course Phy231 ?€?????d?p??h????assignment #003003. Velocity Relationships
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19:31:48 `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?
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RESPONSE --> m/s confidence assessment: 3
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19:32:21 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 --> Did I misread the last question? I could have sworn it said 'ds was meters. self critique assessment: 3
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19:32:48 `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 --> cm confidence assessment: 3
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19:32:52 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 --> self critique assessment: 3
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19:33:12 `q003. Explain the algebra of multiplying the unit cm / sec by the unit sec.
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RESPONSE --> cm/sec * sec (cancel sec) =cm confidence assessment: 3
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19:33:17 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 --> self critique assessment: 3
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19:33:36 `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 --> sec confidence assessment: 3
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19:33:43 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 --> self critique assessment: 3
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19:34:28 `q005. Explain the algebra of dividing the unit km / sec into the unit km.
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RESPONSE --> km/(km/sec) (division is multiplying by reciprical) =km*(sec/km) (cancel km) =sec confidence assessment: 3
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19:34:33 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 --> self critique assessment: 3
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19:35:32 `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 --> (10m-4m)/(5s-2s) 6m/3s vAve=2m/s 'dt=6m `ds=3s confidence assessment: 3
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19:35:36 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 --> self critique assessment: 3
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19:35:58 `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 --> 'ds=s2-s1 'dt=t2-t1 confidence assessment: 3
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19:36:02 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 --> self critique assessment: 3
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19:40:07 `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 --> rise=10m-4m=6m=`ds run=5s-2s=3s=`dt confidence assessment: 3
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19:40:11 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 --> self critique assessment: 3
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19:40:24 `q009. What is the slope of this triangle and what does it represent?
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RESPONSE --> 2m/s=aVe confidence assessment: 3
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19:40:27 The slope of this graph is 6 meters / 3 seconds = 2 meters / second.
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RESPONSE --> self critique assessment: 3
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19:41:17 `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 --> It implies that in less time more displacement happened which implies a higher rate of change, higher velocity. confidence assessment: 3
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19:41:20 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 --> self critique assessment: 3
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19:45:45 `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?
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RESPONSE --> Graph is increasing at an increasing rate. Since the slope is increasing at an increasing rate, the velocity is increasing. confidence assessment: 3
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19:45:48 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 --> self critique assessment: 3
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course Phy231 At clock time t = 7 sec, a ball rolling straight down a hill is moving at 6 m/s and is 47 m from the top of the hill. It accelerates uniformly at a rate of .6 m/s/s until clock time t = 17 sec. What is its velocity at this point and what is its average velocity during this time? How far is it from the starting point at t = 17 sec?`dt=17s-7s=10s
course Phy231 Reason out the quantities v0, vf, v, vAve, a, s and t: If an object? velocity changes at a uniform rate from 8 cm/s to 11 cm/s as it travels 38 cm, then what is the average acceleration of the object?v0=8cm/s
course Phy231 ??F??J€??q???€??assignment #004
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23:08:58 Intro Prob 6 given init vel, accel, `dt find final vel, dist If initial velocity is v0, acceleration is a and time interval is `dt, then in symbols what are the final velocity vf and the displacement `ds?
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RESPONSE --> vf=v0+a*`dt `ds=((vf+v0)/2)*t confidence assessment: 3
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23:09:21 **You would use accel. and `dt to find `dv: a * `dt = `dv. Adding `dv to initial vel. vo you get final vel. Then average initial vel. and final vel. to get ave. vel.: (v0 + vf) / 2 = ave. vel. You would then multiply ave. vel. and `dt together to get the distance. For example if a = 3 m/s^2, `dt = 5 s and v0 = 3 m/s: 3 m/s^2 * 5 s = 15 m/s = `dv 15 m/s + 3 m/s = 18 m/s = fin. vel. (18 m/s + 3 m/s) / 2 = 10.5 m/s = vAve 10.5 m/s * 5 s = 52.5 m = dist. In more abbreviated form: a * `dt = `dv v0 + `dv = vf (vf + v0) /2 = vAve vAve * `dt = `ds so `ds = (vf + v0) / 2 * `dt. **
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RESPONSE --> self critique assessment: 3
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23:09:58 What is the displacement `ds associated with uniform acceleration from velocity v0 to velocity vf in clock time `dt?
