course Phy 121 鎦¦¬»Ò~ˆÕE¡Q¬e׌¿cŒõÛíåÜÃassignment #005
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20:45:37 `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 velocity changes by an increase of 20 m/s. The aAve of the object would be calculated as follows: aAve = (25-5)/4 = 5 m/s The vAve of the object would be calculated as follows: (25-5)/4 = 5 m/s confidence assessment: 3
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20:49:10 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 --> Since the average velocity of the object is 15 m/s we would multiply this number by 4 seconds and that would be 60 meters traveled in 4 seconds. self critique assessment: 3
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21:05:39 `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 --> Say my car was traveling 5 m/s and I speed up to 25 m/s and want to know how far I have traveled. Since 5 m/s is my v0 or initial velocity and 25 m/s is my vf or final velocity we know our 'ds or change in position which with a certain amount of time we can find a certain distance traveled. confidence assessment: 3
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21:11:15 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 --> Acceleration is equal to change in velocity ('dv) divided by a certain time interval and since Final velocity minus Initial velocity gives us change in velocity we find that we can solve acceleration by dividing final velocity minus initial velocty by change in time. self critique assessment: 3
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21:19:00 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 --> Since our initial clock time is 0 our coordinate for this position on our graph is (0, 5) and since our final velocity occurred 4 seconds later our final velocity would occur at clock time 4 seconds and our coordinate on the graph would be (4, 25). self critique assessment: 3
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21:19:34 `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 --> The graph is increasing at a constant rate. confidence assessment: 3
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21:24:31 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 --> Since slope is equal to rise divided by run our slope would be equal to (25-5)/(4-0) which equals 5 m/s^2. This tells us that the object is increasing, which is its acceleration. self critique assessment: 3
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21:29:13 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 --> Since one side is 5 and the other is 25 we would add the two and divide by 2 which equals 15 m/s. I am not sure exactly what this means other than its velocity.
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