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02:32:06 `q001. Note that there are 11 questions in this assignment.
vAve = `ds / `dt, which is the definition of average velocity and which fits well with our intuition about this concept. If displacement `ds is measured in meters and the time interval `dt is measured in seconds, in what units will vAve be obtained?......!!!!!!!!...................................
RESPONSE --> If we measure displacement ('ds) in meters and the time interval ('dt) then we will measure vAve (average velocity) in meters per second.
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02:32:17 vAve = `ds / `dt. The units of `ds are cm and the units of `dt are sec, so the units of `ds / `dt must be cm / sec. Thus vAve is in cm/s.
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RESPONSE --> ok
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02:33:18 `q002. If the definition equation vAve = `ds / `dt is to be solved for `ds we multiply both sides of the equation by `dt to obtain `ds = vAve * `dt. If vAve is measured in cm / sec and `dt in sec, then in what units must `ds be measured?
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RESPONSE --> If we measure vAve in cm/sec and 'dt in seconds, then we would measure 'ds in cm.
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02:33:24 Since vAve is in cm/sec and `dt in sec, `ds = vAve * `dt must be in units of cm / sec * sec = cm.
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RESPONSE --> ok
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02:36:33 `q003. Explain the algebra of multiplying the unit cm / sec by the unit sec.
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RESPONSE --> We would multiply the unit cm/sec by the unit sec in order to find the displacement by cm.
If vAve = 'ds / 'dt then when we multiply both sides of the equation by the 'dt (time) it will give us the 'ds (displaement) on one side of the equation so that we may easily solve for the displacement in cm..................................................
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02:36:49 When we multiply cm/sec by sec we are multiplying the fractions cm / sec and sec / 1. When we multiply fractions we will multiply numerators and denominators. We obtain cm * sec / ( sec * 1). This can be rearranged as (sec / sec) * (cm / 1), which is the same as 1 * cm / 1. Since multiplication or division by 1 doesn't change a quantity, this is just equal to cm.
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RESPONSE --> ok
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02:37:43 `q004. If the definition vAve = `ds / `dt is to be solved for `dt we multiply both sides of the equation by `dt to obtain vAve * `dt = `ds, then divide both sides by vAve to get `dt = `ds / vAve. If vAve is measured in km / sec and `ds in km, then in what units must `dt be measured?
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RESPONSE --> If vAve is measured in km / sec and 'ds is measured in km, then 'dt must be measured in seconds.
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02:37:48 Since `dt = `ds / vAve and `ds is in km and vAve in km/sec, `ds / vAve will be in km / (km / sec) = seconds.
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RESPONSE --> ok
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02:39:01 `q005. Explain the algebra of dividing the unit km / sec into the unit km.
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RESPONSE --> If vAve = 'ds / 'dt then we have km/sec = km / sec
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02:39:08 The division is km / (km / sec). Since division by a fraction is multiplication by the reciprocal of the fraction, we have km * (sec / km). This is equivalent to multiplication of fractions (km / 1) * (sec / km). Multiplying numerators and denominators we get (km * sec) / (1 * km), which can be rearranged to give us (km / km) * (sec / 1), or 1 * sec / 1, or just sec.
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RESPONSE --> ok
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02:43:52 `q006. If an object moves from position s = 4 meters to position s = 10 meters between clock times t = 2 seconds and t = 5 seconds, then at what average rate is the position of the object changing (i.e., what is the average velocity of the object) during this time interval? What is the change `ds in position, what is the change `dt in clock time, and how do we combine these quantities to obtain the average velocity?
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RESPONSE --> If the object moves from s=4 to s=10 between time t=2 and t=5 then the average rate at which the position is changing is
vAve = 6 (change in 'dt) / 3 (change in 'ds) vAve = 2 So the average rate at which the object is changing is 2m/sec. The change in the 'ds position is 6. The change in the 'dt in clock time is 3. We combine these quantities by dividing them in order to obtain the average velocity..................................................
