#$&* course PHY 201 10/04/2011 8:54 PM 008. `query 8*********************************************
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Given Solution: The vertical motion is completely independent of the horizontal. For an ideal projectile we assume that there is zero force and therefore zero acceleration in the horizontal direction, so that the acceleration in the horizontal direction is zero: Gravity has no component in the horizontal direction and therefore does not affect the horizontal motion. Unless otherwise specified we assume that air resistance is small enough to make no difference--i.e., we say that air resistance is negligible. If no other forces act in the horizontal direction, then the force is the horizontal direction is zero. We also assume negligible air resistance in the vertical direction. If we make this assumption, then the net force in the vertical direction is the force exerted by gravity. (We note that if a falling object speeds up enough, air resistance will become a factor, and if it falls far enough any object falling in air will approach a terminal velocity at which air resistance is equal and opposite to the gravitational force; at that speed the net force will be zero and the object will no longer accelerate at all.) Thus our assumptions for an ideal projectile: • For vertical motion the acceleration is the acceleration of gravity, 9.8 m/s^2 downward. • For horizontal motion the acceleration is zero, so that the velocity in the horizontal direction is constant. We first analyze the vertical motion: We have to declare hoose a positive direction for the vertical motion. Since there is no upward motion in this situation, we might as well choose downward as our positive direction. Thus the acceleration in the vertical direction is 9.8 m/s^2 in our chosen positive direction. Since the initial velocity is only in the horizontal direction, the initial vertical velocity is zero. For a 1-second fall we therefore have v0 = 0, a = 9.8 m/s^2 and `dt = 1 second. • we can easily reason that in 1 second the change in velocity will be 9.8 m/s^2 * 1 s = 9.8 m/s • since the initial velocity is zero, the final velocity will therefore be 9.8 m/s • the average velocity will be the average of initial and final velocities, so vAve = (0 m/s + 9.8 m/s) / 2 = 4.9 m/s • in 1 second at average velocity 4.9 m/s the displacement will be 4.9 m/s * 1 s = 4.9 m • (we could also have used ds= v0 `dt + .5 a `dt^2, the third equation of motion. We would have obtained `ds = 0 m/s * `dt + .5 * 9.8 m/s^2 * (1 s)^2 = 4.9 m, the same as the result obtained using direction reasoning based on definitions) For a 20-meter fall we have v0 = 0, a = 9.8 m/s^2 and `ds = 20 meters. • Using the fourth equation of motion we find that vf = +- sqrt( v0^2 + 2 a `ds) = +-sqrt( (0 m/s)^2 + 2 * 9.8 m/s^2 * 20 m) = +-sqrt( 392 m^2 / s^2) = +-19.8 m/s. • Motion is clearly in the downward direction so the negative solution wouldn't make sense in this context. The average vertical velocity is the average of initial and final vertical velocities: vAve = (0 m/s + 19.8 m/2) / 2 = 9.8 m/s, and the time of fall is `dt = `ds / vAve = 20 m / (9.8 m/s) = 2.04 sec. The horizontal motion is then easy to analyze. The acceleration in the horizontal direction is zero, so the velocity in the horizontal direction is constant. We conclude that the horizontal velocity is equal to the original 5 m/s. This is the initial, final and average horizontal velocity for the motion for any interval between leaving the edge of the platform and encountering the ground. For the 1-second fall the horizontal displacement is therefore vAve * `dt = 5 m/s * 1 s = 5 m. For the 20 m fall the time interval is 2.04 sec so the horizontal displacement is 5 m/s * 2.04 s = 10.2 metesr. 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. STUDENT NOTE: The recent math for in assignment 8 and into 9 seems like its drawing on several equations that I have written in several different areas of my notebook. Is there a place that I can basically go through and write down the most simple equations to the several equations regarding uniform acceleration to make this less time consuming and easier? INSTRUCTOR COMMENT: The four equations of uniformly accelerated motion follow from the definitions of velocity and acceleration. One of the Class Notes links (#6, I believe) specifically outlines the equations. This sounds like what you're looking for. The Linked Outline (click on the Overviews button at the top of Physics I main page, then on the Linked Outline button) can also be very helpful. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ 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: vAve1 = (16m - 8 m) / (7s - 3s) = 2 m/s, vAve2 = (19m - 16m) / (10s - 7s) = 1 m/s, vAve3 = (21m - 19m) / (14s - 10s) = 0.5 m/s. The vAve is decreasing. The velocity vs. clock time graph is decreasing at a decreasing rate because as time increases (at a constant rate), velocity 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): ------------------------------------------------ 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: aAve1 = (16 m/s - 8 m/s) / (7s - 3s) = 2 m/s/s, aAve2 = (19 m/s - 16 m/s) / (10s - 7s) = 1 m/s/s, aAve3 = (20 m/s - 19 m/s) / (12s - 10s) = 0.5 m/s/s. The average acceleration is decreasing. ‘ds1 = vAve * ‘dt , ‘ds1 = (16 m/s + 8 m/s)/2 * 4s = 48 m. ‘ds2 = (19 m/s + 16 m/s)/2 * 3s = 52.5 m. ‘ds3 = (20 m/s + 19 m/s)/2 * 2 s = 39 m. I would say these distances are an overestimation of the actual distance covered because I am assuming uniform acceleration when finding the average velocity. The average acceleration of this object is decreasing. 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): ------------------------------------------------ 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) = 10 m/s^2 * 1 s - 5 m/s = 5 m/s v(2) = 10 m/s^2 * 3s - 5 m/s = 25 m/s v(3) = 10 m/s^2 * 5s - 5 m/s = 45 m/s Interval 1: aAve = ‘dv / ‘dt, aAve = (25 m/s - 5 m/s) / (3s - 1s), aAve = 10 m/s/s vAve = (25 m/s + 5 m/s)/2 = 15 m/s, ‘dt = 3s - 1s = 2s, 2s * 15 m/s = 30 m displacement. Interval 2: aAve = ‘dv / ‘dt, aAve = (45 m/s - 25 m/s) / (3s - 1s), aAve = 10 m/s/s vAve = (45 m/s + 25 m/s)/2 = 35 m/s, ‘dt = 5s - 3s = 2s, 2s * 35 m/s = 70 m displacement. If acceleration is constant the distance versus time graph will be a graph with the points (1s, 7m), (3s, 37m), (5s, 107m). The graph will be concave up, increasing at an increasing rate. The graph of velocity versus time will be a graph with the points (1s, 5 m/s), (3s, 25 m/s), (5s, 45 m/s). The graph will be concave down, increasing at a decreasing rate. The graph of velocity versus time will be a linear graph with a slope of zero and a y-intercept of 10 m/s/s. 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. CALCULUS APPLICATION (UNIVERSITY PHYSICS STUDENTS and other interested students): The velocity function is v(t) = 10 m/s^2 * t - 5 m/s. Acceleration is the rate of change of velocity with respect to clock time, so the acceleration function is the derivative of the velocity function: • a(t) = v ' (t) = 10 m/s^2 The graph of the acceleration function is a straight horizontal line. Velocity is the rate of change of position with respect to clock time, so the velocity function is the derivative of the position function. It follows that the position function is an antiderivative of the velocity function. Using x(t) to denote the position function, which is the general antiderivative of the velocity function v(t) = 10 m/s^2 * t - 5 m/s, we write • x(t) = 5 m/s^2 * t^2 - 5 m/s * t + c, where c is an arbitrary constant. We are told that position is 7 m when t = 1 sec, so we have x(1 sec) = 7 m. By the antiderivative function we also know that • x(1 sec) = 5 m/s^2 * (1 sec)^2 - 5 m/s * 1 sec + c, so that • x(1 sec) = 5 m/s^2 * 1 sec^2 - 5 m/s * 1 sec + c = 5 m - 5 m + c = c. Thus 7 m = c. We substitute this value of c into our x(t) function, giving us the position function s(t) = 5 m/s^2 * t^2 - 5 m/s * t + 7 m The graph of this function is a parabola with vertex at (1 sec, 7 m). This function agrees completely with the t = 3 s and t = 5 s points of the graph: x(3 sec) = 5 m/s^2 * (3 s)^2 - 5 m/s * (3 x) + 7 m = 45 m - 15 m + 7 m = 37 m x(5 sec) = 5 m/s^2 * (5 s)^2 - 5 m/s * (5 x) + 7 m = 125 m - 25 m + 7 m = 107 m. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I drew my velocity versus time graph incorrectly. I found my mistake. ------------------------------------------------ Self-critique Rating:3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!