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RESPONSE --> aVe=(vf+v0)/2 `ds=aVe*`dt confidence assessment: 3
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23:10:05 ** Since accel is uniform vAve = (v0 + vf) / 2. Thus displacement is `ds = vAve * `dt = (v0 + vf) / 2 * `dt, which is the first equation of uniformly accelerated motion. **
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RESPONSE --> self critique assessment: 3
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23:11:00 Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.
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RESPONSE --> Calculate average velocity from v0 and vf. Calculate displacement from average velocity and time interval. confidence assessment: 3
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23:11:15 ** The first level in the diagram would contain `dt, v0 and vf. Then v0 and vf would connect to `dv in the second level. The second level would also contain vAve, connected from vf in the first level to v0 in the first level. The third level would contain an a, connected to `dv in the second level and `dt in the first level. The third level would also contain `ds, connected to vAve in the fourth level and `dt in the first level. **
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RESPONSE --> self critique assessment: 3
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23:12:31 Query Add any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> The velocity of the straw decreased in steps vice a smooth deceleration. The thinner rubber band stretched more than the thicker. confidence assessment: 3
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23:12:48 ** Student Response: I think I really confused what information stood for what in the Force and Pendulum Experiment. However, I enjoy doing the flow diagrams. They make you think in a different way than you are used to. INSTRUCTOR NOTE: These diagrams are valuable for most people. Not all--it depends on learning style--but most. **
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RESPONSE --> self critique assessment: 3
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course Phy231 ???r?????????assignment #004004. Acceleration
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19:33:57 `q001 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?
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RESPONSE --> aAc=(0+20m/s^2)/2=10m/s^2 confidence assessment: 3
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19:36:07 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.
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RESPONSE --> I misread the question as asking for the average change in acceleration. average speed change is equal to the `dspeed/`dt. Ok. self critique assessment: 3
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19:37:34 `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 --> Yes. I would guess as long as the power to weight ratio increased with a more powerful engine it would change velocity faster. confidence assessment: 3
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19:37:50 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 --> self critique assessment: 3
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19:38:46 `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 --> rate of change in car's speed =`dv/`dt confidence assessment: 3
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19:38:54 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 --> self critique assessment: 3
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19:39:15 `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 --> m/s^2 confidence assessment: 3
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19:39:29 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 understand m/s^2 self critique assessment: 3
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19:40:13 `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 --> (-5m/s-10m/s)/5s=-3m/s^2 confidence assessment: 3
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19:40:18 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 --> self critique assessment: 3
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19:41:14 `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 --> acceleration confidence assessment: 2
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19:41:43 The average rate would be expressed by [ave rate of velocity change with respect to clock time] = `dv / `dt. The expression [ave rate of velocity change with respect to clock time] is pretty cumbersome so we give it a name. The name we give it is 'average acceleration', abbreviated by the letter aAve. Using a to represent acceleration, write down the definition of average acceleration. The definition of average acceleration is aAve = `dv / `dt. Please make any comments you feel appropriate about your understanding of the process so far.
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RESPONSE --> aAve=average acceleration. Ok. self critique assessment: 3
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19:42:30 `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 --> aAve=(9m/s-6m/s)/(3.5s-1.5s)=1.5m/s^2 confidence assessment: 3
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19:42:35 Before answering, click for the next question
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RESPONSE --> confidence assessment: 3
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19:43:16 `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 --> `dv=9ms-6m/s=3m/s `dt=3.5s-1.5s=2s confidence assessment: 3
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19:43:19 Before answering, click for the next question
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RESPONSE --> confidence assessment: 3
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19:43:28 `q006b. What therefore is the average rate at which the velocity is changing during this time interval?
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RESPONSE --> 1.5m/s^2 confidence assessment: 3
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19:43:33 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 --> self critique assessment: 3
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19:44:22 `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?