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02:44:22 We see that the changes in position and clock time our `ds = 10 meters - 4 meters = 6 meters and `dt = 5 seconds - 2 seconds = 3 seconds. We see also that the average velocity is vAve = `ds / `dt = 6 meters / (3 seconds) = 2 meters / second. Comment on any discrepancy between this reasoning and your reasoning.
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RESPONSE --> There was no discrepancy in answers.
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02:47:32 `q007. Symbolize this process: If an object moves from position s = s1 to position s = s2 between clock times t = t1 and t = t2, when what expression represents the change `ds in position and what expression represents the change `dt in the clock time?
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RESPONSE --> Change in 'ds would be represented by finding the value for the equation s2 - s1.
Change in 'dt would be represented by finding the value for the equation t2 - t1..................................................
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02:48:48 We see that the change in position is `ds = s2 - s1, obtained as usual by subtracting the first position from the second. Similarly the change in clock time is `dt = t2 - t1. What expression therefore symbolizes the average velocity between the two clock times.
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RESPONSE --> The expression that would symbolize the average velocity between the two clock times would be
(s2 - s1) / (t2 - t1) = vAve.................................................
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02:55:20 `q008. On a graph of position s vs. clock time t we see that the first position s = 4 meters occurs at clock time t = 2 seconds, which corresponds to the point (2 sec, 4 meters) on the graph, while the second position s = 10 meters occurs at clock time t = 5 seconds and therefore corresponds to the point (5 sec, 10 meters). If a right triangle is drawn between these points on the graph, with the sides of the triangle parallel to the s and t axes, the rise of the triangle is the quantity represented by its vertical side and the run is the quantity represented by its horizontal side. This slope of the triangle is defined as the ratio rise / run. What is the rise of the triangle (i.e., the length of the vertical side) and what quantity does the rise represent? What is the run of the triangle and what does it represent?
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RESPONSE --> By graphing the triangle, we can see that the rise of the triangle is represented by the change on the s axis which is 6 units. The rise would represent meters (6 meters to be exact).
The run of the triangle would be represented by the change on the t axis which is 3 units. The units represented would be seconds (3 seconds to be exact)..................................................
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02:55:31 The rise of the triangle represents the change in the position coordinate, which from the first point to the second is 10 m - 4 m = 6 m. The run of the triangle represents the change in the clock time coordinate, which is 5 s - 2 s = 3 s.
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RESPONSE --> ok
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02:56:53 `q009. What is the slope of this triangle and what does it represent?
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RESPONSE --> The rise of the triangle was 6m and the run of the triangle is 3 sec.
We find the slope by using the equation of rise / run. 6m / 3 sec = a slope of 2m/sec.................................................
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02:56:58 The slope of this graph is 6 meters / 3 seconds = 2 meters / second.
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RESPONSE --> ok
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03:00:06 `q010. In what sense does the slope of any graph of position vs. clock time represent the velocity of the object? For example, why does a greater slope imply greater velocity?
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RESPONSE --> Given a point on the graph we can see what the velocity of the object was at that given time. As represented by the points we saw in the previous examples.
The greater the slope, the greater the meters per second covered. As the graph increases, so does the velocity..................................................
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03:00:16 Since the rise between two points on a graph of velocity vs. clock time represents the change in `ds position, and since the run represents the change `dt clock time, the slope represents rise / run, or change in position /change in clock time, or `ds / `dt. This is the definition of average velocity.
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RESPONSE --> ok
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03:04:49 `q011. As a car rolls from rest down a hill, its velocity increases. Describe a graph of the position of the car vs. clock time.
If you have not already done so, tell whether the graph is increasing at an increasing rate, increasing at a decreasing rate, decreasing at an increasing rate, decreasing at a decreasing rate, increasing at a constant rate or decreasing at a constant rate. Is the slope of your graph increasing or decreasing? How does the behavior of the slope of your graph indicate the condition of the problem, namely that the velocity is increasing?......!!!!!!!!...................................
