course Phy 121
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12:42:07 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?
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RESPONSE --> Average rate of change = change in speed / time duration Change in speed = 25m - 5m = 20m Time duration = 4s Ave Rate of change = 20m/4s = 5 m/s
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12:42:25 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 --> OK
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12:42:41 `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 --> An automobile with a fast rate of change in velocity will accelerate faster than a vehicle with a slower rate of change in velocity. This is because the faster car can reach a certain speed quicker. So a more powerful engine would be capable of a greater rate of velocity change.
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12:43:56 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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12:45:24 `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 --> The unit m/s/s is a result of dividing the change in velocity, with a unit of meters/second, by the change in time, with a unit of seconds. Therefore, m/s over s is m/s/s.
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12:47:33 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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12:48:12 `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 --> Meter/second multiplied by 1/second gives us meters/second squared: (m/s) (1/s) = m/(s^2)
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12:48:32 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 --> Ok
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12:49:34 `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 --> Ave rate of `dv = `dv/`dt = (-5-10)m/s / 5s = (-15m/s) / 5s = -3m/s^2
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12:49:59 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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12:50:30 `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 --> Average acceleration is equal to the change in velocity over the time duration. Therefore, it is the average rate of change of velocity. Ave rate of `dv = `dv / `dt. Average acceleration = aAve = `dv / `dt. NOTE: Velocity can be positive or negative, as can the change in velocity or the rate at which velocity changes.
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12:54:05 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 --> Ok
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12:57:53 `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 = `dv / `dt = (9 - 6)m/s / (3.5 -1.5)s = 3m/s / 2s = 1.5 m/s^2
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12:58:03 `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 = 9m/s - 6m/s = 3m/s `dt = 3.5s - 1.5s = 2s
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12:58:42 `q006b. What therefore is the average rate at which the velocity is changing during this time interval?
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RESPONSE --> Velocity changes at 3 meters per second every 2 seconds. Which is equl to a rate of 1.5 meters per second per second.
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12:59:57 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 understand this problem.
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13:01:50 `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 --> The run represents the time interval between two points. In this case the run is equal to 2 seconds. The rise represents the change in velocity between two points. The rise in this case is equal to 3m/s (9m/s - 6m/s). The slops is rise over run. So slope equals 3m/s divided by 2 seconds, which is 1.5m/s^2. The slope is equal to the average acceleration.
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13:03:01 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 understand this problem.
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13:04:19 `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 --> When the slope is steeper, it represents a greater change in velocity or acceleration. When the slope is more horizontal, it represents a slow changing rate of velocity. Because rise represents the change in velocity, a steep slope would have a greater change in velocity. Since run represents the time interval, a flatter slope would represent a slower change in velocity over time.
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13:07:44 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 --> Ok
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13:08:16 `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 --> This graph would first be increasing at a constant rate. When it gets to the point where air resistance becomes significant, the graph begins increasing at a decreasing rate since the velocity is still increasing but not as quickly as it had been. This means the slope would be steep in the beginning and begin to level off after that point where air resistance becomes significant.
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13:10:51 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 remembered from a previous assignment, that air resistance can cause a significant change in velocity. So I knew the rate at which velocity increases, which is shown in the graph as the slope, would become more horizontal.
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13:12:44 `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 --> This graph is of acceleration vs. clock time. So even though velocity is increasing, the rate at which it increases (acceleration) is constant. The graph would therefore, be flat at the beginning of the interval. When air resistance begins slowing the car down, the graph will exhibit a decreasing curve because acceleration is no longer constant, but decreasing.
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13:13:35 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 --> The graph will at first decrease more and more rapidly, and then less and less rapidly as it approaches the t axis. NOTES: 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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D?????€????????? assignment #003 ???m????S???}? Physics I 06-10-2006
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15:49:41 Query Principles of Physics and General College Physics: Summarize your solution to Problem 1.19 (1.80 m + 142.5 cm + 5.34 `micro m to appropriate # of significant figures)
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RESPONSE --> First I converted centimeters and micrometers to meters. Then I added the numbers and rounded to 3 significant figures because three was the smallest number of significant figures in any of the components. THE ORIGINAL PROBLEM: 1.80m + 142.5 cm + 5.34 x 10^5 micro-m = 1.80m + (142.5cm/100cm) + (5.34x10^5mic-m/10^6)= 1.80m + 1.425m + 0.534m = 3.76 meters
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15:52:57 ** 1.80 m has three significant figures (leading zeros don't count, neither to trailing zeros unless there is a decimal point; however zeros which are listed after the decimal point are significant; that's the only way we have of distinguishing, say, 1.80 meter (read to the nearest .01 m, i.e., nearest cm) and 1.000 meter (read to the nearest millimeter). Therefore nothing below .01 m can be distinguished. 142.5 cm is .01425 m, good to within .00001 m. 5.34 * `micro m means 5.34 * 10^-6 m, or .00000534 m, good to within .00000001 m. Then theses are added you get 1.81425534 m; however the 1.80 m is only good to within .01 m so the result is 1.81 m. The rest of the number is meaningless, since the first number itself could be off by as much as .01 m. **
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RESPONSE --> The problem in the book is different from the problem that was in this question. I described the process I used in solving the book problem. From QUERY: 1.80 m + 142.5 cm + 5.34 `micro m From BOOK: 1.80m + 142.5 cm + (5.34 x 10^5) micro-m
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15:54:14 University Physics #34: Summarize your solution to Problem 1.34 (4 km on line then 3.1 km after 45 deg turn by components, verify by scaled sketch).
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RESPONSE --> I am a Principle of Physics student. I am not required to answer this question.
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15:54:23 ** THE FOLLOWING CORRECT SOLUTION WAS GIVEN BY A STUDENT: The components of vectors A (2.6km in the y direction) and B (4.0km in the x direction) are known. We find the components of vector C(of length 3.1km) by using the sin and cos functions. }Cx was 3.1 km * cos(45 deg) = 2.19. Adding the x component of the second vector, 4.0, we get 6.19km. Cy was 2.19 and i added the 2.6 km y displacement of the first vector to get 4.79. So Rx = 6.19 km and Ry = 4.79 km. To get vector R, i used the pythagorean theorem to get the magnitude of vector R, which was sqrt( (6.29 km)^2 + (4.79 km)^2 ) = 7.3 km. The angle is theta = arctan(Ry / Rx) = arctan(4.79 / 6.19) = 37.7 degrees. **
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RESPONSE -->
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