course Phy 202 My mouse was sticking on some of the problems so instead of the next question appearing the next question and the answer to that question would appear. I will try to have it fixed so it does not happening again. ɾxēVKassignment #002
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16:59:28 `q001. You will frequently need to describe the graphs you have constructed in this course. This exercise is designed to get you used to some of the terminology we use to describe graphs. Please complete this exercise and email your work to the instructor. Note that you should do these graphs on paper without using a calculator. None of the arithmetic involved here should require a calculator, and you should not require the graphing capabilities of your calculator to answer these questions. Problem 1. We make a table for y = 2x + 7 as follows: We construct two columns, and label the first column 'x' and the second 'y'. Put the numbers -3, -2, -1, -, 1, 2, 3 in the 'x' column. We substitute -3 into the expression and get y = 2(-3) + 7 = 1. We substitute -2 and get y = 2(-2) + 7 = 3. Substituting the remaining numbers we get y values 5, 7, 9, 11 and 13. These numbers go into the second column, each next to the x value from which it was obtained. We then graph these points on a set of x-y coordinate axes. Noting that these points lie on a straight line, we then construct the line through the points. Now make a table for and graph the function y = 3x - 4. Identify the intercepts of the graph, i.e., the points where the graph goes through the x and the y axes.
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RESPONSE --> x intercept = 4/3 y intercept = -4 Table x y -2 -10 -1 -7 0 -4 1 -1 2 2 y = 3x-4 0 = 3x-4 4 = 3x x = 4/3 By substituting 0 for y can solve for the x intercept. y = 3x-4 y = 3(0)-4 y = -4 By substituting 0 for x you can solve for the y intercept. As well you know its the intercept from the y=mx+b equation. confidence assessment: 3
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17:00:33 The graph goes through the x axis when y = 0 and through the y axis when x = 0. The x-intercept is therefore when 0 = 3x - 4, so 4 = 3x and x = 4/3. The y-intercept is when y = 3 * 0 - 4 = -4. Thus the x intercept is at (4/3, 0) and the y intercept is at (0, -4). Your graph should confirm this.
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RESPONSE --> I did not use proper notation. I should have state the answers as: (4/3, 0) and (0,-4) self critique assessment: 2
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17:01:22 `q002. Does the steepness of the graph in the preceding exercise (of the function y = 3x - 4) change? If so describe how it changes.
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RESPONSE --> The steepness does not change. It has a constant slope of 3. confidence assessment: 3
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17:01:38 The graph forms a straight line with no change in steepness.
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RESPONSE --> ok self critique assessment: 3
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17:05:59 `q003. What is the slope of the graph of the preceding two exercises (the function ia y = 3x - 4;slope is rise / run between two points of the graph)?
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RESPONSE --> m = (y1-y2) / (x1-x2) m = (0 - -4) / (4/3 - 0) m = 4/ (4/3) m = 3 = slope confidence assessment: 3
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17:07:55 Between any two points of the graph rise / run = 3. For example, when x = 2 we have y = 3 * 2 - 4 = 2 and when x = 8 we have y = 3 * 8 - 4 = 20. Between these points the rise is 20 - 2 = 18 and the run is 8 - 2 = 6 so the slope is rise / run = 18 / 6 = 3. Note that 3 is the coefficient of x in y = 3x - 4. Note the following for reference in subsequent problems: The graph of this function is a straight line. The graph increases as we move from left to right. We therefore say that the graph is increasing, and that it is increasing at constant rate because the steepness of a straight line doesn't change.
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RESPONSE --> I should have noted the fact that 3 is the coefficient and thus is the slope. y = mx+b where m= slope and b= y intercept. self critique assessment: 2
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17:12:27 `q004. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> The graph is increasing. The steepness changes. It gets steeper as it increases. I feel the graph is increasing at an increasing rate because the slope rises at a increasing interval. confidence assessment: 2
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17:13:38 Graph points include (0,0), (1,1), (2,4) and (3,9). The y values are 0, 1, 4 and 9, which increase as we move from left to right. The increases between these points are 1, 3 and 5, so the graph not only increases, it increases at an increasing rate STUDENT QUESTION: I understand increasing...im just not sure at what rate...how do you determine increasing at an increasing rate or a constant rate? INSTRUCTOR RESPONSE: Does the y value increase by the same amount, by a greater amount or by a lesser amount every time x increases by 1? In this case the increases get greater and greater. So the graph increases, and at an increasing rate. *&*&.
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RESPONSE --> I answered correctly but could have supported my answer better by including data points from 0 to 3 and the corresponding slopes . self critique assessment: 2
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17:18:38 `q005. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = -3 and x = 0. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> Data points: (-3,9) (-2,4) (-1,1) (0,0) The graph is decreasing. The steepness of the graph is decreasing at an increasing rate. The slope goes from -5 to -3 to -1. confidence assessment: 2
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17:20:27 From left to right the graph is decreasing (points (-3,9), (-2,4), (-1,1), (0,0) show y values 9, 4, 1, 0 as we move from left to right ). The magnitudes of the changes in x from 9 to 4 to 1 to 0 decrease, so the steepness is decreasing. Thus the graph is decreasing, but more and more slowly. We therefore say that the graph is decreasing at a decreasing rate.
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RESPONSE --> I was incorrect about the rate of decrease. I understand that it is decreasing more slowly because the slopes are less extreme negatitive values. self critique assessment: 2
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17:26:06 `q006. Make a table of y vs. x for y = `sqrt(x). [note: `sqrt(x) means 'the square root of x']. Graph y = `sqrt(x) between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> Points are: (0,0), (1,1), (2, 1.414), (3, 1.732) The graph is increasing. The steepness of the graph changes and is decreasing. The graph is increasing at a decreasing rate. confidence assessment: 2
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17:26:57 If you use x values 0, 1, 2, 3, 4 you will obtain graph points (0,0), (1,1), (2,1.414), (3. 1.732), (4,2). The y value changes by less and less for every succeeding x value. Thus the steepness of the graph is decreasing. The graph would be increasing at a decreasing rate. If the graph respresents the profile of a hill, the hill starts out very steep but gets easier and easier to climb. You are still climbing but you go up by less with each step, so the rate of increase is decreasing. If your graph doesn't look like this then you probably are not using a consistent scale for at least one of the axes. If your graph isn't as desribed take another look at your plot and make a note in your response indicating any difficulties.
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RESPONSE --> Ok self critique assessment: 3
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17:33:04 06-03-2007 17:33:04 `q007. Make a table of y vs. x for y = 5 * 2^(-x). Graph y = 5 * 2^(-x) between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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NOTES -------> Points include: (0,5), (1, 2.5), (2, 1.25), (3, .625) The graph is decreasing. The steepness of the graph changes and is becoming less steep (decreasing). Decreasing at a constant rate because each slope is one half its previous. So it is constantly decresing by half.
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17:35:55 ** From basic algebra recall that a^(-b) = 1 / (a^b). So, for example: 2^-2 = 1 / (2^2) = 1/4, so 5 * 2^-2 = 5 * 1/4 = 5/4. 5* 2^-3 = 5 * (1 / 2^3) = 5 * 1/8 = 5/8. Etc. The decimal equivalents of the values for x = 0 to x = 3 will be 5, 2.5, 1.25, .625. These values decrease, but by less and less each time. The graph is therefore decreasing at a decreasing rate. **
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RESPONSE --> I found the slopes correctly but made the wrong association. The slopes are decreasing at a decreasing rate because each slope is lower than the previous but it is less each time because each half is smaller than the one before. So though they are decreasing by .5m it is not a constant decrease. self critique assessment: 2
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17:39:33 `q008. Suppose you stand still in front of a driveway. A car starts out next to you and moves away from you, traveling faster and faster. If y represents the distance from you to the car and t represents the time in seconds since the car started out, would a graph of y vs. t be increasing or decreasing? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> The graph would be increasing at an increasing rate. Though time increases at a constant rate the speed of the car is increasing at an increasing rate so as time increased it would be further and further away because it is continuing to increase speed. confidence assessment: 2
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17:39:48 ** The speed of the car increases so it goes further each second. On a graph of distance vs. clock time there would be a greater change in distance with each second, which would cause a greater slope with each subsequent second. The graph would therefore be increasing at an increasing rate. **
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RESPONSE --> ok self critique assessment: 3
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cKz{x߃q assignment #006 006. Physics qa initial problems 06-03-2007
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17:52:17 `q001. There are two parts to this problem. Reason them out using common sense. If the speed of an automobile changes by 2 mph every second, then how long will it take the speedometer to move from the 20 mph mark to the 30 mph mark? Given the same rate of change of speed, if the speedometer initially reads 10 mph, what will it read 7 seconds later?