RESPONSE --> The graph would be one that will increase at an increasing rate. Initially the graph would start with a slow rise as the car would just be starting its ascent from rest. As the car gains speed it will continue to increase in speed also causing an increase in distance covered in a smaller amount of time.
The slope of the graph would be increasing as the rise and run would continue to increase. An increasing slope would lend to the fact that both the speed and distance being covered were both increasing causing the rise and run to also increase..................................................
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03:05:02 The graph should have been increasing, since the position of the car increases with time (the car gets further and further from its starting point). The slope of the graph should have been increasing, since it is the slope of the graph that indicates velocity. An increasing graph within increasing slope is said to be increasing at an increasing rate.
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RESPONSE --> ok
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19:47:23 query Note that there are 10 questions in this assignment.
At a certain instant the speedometer of a car reads 5 meters / second (of course cars in this country generally read speeds in miles per hour and km per hour, not meters / sec; but they could easily have their faces re-painted to read in meters/second, and we assume that this speedometer has been similarly altered). Exactly 4 seconds later the speedometer reads 25 meters/second (that, incidentally, indicates very good acceleration, as you will understand later). At what average rate is the speed of the car changing with respect to clock time?......!!!!!!!!...................................
RESPONSE --> First we would subtract the original rate of 5 m/s from the final rate of 25 m/s. This would give us a change of 20 m/s. The time it took to make the change of 20 m/s was 4 seconds. So we divide 20 m/s by 4 seconds, this gives us a average rate of 5 m/s change in speed with respect to the clock time.
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19:47:34 The rate of change of the speed with respect clock time is equal to the change in the speed divided by the change in the clock time. So we must ask, what is the change in the speed, what is the change in the clock time and what therefore is the rate at which the speed is changing with respect to clock time?
The change in speed from 5 meters/second to 25 meters/second is 20 meters/second. This occurs in a time interval lasting 4 seconds. The average rate of change of the speed is therefore (20 meters/second)/(4 seconds) = 5 meters / second / second. This means that on the average, per second, the speed changes by 5 meters/second.......!!!!!!!!...................................
RESPONSE --> ok
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19:49:22 `q002. Explain in commonsense terms of the significance for an automobile of the rate at which its velocity changes. Do you think that a car with a more powerful engine would be capable of a greater rate of velocity change?
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RESPONSE --> The change in velocity is significant as it measures the acceleration of the car. Yes, I believe that a car with a more powerful engine would have a greater rate of velocity change.
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19:49:33 A car whose velocity changes more rapidly will attain a given speed in a shorter time, and will be able to 'pull away from' a car which is capable of only a lesser rate of change in velocity. A more powerful engine, all other factors (e.g., weight and gearing) being equal, would be capable of a greater change in velocity in a given time interval.
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RESPONSE --> ok
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19:52:26 `q003. Explain how we obtain the units meters / second / second in our calculation of the rate of change of the car's speed.
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RESPONSE --> It is double the average velocity (vAve).
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19:53:18 When we divide the change in velocity, expressed in meters/second, by the duration of the time interval in seconds, we get units of (meters / second) / second, often written meters / second / second.
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RESPONSE --> ok
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19:54:37 `q004. The unit (meters / second) / second is actually a complex fraction, having a numerator which is itself a fraction. Such a fraction can be simplified by multiplying the numerator by the reciprocal of the denominator. We thus get (meters / second) * (1/ second). What do we get when we multiply these two fractions?
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RESPONSE --> When we multiply the two fractions we would get 1m / second^2.
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19:56:03 Multiplying the numerators we get meters * 1; multiplying the denominators we get second * second, which is second^2. Our result is therefore meters * 1 / second^2, or just meters / second^2. If appropriate you may at this point comment on your understanding of the units of the rate of change of velocity.
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RESPONSE --> I am still somewhat confused as to acceleration and average velocity, I am continuing to work on this.
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19:58:16 `q004. If the velocity of an object changes from 10 m/s to -5 m/s during a time interval of 5 seconds, then at what average rate is the velocity changing?
