course Phy 201 z{yZqgassignment #005
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21:23:34 `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 = vf - vo `dv = 25 m/s - 5m/s `dv = 20 m/s aAve = `dv / `dt aAve = (20m/s) / 4s aAve = 5m/s/s vAve = (vf + vo) / 2 vAve = (25 m/s + 5 m/s) / 2 vAve = 15m/s confidence assessment: 2
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21:23:52 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 --> ok self critique assessment: 3
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21:31:27 `q002. How far does the object of the preceding problem travel in the 4 seconds?
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RESPONSE --> `ds = `dt * vAve `ds = 4 s * 15m/s `ds = 60m confidence assessment: 2
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21:31:38 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 --> ok self critique assessment: 3
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21:34:26 `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 --> Subtract v0 from vf to find the `dv. Divide the `dt by `dv to find the aAve. Multiply the vAve by `dt to find the distance traveled. confidence assessment: 2
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21:34:36 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 --> ok self critique assessment: 3
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21:38:25 `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 = ((vf + v0) / 2)* `dt confidence assessment: 2
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21:38:46 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 --> OK self critique assessment: 3
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21:39:02 `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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21:39:35 When this average velocity is multiplied by `dt we get the displacement, which is `ds = (v0 + vf) / 2 * `dt.
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RESPONSE --> OK self critique assessment: 3
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21:46:32
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RESPONSE --> The graph would be a straight line that increases from L to R. The final velocity is attained at t=4s. v coordinate = (0s, 5m/s) confidence assessment: 2
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21:48:16 The initial velocity of 5 m/s occurs at t = 0 s so the corresponding graph point is (0 s, 5 m/s). The final velocity of 25 meters/second occurs after a time interval of `dt = 4 seconds; since the time interval began at t = 0 sec it ends at at t = 4 seconds and the corresponding graph point is ( 4 s, 25 m/s).
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RESPONSE --> I didn't realize we were supposed to five the (4s, 25 m/s) coordinate, but I would have easily been able to do so. self critique assessment: 2
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21:49:02 `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. confidence assessment: 3
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21:49:12 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 --> ok self critique assessment: 3
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21:50:31 `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 --> rise = 20 run = 4 slope = 5m/s/s The slope corresponds to the aAve confidence assessment: 2
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21:50:47 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 --> ok self critique assessment: 3
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21:54:11 `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 --> Ave. altitude = 15 m/s which represents vAve. Area = 60 m which represents the change in position. confidence assessment: 2
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21:54:25 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 --> ok self critique assessment: 3
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