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RESPONSE --> 1) If speed changes by 2mph/s it will take 5 seconds to change speed by 10 mph. 30mph - 20mph = 10mph 10mph/ 2mph/s = 5s 2) If speed changes by 2mph/s after 7s it would have increased by 14mph which added to the original speed would equal 24mph. 2mph/s * 7 s = 14mph 14mph + 10mph = 24mph confidence assessment: 3
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17:52:29 It will take 5 seconds to complete the change. 30 mph - 20 mph = 10 mph change at 2 mph per second (i.e., 2 mph every second) implies 5 seconds to go from 20 mph to 30 mph Change in speed is 2 mph/second * 7 seconds = 14 mph Add this to the initial 10 mph and the speedometer now reads 24 mph.
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RESPONSE --> ok self critique assessment: 3
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18:44:11 `q002. An automobile traveling down a hill passes a certain milepost traveling at a speed of 10 mph, and proceeds to coast to a certain lamppost further down the hill, with its speed increasing by 2 mph every second. The time required to reach the lamppost is 10 seconds. It then repeats the process, this time passing the milepost at a speed of 20 mph. Will the vehicle require more or less than 10 seconds to reach the lamppost? Since its initial speed was 10 mph greater than before, does it follow that its speed at the lamppost will be 10 mph greater than before?
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RESPONSE --> It will require less than 10s to reach the lamppost. I think it will be greater than 10mph more than the orginial speed. confidence assessment: 2
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18:45:55 If it starts coasting down the same section of road at 20 mph, and if velocity changes by the same amount every second, the automobile should always be traveling faster than if it started at 10 mph, and would therefore take less than 10 seconds. The conditions here specify equal distances, which implies less time on the second run. The key is that, as observed above, the automobile has less than 10 seconds to increase its speed. Since its speed is changing at the same rate as before and it has less time to change it will therefore change by less.
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RESPONSE --> The first answer was correct but in the second I failed to recognize that the decrease in time will in turn decrease the amount of speed gained by the time it passes the lamppost; thus changing by less instead of more. self critique assessment: 2
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18:53:09 4. If an automobile of mass 1200 kg is pulled by a net force of 1800 Newtons, then the number of Newtons per kg is 1800 / 1200 = 1.5. The rate at which an automobile speeds up is determined by the net number of Newtons per kg. Two teams pulling on ropes are competing to see which can most quickly accelerate their initially stationary automobile to 5 mph. One team exerts a net force of 3000 Newtons on a 1500 kg automobile while another exerts a net force of 5000 Newtons on a 2000 kg automobile. Which team will win and why? If someone pulled with a force of 500 Newtons in the opposite direction on the automobile predicted to win, would the other team then win?
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RESPONSE --> Team 1 = 3000N/1500kg = 2N/kg Team 2 = 5000N/2000kg = 2.5N/kg Team 2 will win because they are exerting more N per kg thus pulling with a greater force per kg will be able to reach 5 mph sooner. No. Team 2 = (5000N-500N)/2000kg = 2.25N/kg Team 2 will still have more force per kg. self critique assessment: 2
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18:56:54 Greater speed and greater mass both provide advantages. In this case the player with the greater mass has less speed, so we have to use some combination of speed and mass to arrive at a conclusion. It turns out that if we multiply speed by mass we get the determining quantity, which is called momentum. 250 lb * 10 ft/sec = 2500 lb ft / sec and 200 lb * 20 ft/sec = 4000 lb ft / sec, so the second player will dominate the collision. In this course we won't use pounds as units, and in a sense that will become apparent later on pounds aren't even valid units to use here. However that's a distinction we'll worry about when we come to it.
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RESPONSE --> ok self critique assessment: 3
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19:00:23 `q006. Two climbers eat Cheerios for breakfast and then climb up a steep mountain as far as they can until they use up all their energy from the meal. All other things being equal, who should be able to climb further up the mountain, the 200-lb climber who has eaten 12 ounces of Cheerios or the 150-lb climber who has eaten 10 ounces of Cheerios?
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RESPONSE --> 200 lb / 12 oz = 16.67 lb/oz 150 lb/ 10 oz = 15 lb/ oz The 200lb climber who climb further up the mountain. confidence assessment: 2
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19:04:09 The comparison we make here is the number of ounces of Cheerios per pound of body weight. We see that the first climber has 12 oz / (200 lb) = .06 oz / lb of weight, while the second has 10 0z / (150 lb) = .067 oz / lb. The second climber therefore has more energy per pound of body weight. It's the ounces of Cheerios that supply energy to lift the pounds of climber. The climber with the fewer pounds to lift for each ounce of energy-producing Cheerios will climb further.
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RESPONSE --> I should have divided the other way to get oz of energy/lb instead of lb/oz of energy. self critique assessment: 1
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19:07:12 It turns out that, neglecting air resistance, since the slope is the same for both, both automobiles will change velocity at the same rate. So in this case the second would require exactly twice as long. If you include air resistance the faster car experiences more so it actually takes a bit less than twice as long as the slower. For the same reasons as before, and because velocity would change at a constant rate (neglecting air resistance) it would be exactly twice as great if air resistance is neglected. Interestingly if it takes twice as much time and the average velocity is twice as great the faster car travels four times as far. If there is air resistance then it slows the faster car down more at the beginning than at the end and the average velocity will be a bit less than twice as great and the coasting distance less than four times as far.
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RESPONSE --> ok self critique assessment: 3
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19:18:25 `q008. When a 100 lb person hangs from a certain bungee cord, the cord stretches by 5 feet beyond its initial unstretched length. When a person weighing 150 lbs hangs from the same cord, the cord is stretched by 9 feet beyond its initial unstretched length. When a person weighing 200 lbs hangs from the same cord, the cord is stretched by 12 feet beyond its initial unstretched length. Based on these figures, would you expect that a person of weight 125 lbs would stretch the cord more or less than 7 feet beyond its initial unstretched length?
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RESPONSE --> I expect a person would stretch the cord less than 7 feet beyond its initial unstretched length. With the 100 lb person the cord stretches 1 foot for every 20 lb. However in both the 150 and 200 lb people it stretches 1 foot for every 16.7 lb. So if you take an average of the two you get 18.35 lb/ft and if you divide 125 lb by 18.35 lb/ft you get 6.8 ft. confidence assessment: 1
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19:19:23 From 100 lbs to 150 lbs the stretch increased by 4 feet, from 150 lbs to 200 lbs the increase was only 3 feet. Thus it appears that at least in the 100 lb - 200 lb rands each additional pound results in less increase in length than the last and that there would be more increase between 100 lb and 125 lb than between 125 lb and 150 lb. This leads to the conclusion that the stretch for 125 lb would be more than halfway from 5 ft to 9 ft, or more than 7 ft. A graph of stretch vs. weight would visually reveal the nature of the nonlinearity of this graph and would also show that the stretch at 125 lb must be more than 7 feet (the graph would be concave downward, or increasing at a decreasing rate, so the midway stretch would be higher than expected by a linear approximation).