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RESPONSE --> We would divide the change in velocity (-15 m/s) by the time interval of 5 sec. This would give us an answer of -3. So the velocity is changing at an average rate of -3 m/sec.
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19:58:24 We see that the velocity changes from 10 meters/second to -5 meters/second, a change of -15 meters / second, during a five-second time interval. A change of -15 m/s during a 5 s time interval implies an average rate of -15 m/s / (5 s) = -3 (m/s)/ s = -3 m/s^2. This is the same as (-3 m/s) / s, as we saw above. So the velocity is changing by -3 m/s every second.
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RESPONSE --> ok
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20:00:33 `q005. You should have noted that velocity can be positive or negative, as can the change in velocity or the rate at which velocity changes. The average rate at which a quantity changes with respect to time over a given time interval is equal to the change in the quantity divided by the duration of the time interval. In this case we are calculating the average rate at which the velocity changes. If v represents velocity then we we use `dv to represent the change in velocity and `dt to represent the duration of the time interval. What expression do we therefore use to express the average rate at which the velocity changes?
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RESPONSE --> Change in Velocity = 'dv2 - 'dv1 / 'dt2 - 'dt1
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20:07:34 `q006. If a runner is moving at 6 meters / sec at clock time t = 1.5 sec after starting a race, and at 9 meters / sec at clock time t = 3.5 sec after starting, then what is the average acceleration of the runner between these two clock times?
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RESPONSE --> The definition of acceleration is the change in velocity divided by the change in time.
I am still having a small struggle with the velocity and acceleration but I am working on it. If a runner were to accelerate from 6 m/s at t=1.5 and at 9 m/s at t= 3.5 sec, then the average acceleration would be change in velocity / change in time 3 m/s / 2 sec or 1.5 m/s/s.................................................
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20:07:39 click for the next question
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RESPONSE --> ok
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20:08:09 `q006a. What is the change `dv in the velocity of the runner during the time interval, and what is the change `dt in clock time during this interval?
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RESPONSE --> The change in velocity was 3 m/s and the change in clock time was 2 seconds.
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20:08:11 09-18-2005 20:08:11 click for the next question
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NOTES ------->
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20:08:44 `q006b. What therefore is the average rate at which the velocity is changing during this time interval?
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RESPONSE --> The average rate that the velocity was changing during this time is 1.5 m/s/s.
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20:09:49 We see that the runner's velocity changes from 6 meters/second to 9 meters/second, a change of `dv = 9 m/s - 6 m/s = 3 m/s, during a time interval their runs from t = 1.5 sec to t = 3.5 sec so that the duration of the interval is `dt = 3.5 s - 1.5 s = 2.0 s. The rate at which the velocity changes is therefore 3 m/s / (2.0 s) = 1.5 m/s^2. Please comment if you wish on your understanding of the problem at this point.
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RESPONSE --> I think I had the correct answer on this question. Is m/s/s the same as m/s^2? If so, which way would you prefer the answer be written?
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20:13:01 `q007. On a graph of velocity vs. clock time, we can represent the two events of this problem by two points on the graph. The first point will be (1.5 sec, 6 meters/second) and the second point will be (3.5 sec, 9 meters / sec). What is the run between these points and what does it represent?
What is the rise between these points what does it represent? What is the slope between these points what does it represent?......!!!!!!!!...................................
RESPONSE --> The run between these points is 2. It represents the change in clock time.
The rise between these point is 3. It represents the change in velocity. The slope would be 1.5 and it represents the change in velocity between these 2 points and makes an estimate of other velocity vs time points..................................................
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20:13:22 The rise from the first point to the second is from 6 meters/second to 9 meters/second, or 3 m/s. This represents the change `dv in velocity. The run is from 1.5 seconds to 3.5 seconds, or 2 seconds, and represents the change `dt in clock time. The slope, being the rise divided by the run, is 3 m/s / (2 sec) = 1.5 m/s^2. This slope represents `dv / `dt, which is the average acceleration during the time interval. You may if you wish comment on your understanding to this point.
