course Phy 231 6/10 6:30 Question: `q001. Note that there are 9 questions in this assignment. 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.
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Given Solution: 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). Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: `q002. How far does the object of the preceding problem travel in the 4 seconds? Your solution: Displacement = average velocity*time = 15m/s * 4s = 60m confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: `q003. Explain in commonsense terms how we determine the acceleration and distance traveled if we know the initial velocity v0, and final velocity vf and the time interval `dt. Your solution: To determine the average acceleration you divide the change in velocity (vf-v0) by the change in time ‘dt. The distance travelled is found by calculating the average velocity of the object (vf+v0 / 2) and multiplying it by the time interval. You can also take the integral of the function with the limits of integration being the upper and lower limits of the time interval. Confidence Assessment: 3
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Given Solution: In commonsense terms, we find the change in velocity since we know the initial and final velocities, and we know the time interval, so we can easily calculate the acceleration. Again since we know initial and final velocities we can easily calculate the average velocity, and since we know the time interval we can now determine the distance traveled. Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: `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. Your solution: aAve = (vf-v0)/’dt vAve = (vf+v0)/2 Displacement = ((vf+v0)/2)*’dt confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The acceleration is equal to the change in velocity divided by the time interval; since the change in velocity is vf - v0 we see that the acceleration is a = ( vf - v0 ) / `dt. The average velocity is the average of the initial and final velocities, which is expressed as (vf + v0) / 2. When this average velocity is multiplied by `dt we get the displacement, which is `ds = (v0 + vf) / 2 * `dt. Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: `q006. This situation is identical to the previous, and the conditions implied by uniformly accelerated motion apply. 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? Your solution: The velocity versus clock time graph originates at (0, 5m/s) because the initial velocity is 5m/s. It is a straight, increasing line. The slope of the graph is 5m/s^2. The final velocity, 25m/s, is attained at t=4. The coordinate for the initial velocity is (0sec, 5m/s). The coordinate for the final velocity is (4s, 25m/s). confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The initial velocity of 5 m/s occurs at t = 0 s so the corresponding graph point is (0 s, 5 m/s). The final velocity of 25 meters/second occurs after a time interval of `dt = 4 seconds; since the time interval began at t = 0 sec it ends at at t = 4 seconds and the corresponding graph point is ( 4 s, 25 m/s). Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: `q007. Is the v vs. t 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? Your solution: The velocity versus time graph is increasing between the two points and it increases at a constant rate because this is a constant acceleration problem. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since the acceleration is uniform, the graph is a straight line. The graph therefore increases at a constant rate from the point (0, 5 m/s) to the point (4 s, 25 m/s). Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: `q008. What is the slope of the graph between the two given points, and what is the meaning of this slope? Your solution: The slope of the graph is 25m/s-5m/s / 4s = 20m/s / 4s = 5m/s^2, which is the same as the acceleration of the object. It is the change in velocity divided by the change in clock time. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The rise of the graph is from 5 m/s to 25 m/s and is therefore 20 meters/second, which represents the change in the velocity of the object. The run of the graph is from 0 seconds to 4 seconds, and is therefore 4 seconds, which represents the time interval during which the velocity changes. The slope of the graph is rise / run = ( 20 m/s ) / (4 s) = 5 m/s^2, which represents the change `dv in the velocity divided by the change `dt in the clock time and therefore represents the acceleration of the object. Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: `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? Your solution: The average altitude is (5m/s+25m/s)/2 = 30m/s / 2 = 15m/s. This is the average velocity of the object during the 4 second time interval. The area of the trapezoid is the average altitude*width = 15m/s*4s = 60m. This represents the displacement of the object during the interval. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The two altitudes are 5 meters/second and 25 meters/second, and their average is 15 meters/second. This represents the average velocity of the object on the time interval. The area of the trapezoid is equal to the product of the average altitude and the base, which is 15 m/s * 4 s = 60 meters. This represents the product of the average velocity and the time interval, which is the displacement during the time interval. Self-critique (if necessary): OK Self-critique rating #$&* OK 005. Uniformly Accelerated Motion "