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RESPONSE --> ok. since its nonlinear you cannot apply the approach i took. self critique assessment: 1
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19:24:19 `q010. Two identical light bulbs are placed at the centers of large and identically frosted glass spheres, one of diameter 1 foot and the other of diameter 2 feet. To a moth seeking light from half a mile away, unable to distinguish the difference in size between the spheres, will the larger sphere appear brighter, dimmer or of the same brightness as the first? To a small moth walking on the surface of the spheres, able to detect from there only the light coming from 1 square inch of the sphere, will the second sphere appear to have the same brightness as the first, twice the brightness of the first, half the brightness of the first, more than twice the brightness of the first, or less than half the brightness of the first?
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RESPONSE --> The larger sphere will appear brighter due to the larger surface area. The moth on the surface would preceive the same brightness in both spheres. confidence assessment: 0
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19:31:36 `q011. The water in a small container is frozen in a freezer until its temperature reaches -20 Celsius. The container is then placed in a microwave oven, which proceeds to deliver energy at a constant rate of 600 Joules per second. After 10 seconds the ice is still solid and its temperature is -1 Celsius. After another 10 seconds a little bit of the cube is melted and the temperature is 0 Celsius. After another minute most of the ice is melted but there is still a good bit of ice left, and the ice and water combination is still at 0 Celsius. After another minute all the ice is melted and the temperature of the water has risen to 40 degrees Celsius. Place the following in order, from the one requiring the least energy to the one requiring the most: Increasing the temperature of the ice by 20 degrees to reach its melting point. Melting the ice at its melting point. Increasing the temperature of the water by 20 degrees after all the ice melted. At what temperature does it appear ice melts, and what is the evidence for your conclusion?
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RESPONSE --> ok Least Increasing the temp of the water by 20 degrees after the ice melted. Increasing the temp of the ice to the melting point. Melting the ice at its melting point The ice melts above 0 degrees celsius. At -1 degree very little ice at began to melt however it was still at 0 when a significant amount of ice had melted and was in a water/ice mix. confidence assessment: 1
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19:32:07 Since the temperature is the same when a little of the ice is melted as when most of it is melted, melting takes place at this temperature, which is 0 Celsius. The time required to melt the ice is greater than any of the other times so melting at 0 C takes the most energy. Since we don't know how much ice remains unmelted before the final minute, it is impossible to distinguish between the other two quantities, but it turns out that it takes less energy to increase the temperature of ice than of liquid water.
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RESPONSE --> ok self critique assessment: 3
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21:35:50 If the two 6-inch peaks meet and reinforce one another completely, the height of the 'combined' peak will be 6 in + 6 in = 12 in. If for example you move 3 ft closer to one end you move 3 ft further from the other and peaks, which are 6 ft apart, will still be meeting peaks. [ Think of it this way: If you move 3 ft closer to one end you move 3 ft further from the other. This shifts your relative position to the two waves by 6 feet (3 feet closer to the one you're moving toward, 3 feet further from the other). So if you were meeting peaks at the original position, someone at your new position would at the same time be meeting valleys, with two peaks closing in from opposite directions. A short time later the two peaks would meet at that point. ] However if you move 1.5 ft the net 'shift' will be 3 ft and peaks will be meeting valleys so you will be in the calmest water.
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RESPONSE --> If you receive both peaks at the center point you simutaneously you would bob up and down 12 inches. Ok self critique assessment: 3
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xE· assignment #001 001. Rates qa rates 06-03-2007
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21:54:42 Your answer has been noted. Enter 'ok' in the Response Box and click on Enter Response, then click on Next Question/Answer for the first real question.