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RESPONSE --> I think I am ok here.
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20:15:11 `q008. In what sense does the slope of any graph of velocity vs. clock time represent the acceleration of the object? For example, why does a greater slope imply greater acceleration?
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RESPONSE --> It gives us an idea of the intervals at other given time intervals and velocity points for the given points. The greater slope implies that the object is accelerating more rapidly and therefore the velocity is changing faster than the clock time. It would be increasing at an increasing rate.
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20:16:02 Since the rise between two points on a graph of velocity vs. clock time represents the change in `dv velocity, and since the run represents the change `dt clock time, the slope represents rise / run, or change in velocity /change in clock time, or `dv / `dt. This is the definition of average acceleration.
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RESPONSE --> I have made these notes.
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20:20:04 `q009. This is the same situation as in the preceding problem: An automobile coasts down a hill with a constant slope. At first its velocity increases at a very nearly constant rate. After it attains a certain velocity, air resistance becomes significant and the rate at which velocity changes decreases, though the velocity continues to increase. Describe a graph of velocity vs. clock time for this automobile (e.g., neither increasing nor decreasing; increasing at an increasing rate, a constant rate, a decreasing rate; decreasing at an increasing, constant or decreasing rate; the description could be different for different parts of the graph).
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RESPONSE --> Initially the graph would have a greater slope with the change in velocity being faster than the clock time. Once the car reached the rate where the air resistance became a hinderance the slope would begin to take a more sideways approach, showing us that the velocity is still increasing, just not at a rate as fast as previous observations.
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20:21:19 Your graph should have velocity as the vertical axis and clock time as the horizontal axis. The graph should be increasing since the velocity starts at zero and increases. At first the graph should be increasing at a constant rate, because the velocity is increasing at a constant rate. The graph should continue increasing by after a time it should begin increasing at a decreasing rate, since the rate at which the velocity changes begins decreasing due to air resistance. However the graph should never decrease, although as air resistance gets greater and greater the graph might come closer and closer to leveling off. Critique your solution by describing or insights you had or insights you had and by explaining how you now know how to avoid those errors.
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RESPONSE --> I had the correct answer. The velocity could not have decreased as it was stated that the velocity stayed constant, just the clock time was increased.
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20:23:08 `q010. An automobile coasts down a hill with a constant slope. At first its velocity increases at a very nearly constant rate. After it attains a certain velocity, air resistance becomes significant and the rate at which velocity changes decreases, though the velocity continues to increase. Describe a graph of acceleration vs. clock time for this automobile (e.g., neither increasing nor decreasing; increasing at an increasing rate, a constant rate, a decreasing rate; decreasing at an increasing, constant or decreasing rate; the description could be different for different parts of the graph).
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RESPONSE --> Initially the graph would have a nearly linear slope, as the graph approached where the car began to take on air resistance the slope would gradually level off signaling the decrease in velocity caused by the air.
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20:25:13 Your graph should have acceleration as the vertical axis and clock time as the horizontal axis. At first the graph should be neither increasing nor decreasing, since it first the acceleration is constant. Then after a time the graph should begin decreasing, which indicates the decreasing rate at which velocity changes as air resistance begins having an effect. An accurate description of whether the graph decreases at a constant, increasing or decreasing rate is not required at this point, because the reasoning is somewhat complex and requires knowledge you are not expected to possess at this point. However it is noted that the graph will at first decrease more and more rapidly, and then less and less rapidly as it approaches the t axis.
In answer to the following question posed at this point by a student: Can you clarify some more the differences in acceleration and velocity? ** Velocity is the rate at which position changes and the standard units are cm/sec or m/sec. Acceleration is the rate at which velocity changes and its standard units are cm/s^2 or m/s^2. Velocity is the slope of a position vs. clock time graph. Acceleration is the slope of a velocity vs. clock time graph. **......!!!!!!!!...................................
RESPONSE --> ok
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