course Phy 121 008. `query 8*********************************************
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Given Solution: In summary: During the 1-second free fall the vertical displacement is 4.9 meters in the downward direction and 5 meters in the horizontal direction. During the 20-meter free fall the time of fall is 2.04 seconds and the horizontal displacement is about 10.4 meters.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: Four points of a position vs. clock time graph are (8 m, 3 sec), (16 m, 7 sec), (19 m, 10 sec) and (21 m, 14 sec). What is the average velocity on each of the three intervals? Is the average velocity increasing or decreasing? Do you expect that the velocity vs. clock time graph is increasing at an increasing rate, increasing at a decreasing rate, increasing at a constant rate, decreasing at an increasing rate, decreasing at a decreasing rate or decreasing at a constant rate, and why? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 8/4=2m/s; 3/3=1m/s; 2/4=.5m/s. The average velocity is decreasing. The graph is decreasing at a decreasing rate. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: On the interval between the first two points the position changes by `ds 16 meters - 8 meters = 8 meters and the time by `dt 7 s - 3 s = 4 s so the average velocity (ave rate of change of position with respect to clock time) is vAve = `ds / `dt = 8 meters / (4 s) = 2 m/s. On the second interval we reason similarly to obtain vAve = `ds / `dt = 3 m / (3 s) = 1 m/s. Between these two intervals it is clear that the average velocity decreases. On the third interval we get vAve = 2 m / (4 s) = .5 m/s. Based on this evidence the velocity seems to be decreasing, and since the decrease from 1 m/s to .5 m/s is less than the decrease from 2 m/s to 1 m/s, it appears to be decreasing at a decreasing rate. However, the question asked about the velocity vs. clock time graph, so we had better sketch the graph. Using the average velocity on each interval vs. the midpoint clock time of that interval, we obtain the graph depicted below (it is recommended that you hand-sketch simple graphs like this; you learn more by hand-sketching and with a little practice it can be done in less time than it takes to create the graph on a calculator or spreadsheet, which should be reserved for situations where you have extensive data sets or require more precision than you can achieve by hand). If you try to fit a straight line to the three points you will find that it doesn't quite work. It becomes clear that the graph is decreasing but at a decreasing rate (i.e., that is is decreasing and concave up). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: If four points of a velocity vs. clock time graph are (3 sec, 8 meters/sec), (7 sec, 16 meters/sec), (10 sec, 19 meters/sec) and (12 sec, 20 meters / sec), then: What is the average acceleration on each of the two intervals? Is the average acceleration increasing or decreasing? Approximately how far does the object move on each interval? (General College Physics and University Physics students in particular: Do you think your estimates of the distances are overestimates or underestimates?) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 8/4=4m/s^2; 3/3=1m/s^2; ½=.5m/s^2. The average acceleration is decreasing. At a decreasing rate. The estimated distances would be 12*4=48m; 17.5*3=52.5m; and 19.5*2=39m. Since the acceleration isnt constant these estimates arent completely accurate. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We first analyze accelerations: On the interval between the first two points the velocity changes by `dv = 16 meters / sec - 8 meters / sec = 8 meters and the time by `dt 7 s - 3 s = 4 s so the average acceleration (ave rate of change of velocity with respect to clock time) is aAve = `dv / `dt = 8 meters / sec / (4 s) = 2 m/s^2. On the second interval we reason similarly to obtain aAve = `dv / `dt = 3 m / s / (3 s) = 1 m/s^2. On the third interval we get aAve = `dv / `dt = 1 m / s / (2 s) = .5 m/s^2. Note that the accelerations are not the same, so in subsequent analysis we cannot assume that acceleration is constant. Now we determine the approximate displacement on each interval: On the first interval the average of initial and final velocities is (8 m/s + 16 m/s) / 2 = 12 m/s, and the time interval is `dt = 7 s - 3 s = 4 s. The acceleration on this interval cannot be assumed constant, so 12 m/s is only an approximation to the average velocity on the interval. Using this approximation we have `ds = 12 m/s * 4 s = 48 meters. Similar comments apply to the second and third intervals. On the second we estimate the average velocity to be (16 m/s + 19 m/s) / 2 = 17.5 m/s, and the time interval is (10 s - 7 s) = 3 s so that the approximate displacement is `ds = 17.5 m/s * 3 s = 52.5 m. On the third we estimate the average velocity to be (19 m/s + 20 m/s) / 2 = 19.5 m/s, and the time interval is (12 s - 10 s) = 2 s so that the approximate displacement is `ds = 19.5 m/s * 2 s = 39 m. A graph of average acceleration vs. midpoint clock time: If you try to fit a straight line to the three points you will find that it doesn't quite work. It becomes clear that the graph is decreasing but at a decreasing rate (i.e., that is is decreasing and concave up). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: If the velocity of a falling object is given by the velocity function v(t) = 10 m/s^2 * t - 5 m/s, then Find the velocities at t = 1, 3 and 5 seconds. Sketch a velocity vs. clock time graph, showing and labeling the three corresponding points. Estimate the displacement and acceleration on each of the two intervals. Assuming that the t = 1 sec position is 7 meters, describe the position vs. clock time graph, the velocity vs. clock time graph and the acceleration vs. clock time graph for the motion between t = 1 and t = 5 seconds. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: V(1)=5m/s; v(3)=25m/s; v(5)-=45m/s. A=10m/s^2 The average velocities times the time interval will equal the distance. The points would be (1, 7); (3, 37); (5, 107). confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The velocities are v(1 s) = 10 m/s^2 * (1 s) - 5 m/s = 10 m/s - 5 m/s = 5 m/s v(3 s) = 10 m/s^2 * (3 s) - 5 m/s = 30 m/s - 5 m/s = 25 m/s v(5 s) = 10 m/s^2 * (5 s) - 5 m/s = 50 m/s - 5 m/s = 45 m/s The acceleration on the first interval is aAve = `dv / `dt = (25 m/s - 5 m/s) (3 s - 1 s) = 10 m/s^2 and the acceleration on the second interval is also 10 m/s^2. If the acceleration turns out to be constant then the average velocity on each interval will be the average of the initial and final velocities on the interval and we will have first interval: vAve = (25 m/s + 5 m/s) / 2 = 15 m/s, `dt = (3 s - 1 s) = 2 s so that `ds = vAve * `dt = 15 m/s * 2 s = 30 m second interval: vAve = (45 m/s + 25 m/s) / 2 = 35 m/s, `dt = (5 s - 3 s) = 2 s so that `ds = vAve * `dt = 35 m/s * 2 s = 70 m The v vs. t graph appears to be a straight line through the three corresponding points. The slope of the line is 10 m/s^2 and it intercepts the t axis at (.5 s, 0), and the v axis at (0, -5 m/s). The acceleration vs. t graph appears to be horizontal, with constant acceleration 10 m/s^2. The position vs. t graph passes through the given point (1 s, 7 m), consistent with the information that position is 7 m at clock time t = 1 s. During the first interval, which extends from t = 1 s to t = 3 s, we have seen that the position changes by 30 m. Thus at clock time t = 3s the new position will be the original 7 m position, plus the 30 m change in position, so the position is 37 m.. Thus and the graph includes the point (3 s, 37 m). During the next 2-second interval, between t = 3 s and t = 5 s, the position changes by another 70 m, as calculated previously. Adding the new 70 m displacement to the 37 m position we find that at t = 5 s the position is 37 m + 70 m = 107 m. The graph therefore passes through the point (5 s, 107 m). The position vs. clock time points are depicted on the first graph below, and a smooth curve through these points in the second graph. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK "