 cm/s
• Can you use one of your answers, with the given area, to determine the base of the trapezoid? &$&$
Yes, it is the area divided by the average velocity that gives us the change in time; also known as the base.
140 cm / (40 cm/s) = `dt = 3.5 sec
• Can you now find the slope of the trapezoid? &$&$
Yes, (vf – v0) / `dt = a = (20 -50 cm/sec) / 3.5sec = -8.57 cm/sec^2
Introductory Problem Sets
Work through Introductory Problem Set 1 (http://vhmthphy.vhcc.edu/ph1introsets/default.htm > Set 1). You should find these problems to be pretty easy, but be sure you understand everything in the given solutions.
You should also preview Introductory Problem Set 2 (http://vhmthphy.vhcc.edu/ph1introsets/default.htm > Set 2). These problems are a bit more challenging, and at this point you might or might not understand everything you see. If you don't understand everything, you should submit at least one question related to something you're not sure you understand.
Lego toy car:
As shown in class on 090831, a toy car which moves through displacement 30 cm in 1.2 seconds, ending up at rest at the end of this time interval, has an average rate of change of position with respect to clock time of 25 cm / s, and by the definition of average velocity, this is its average velocity. If its v vs. t graph is a straight line, we conclude that its velocity changes from 50 cm/s to 0 cm/s during the 1.2 seconds, and the average rate of change of its velocity with respect to clock time is therefore about -41.7 cm/s.
The same toy car, given an initial push in the opposite direction, moves through displacement -60 cm in 1.5 seconds as it comes to rest. If you previously submitted the correct solution to this situation you found that the acceleration of this car was + 53.3 cm/s^2, approx.. If you didn't get this result, then you should answer the following questions (if you got -53.3 cm/s^2 and know what you did wrong to get the negative sign, you can just explain that): &$&$
I know what I did wrong, it was because I did not choose a positive direction or I either I just didn’t change the sign indicating the positive and negative direction.
• Using the definitions of average velocity and average rate of change, determine the average velocity of the car during this interval. Explain completely how you got your results. &$&$
Positive direction: average velocity = (sf – s0)/`dt = (0 – 30)/1.2 = -25 cm/s
Negative direction: average velocity = (sf – s0)/`dt = (0 - -60)/1.5 = +40 cm/s
• Describe your graph of velocity vs. clock time for this interval, give the altitudes of the corresponding v vs. t trapezoid and verify that the average altitude of this trapezoid is equal to the average velocity you obtained in the preceding step. &$&$
Positive direction altitudes: v0 = 50 cm/s, vf = 0 cm/s
Negative direction altitudes: v0= -80 cm/s, vf = 0cm/s
• What is the car's initial velocity, its final velocity, and the change in its velocity on this interval? &$&$
Positive directio: v0 = 50 cm/s, vf = 0 cm/s
Negative direction: v0= -80 cm/s, vf = 0cm/s
• What therefore is its acceleration on this interval? &$&$
Positive direction acceleration: (vf - v0) / `dt = (0 - 50 cm/s) /1.2s ~ -41.67 cm/s^2 = a
Negative direction acceleration: (vf - v0)/ `dt = (0 – (-80 cm/s)) / 1.5s ~ +53.3 cm/s^2 = a
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&#Your work looks good. See my notes. Let me know if you have any questions. &#