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RESPONSE --> run=3.5s-1.5s=2s rise=9m/s-6m/s=3m/s slope=1.5m/s^2, represents accleration. confidence assessment: 3
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19:44:32 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 --> self critique assessment: 3
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19:46:43 `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 --> The graph represents the relationship between velocity and its value over time, which is acceleration. Greater slope means a greater change in velocity with respect to time, thus greater acceleration. confidence assessment: 3
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19:47:00 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 --> self critique assessment: 3
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19:47:55 `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 --> A straight line that eventually becomes increasing at a decreasing rate. confidence assessment: 3
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19:48:13 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 --> self critique assessment: 3
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19:51:06 `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 --> Graph would be acceleration Y, time X. A straight horizontal line until the air resistance becomes significant at which time it becomes decreasing at a decreasing rate. confidence assessment: 3
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19:51:38 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. **
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RESPONSE --> self critique assessment: 3
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course Phy231 ??????|y???????assignment #005
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11:15:05 `q001. Note that there are 9 questions in this assignment. If the acceleration of an object is uniform, then the following statements apply: 1. A graph of velocity vs. clock time forms a straight line, either level or increasing at a constant rate or decreasing at a constant rate. 2. The average velocity of the object over any time interval is equal to the average of its velocity at the beginning of the time interval (called its initial velocity) and its velocity at the end of the time interval (called its final velocity). 3. The velocity of the object changes at a constant rate (this third statement being obvious since the rate at which the velocity changes is the acceleration, which is assumed here to be constant). 4. The acceleration of the object at every instant is equal to the average acceleration of the object. Suppose that an object increases its velocity at a uniform rate, from an initial velocity of 5 m/s to a final velocity of 25 m/s during a time interval of 4 seconds. By how much does the velocity of the object change? What is the average acceleration of the object? What is the average velocity of the object?
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RESPONSE --> `dv=25m/s-5m/s=20m/s aAc=20m/s/4s=5m/s^2 aVe=(5m/s+20m/s)/2=12.5m/s confidence assessment: 3
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11:15:47 The velocity of the object changes from 5 meters/second to 25 meters/second so the change in velocity is 20 meters/second. The average acceleration is therefore (20 meters/second) / (4 seconds) = 5 m / s^2. The average velocity of the object is the average of its initial and final velocities, as asserted above, and is therefore equal to (5 meters/second + 25 meters/second) / 2 = 15 meters/second (note that two numbers are averaged by adding them and dividing by 2).
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RESPONSE --> I had an error in my addition for calculating the average velocity. self critique assessment: 2
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11:16:10 `q002. How far does the object of the preceding problem travel in the 4 seconds?
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RESPONSE --> 15m/s*4s=60m confidence assessment: 3
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11:16:14 The displacement `ds of the object is the product vAve `dt of its average velocity and the time interval, so this object travels 15 m/s * 4 s = 60 meters during the 4-second time interval.
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RESPONSE --> self critique assessment: 3
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11:17:41 `q003. Explain in commonsense terms how we determine the acceleration and distance traveled if we know the initial velocity v0, and final velocity vf and the time interval `dt.
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RESPONSE --> Acceleration is equal to the slope of the graph of velocity vs time. So we take the difference of final and initial velocity divided by change in time. Distance traveled is equal to the average velocity for a given time interval times the time interval. confidence assessment: 3
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11:17:44 In commonsense terms, we find the change in velocity since we know the initial and final velocities, and we know the time interval, so we can easily calculate the acceleration. Again since we know initial and final velocities we can easily calculate the average velocity, and since we know the time interval we can now determine the distance traveled.
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RESPONSE --> self critique assessment: 3
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11:18:59 `q004. Symbolize the situation by first giving the expression for the acceleration in terms of v0, vf and `dt, then by giving the expression for vAve in terms of v0 and vf, and finally by giving the expression for the displacement in terms of v0, vf and `dt.
