course Phy201 ??????z??????r??assignment #005005. Uniformly Accelerated Motion
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16:29:12 `questionNumber 50000 `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 --> to find the'dV all you do is take the vf-v0 which is 25m/s-5m/s=20m/s a='dv/'dt (20m/s)/(4s)=5m/s/s Vavg=(25m/s+5m/s)/2=15m/s confidence assessment:
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16:33:17 `questionNumber 50000 `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 --> take (finalv+intialv/2)*'dt confidence assessment: 3
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16:36:41 `questionNumber 50000 `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='dv/'dt Vavg=Vf+V0/2 'ds=Vavg*'dt confidence assessment: 3
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16:48:02 `questionNumber 50000 `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 --> So assuming this you can create a scenario. A ball rolling at point A rolls at a velocity of 5m/s this then increases at point B to 15m/s in a time of 4 s. Since it is uniform you can assume the graph to be increasing at a constant rate. V1=5m/s V2=15m/s t=4s Vavg=10m/s slope of the graph also acceleration=10m/s/4s=2.5m/s/s confidence assessment: 3
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16:48:57 `questionNumber 50000 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 --> My fault i assumed you were supposed to create a scenario. self critique assessment: 3
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16:49:18 `questionNumber 50000 `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 --> Increase constant confidence assessment: 3
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16:51:04 `questionNumber 50000 `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 --> A=25m/s-5m/s/4s=5m/s/s the slope is also the acceleration confidence assessment: 3
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16:58:19 `questionNumber 50000 `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 --> basically area is translating into displacement. Assuming this we obtain 15m/s(4s+4s/2) and gives you an area of 60m which is also the 'ds confidence assessment: 3
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