#$&* course phy 121 I sent this to you in an email earlier when this page was down but im also sending a copy here If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
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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 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): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: Four points of a position vs. clock time graph are (3 sec, 8 m), (7 sec, 16 m), (10 sec, 19 m) and (14 sec, 21 m). 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: 2.35m/s 1.95m/s 1.7m/s The average velocity is decreasing Increasing at a decreasing rate. The time continues at an increasing rate but corresponds with a change in position that is decreasing. 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): I calculated the average velocity like acceleration was constant when that had nothing to do with the problem nor should it ever have been inferred. My calculations do still show that the average velocity is decreasing, however the straight line through my graph did not depict decreasing at a decreasing rate. ------------------------------------------------ Self-critique rating: ********************************************* 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: There are 3 intervals, 2m/s^2 1m/s^2 0.5m/s^2 The average acceleration is decreasing. The only method I am aware of for finding the change in position is vAve * `dt, I cant find vAve using the traditional method of `ds/`dt since I dont have `ds, and the average of the initial and final velocities is only the average velocity when the acceleration is uniform, in this case it is not. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Ah so we can still find the average velocity within an accurate enough range to still be useful even when the acceleration is not presumed to be uniform. We therefore say that the average velocity is an approximation of the actual value. ------------------------------------------------ Self-critique Rating: ok
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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): With the information we were given it should not have been difficult for me to sketch the corresponding graphs after calculating the quantities. The 1st interval increased by 30 m, therefore if the graph of position vs. clock time were to start at 7m, in a 2 second interval it would increase by 30m to 37m. The next 2 second interval showed an increase of 70 meters, which would then but us at 107 m. The acceleration remained constant so on a graph of acceleration vs. clock time the points would have formed a straight line. ------------------------------------------------ Self-critique Rating: My usage of graphs had been previously limited, I am getting better at being able to visualize certain aspects of graphs in my head without having to actually look at whatever sketch Ive made. All the quantities above were calculated on paper depicted on a sketched graph, which I was able to perform without having to look at the solution, so im getting better, however I did have to refer to the solution for the position vs. clock time graph but I am clear on this matter now.
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Given Solution: `aAt a rate of -4 cm/s, for the 1-second interval between t = 20 s and t = 21 s the change in depth would be -4 cm/s * 1 sec = -4 cm. If the depth was 80 cm at t = 20 sec, the depth at t = 21 sec would be 80 cm - 4 cm/ = 76 cm. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q002. Using the same information, what the you expect the depth will be depth at the t = 30 sec instant? Do you think this estimate is more or less accurate than the estimate you made for the t = 21 second instant? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If you use the same information, then the rate is still -4 cm/s. The difference between 30 seconds and 20 seconds is 10 seconds. With the same information, you multiply the -4 cm/s and the 10 seconds to figure out how many cm change there is. When you do this, you get a result of -40. Since the depth at 20 seconds is 80 cm, then you simply take 80 cm and subtract 40 from it, which gives you a depth of 40 cm. Even though the time difference is larger, you should still expect this to be the exact value, so the accuracy should be about the same. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aAt - 4 cm/s, during the 10-second interval between t = 20 sec and t = 30 sec we would expect a depth change of -4 cm/sec * 10 sec = -40 cm, which would result in a t = 30 sec depth of 80 cm - 40 cm = 40 cm. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q003. If you know that the depth in the preceding example is changing at the rate of -3 cm/s at the t = 30 sec instant, how will this change your estimate for the depth at t = 30 seconds--i.e., will your estimate be the same as before, will you estimate a greater change in depth or a lesser change in depth? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If the rate changes to -3 cm/s compared to a rate of -4 cm/s, then you can expect a lesser change in depth since it is changing by 1 less cm per second. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aSince the rate of depth change has changed from -4 cm / s at t = 20 s to -3 cm / s at t = 30 s, we conclude that the depth probably wouldn't change as much has before. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. What is your specific estimate of the depth at t = 30 seconds? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If the rate at t=20 is -4 cm/s and at t=30 the rate is -3 cm/s then we can infer that the average rate of change is -3.5 cm/s (combine then divide by 2). Since the difference between the two times is 10 seconds, then you simply multiply -3.5 * 10 = -35 cm/s. Since the rate at t=20 is 80 cm, then you can subtract the 35 from 80 which will give you depth of 45 cm at t=30. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aKnowing that at t = 20 sec the rate is -4 cm/s, and at t = 30 sec the rate is -3 cm/s, we could reasonably conjecture that the approximate average rate of change between these to clock times must be about -3.5 cm/s. Over the 10-second interval between t = 20 s and t = 30 s, this would result in a depth change of -3.5 cm/s * 10 sec = -35 cm, and a t = 30 s depth of 80 cm - 35 cm = 45 cm. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q005. If we have a uniform cylinder with a uniformly sized hole from which water is leaking, so that the quadratic model is very nearly a precise model of what actually happens, then the prediction that the depth will change and average rate of -3.5 cm/sec is accurate. This is because the rate at which the water depth changes will in this case be a linear function of clock time, and the average value of a linear function between two clock times must be equal to the average of its values at those to clock times. If y is the function that tells us the depth of the water as a function of clock time, then we let y ' stand for the function that tells us the rate at which depth changes as a function of clock time. If the rate at which depth changes is accurately modeled by the linear function y ' = .1 t - 6, with t in sec and y in cm/s, verify that the rates at t = 20 sec and t = 30 sec are indeed -4 cm/s and -3 cm/s. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Y = 0.1t - 6 Y(20) = 0.1(20) - 6 Y(20) = -4 cm/s Y(30) = 0.1(30)-6 Y(30) = -3 cm/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aAt t = 20 sec, we evaluate y ' to obtain y ' = .1 ( 20 sec) - 6 = 2 - 6 = -4, representing -4 cm/s. At t = 30 sec, we evaluate y' to obtain y' = .1 ( 30 sec) - 6 = 3 - 6 = -3, representing -3 cm/s. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q006. For the rate function y ' = .1 t - 6, at what clock time does the rate of depth change first equal zero? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If at t=20 the result of the rate function is y = .1t - 6 is -4 and at t=30 the result of the rate function is -3, then you would simply follow the trend. If the rate is changing by a value of -1 for every 10 seconds, then you would simply continue along the trend. Its going to take an additional 30 seconds in order to reach the point of 0, therefore, the clock time at which the rate will first equal 0 is 6 seconds. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aThe rate of depth change is first equal to zero when y ' = .1 t - 6 = 0. This equation is easily solved to see that t = 60 sec. STUDENT COMMENT This was pretty straight forward, I could look at it and figure out the time to find zero, but if the times were not spaced out by ten second intervals the finding of zero would be hard to do. INSTRUCTOR RESPONSE If you write down the equation and solve it, it works out easily enough. For example if the equation was .07 t - 12 = 0, you would add 12 to both sides then divide by .07 to get t = 12 / .07, which is approximately 170. Of course if the equation is more difficult (e.g an equation like .02 t^2 + 4.2 t^2 - t + 9 = 0) it gets harder to solve for t, and it doesn't take much to come up the an equation that's impossible to solve. But the linear equation of this problem wouldn't be difficult. You will in any case be expected to be able to solve linear and quadratic equations. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q007. How much depth change is there between t = 20 sec and the time at which depth stops changing? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If the depth at t = 20 seconds is -4 cm and the depth stops changing at t = 60 seconds which is 0 cm, then you would simply take the two depth changes in the 40 second period and figure out what the average is. -4 cm/s and 0 cm/s which is -2 cm/s. After you do this, then you have to take the 40 second time difference and multiply it by the -2 cm/s to find the rate. This gives you a result of -80 cm. At t=20 seconds the rate is 80 cm. When you take 80 cm - 80 cm, your result is 0, which is where the depth stops changing. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: `aThe rate of depth change at t = 20 sec is - 4 cm/s; at t = 60 sec the rate is 0 cm/s. The average rate at which depth changes during this 40-second interval is therefore the average of -4 cm/s and 0 cm/s, or -2 cm/s. At an average rate of -2 cm/s for 40 s, the depth change will be -80 cm. Starting at 80 cm when t = 20 sec, we see that the depth at t = 60 is therefore 80 cm - 80 cm = 0. STUDENT QUESTION I dont totally understand where the 2 cm/s comes from. INSTRUCTOR RESPONSE The two rates -4 cm/s and 0 cm/s, calculated from the given rate function, are applicable to the interval between t = 20 sec and t = 60 sec. The first is the rate at the beginning of the interval, and the second is the rate at the end of the interval. Without additional information, our first conjecture would be that the average rate is the average of the initial and final rates. For different situations this conjecture might be more or less valid; in this case since the rate function is linear, it turns out that it is completely valid. The average of the two rates -4 cm/s and 0 cm/s is -2 cm/s, and this is the rate we apply to our analysis of this interval. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): The first time I read through this question, I had a different method planned out, but when I reread it, I got the same result as the given solution. ------------------------------------------------ Self-critique Rating: 3 "