course Phy 231 züýÚÒŽ„¾TÝÝO®z`Íã¡à÷•íâšàassignment #005
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15:22:01 `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 --> The given values are: v0= 5m/s, vf=25m/s, and `dt = 4s `dv = 25m/s - 5m/s = 20m/s a = `dv/`dt = 20m/s/4s = 5m/s^2 vAve = (25m/s + 5m/s)/2 = 30m/s/2 = 15m/s confidence assessment: 3
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15:22:14 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 --> self critique assessment: 3
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15:23:43 `q002. How far does the object of the preceding problem travel in the 4 seconds?
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RESPONSE --> `ds = vAve * `dt `ds = 15m/s * 4s = 60m confidence assessment: 3
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15:24:04 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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15:29:18 `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 a rate of change in velocity with respect to time. By knowing the initial and final velocity, we can determine how much the velocity changed. We need the time interval to complete this picture of acceleration. Distance can be determined by knowing what the average velocity an object is moving over the time interval. To do this, we need to know the average velocity, which requires knowing the initial and final velocities and the time interval. confidence assessment: 3
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15:29:35 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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15:31:16 `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 --> a = (vf - v0)/`dt vAve = (vf + v0)/2 `ds = (v0+vf)/2 * `dt confidence assessment: 3
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15:31:32 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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15:31:59 `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 --> confidence assessment: 3
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15:32:09 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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