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RESPONSE --> aAve=(vf-v0)/`dt vAve=(vf+v0)/2 `ds=vAve*`dt confidence assessment: 3
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11:19:04 The acceleration is equal to the change in velocity divided by the time interval; since the change in velocity is vf - v0 we see that the acceleration is a = ( vf - v0 ) / `dt.
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RESPONSE --> self critique assessment: 3
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11:19:14 `q005. The average velocity is the average of the initial and final velocities, which is expressed as (vf + v0) / 2.
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RESPONSE --> Ok confidence assessment: 3
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11:19:21 When this average velocity is multiplied by `dt we get the displacement, which is `ds = (v0 + vf) / 2 * `dt.
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RESPONSE --> self critique assessment: 3
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11:22:22 `q006. This situation is identical to the previous, and the conditions implied by uniformly accelerated motion are repeated here for your review: If the acceleration of an object is uniform, then the following statements apply: 1. A graph of velocity vs. clock time forms a straight line, either level or increasing at a constant rate or decreasing at a constant rate. 2. The average velocity of the object over any time interval is equal to the average of its velocity at the beginning of the time interval (called its initial velocity) and its velocity at the end of the time interval (called its final velocity). 3. The velocity of the object changes at a constant rate (this third statement being obvious since the rate at which the velocity changes is the acceleration, which is assumed here to be constant). 4. The acceleration of the object at every instant is equal to the average acceleration of the object. Describe a graph of velocity vs. clock time, assuming that the initial velocity occurs at clock time t = 0. At what clock time is the final velocity then attained? What are the coordinates of the point on the graph corresponding to the initial velocity (hint: the t coordinate is 0, as specified here; what is the v coordinate at this clock time? i.e., what is the velocity when t = 0?). What are the coordinates of the point corresponding to the final velocity?
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RESPONSE --> There is no data provided for this problem. I'm assuming I should use the data from the earlier problems. time of vf=4s 0,5 0,25 confidence assessment: 2
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11:23:10 `q007. Is the graph increasing, decreasing or level between the two points, and if increasing or decreasing is the increase or decrease at a constant, increasing or decreasing rate?
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RESPONSE --> Increasing at a constant rate since acceleration is uniform. confidence assessment: 3
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11:23:16 Since the acceleration is uniform, the graph is a straight line. The graph therefore increases at a constant rate from the point (0, 5 m/s) to the point (4 s, 25 m/s).
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RESPONSE --> self critique assessment: 3
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11:23:52 `q008. What is the slope of the graph between the two given points, and what is the meaning of this slope?
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RESPONSE --> (25-5)/(4-0)=5. Slope is acceleartion. confidence assessment: 3
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11:23:57 The rise of the graph is from 5 m/s to 25 m/s and is therefore 20 meters/second, which represents the change in the velocity of the object. The run of the graph is from 0 seconds to 4 seconds, and is therefore 4 seconds, which represents the time interval during which the velocity changes. The slope of the graph is rise / run = ( 20 m/s ) / (4 s) = 5 m/s^2, which represents the change `dv in the velocity divided by the change `dt in the clock time and therefore represents the acceleration of the object.
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RESPONSE --> self critique assessment: 3
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11:25:21 `q009. The graph forms a trapezoid, starting from the point (0,0), rising to the point (0,5 m/s), then sloping upward to (4 s, 25 m/s), then descending to the point (4 s, 0) and returning to the origin (0,0). This trapezoid has two altitudes, 5 m/s on the left and 25 m/s on the right, and a base which represents a width of 4 seconds. What is the average altitude of the trapezoid and what does it represent, and what is the area of the trapezoid and what does it represent?
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RESPONSE --> avg alt=(5+25)/2=30 =vAve area=30*4=120=`ds confidence assessment: 3
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11:25:31 The two altitudes are 5 meters/second and 25 meters/second, and their average is 15 meters/second. This represents the average velocity of the object on the time interval. The area of the trapezoid is equal to the product of the average altitude and the base, which is 15 m/s * 4 s = 60 meters. This represents the product of the average velocity and the time interval, which is the displacement during the time interval.
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RESPONSE --> self critique assessment: 3
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