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RESPONSE --> Ok confidence assessment: 3
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21:58:00 `q002. Note that there are 10 questions in this assignment. The questions are of increasing difficulty--the first questions are fairly easy but later questions are very tricky. The main purposes of these exercises are to refine your thinking about rates, and to see how you process challenging information. Continue as far as you can until you are completely lost. Students who are prepared for the highest-level math courses might not ever get lost. If you make $50 in 5 hr, then at what rate are you earning money?
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RESPONSE --> You are earing $10/hour. $50/5hr = $10/hr confidence assessment: 3
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21:58:20 The rate at which you are earning money is the number of dollars per hour you are earning. You are earning money at the rate of 50 dollars / (5 hours) = 10 dollars / hour. It is very likely that you immediately came up with the $10 / hour because almosteveryone is familiar with the concept of the pay rate, the number of dollars per hour. Note carefully that the pay rate is found by dividing the quantity earned by the time required to earn it. Time rates in general are found by dividing an accumulated quantity by the time required to accumulate it. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> ok confidence assessment: 3
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22:04:27 `q003.If you make $60,000 per year then how much do you make per month?
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RESPONSE --> $60,000/year / 12mo/yr = $5,000/mo confidence assessment: 3
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22:05:09 Most people will very quickly see that we need to divide $60,000 by 12 months, giving us 60,000 dollars / (12 months) = 5000 dollars / month. Note that again we have found a time rate, dividing the accumulated quantity by the time required to accumulate it. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> ok confidence assessment: 3
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22:10:05 `q004. Suppose that the $60,000 is made in a year by a small business. Would be more appropriate to say that the business makes $5000 per month, or that the business makes an average of $5000 per month?
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RESPONSE --> It would be more appropriate to say that the business makes an average of $5000. confidence assessment: 3
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22:10:24 Small businesses do not usually make the same amount of money every month. The amount made depends on the demand for the services or commodities provided by the business, and there are often seasonal fluctuations in addition to other market fluctuations. It is almost certain that a small business making $60,000 per year will make more than $5000 in some months and less than $5000 in others. Therefore it is much more appropriate to say that the business makes and average of $5000 per month. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> Ok confidence assessment: 3
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22:31:16 `q005. If you travel 300 miles in 6 hours, at what average rate are you covering distance, and why do we say average rate instead of just plain rate?
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RESPONSE --> 300 miles/ 6 hours = 50 miles/hour It would be an average rate because you cannot maintain a perfect speed over time. confidence assessment: 3
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22:31:42 The average rate is 50 miles per hour, or 50 miles / hour. This is obtained by dividing the accumulated quantity, the 300 miles, by the time required to accumulate it, obtaining ave rate = 300 miles / ( 6 hours) = 50 miles / hour. Note that the rate at which distance is covered is called speed. The car has an average speed of 50 miles/hour. We say 'average rate' in this case because it is almost certain that slight changes in pressure on the accelerator, traffic conditions and other factors ensure that the speed will sometimes be greater than 50 miles/hour and sometimes less than 50 miles/hour; the 50 miles/hour we obtain from the given information is clearly and overall average of the velocities. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> ok confidence assessment: 3
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22:33:52 `q006. If you use 60 gallons of gasoline on a 1200 mile trip, then at what average rate are you using gasoline, with respect to miles traveled?
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RESPONSE --> 60 gal / 1200 miles =.05gal/mile It asks for the rate of gal of gas use in respect to miles traveled. Not how many miles you can travel to each gallon. confidence assessment: 3
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22:34:03 The rate of change of one quantity with respect to another is the change in the first quantity, divided by the change in the second. As in previous examples, we found the rate at which money was made with respect to time by dividing the amount of money made by the time required to make it. By analogy, the rate at which we use fuel with respect to miles traveled is the change in the amount of fuel divided by the number of miles traveled. In this case we use 60 gallons of fuel in 1200 miles, so the average rate it 60 gal / (1200 miles) = .05 gallons / mile. Note that this question didn't ask for miles per gallon. Miles per gallon is an appropriate and common calculation, but it measures the rate at which miles are covered with respect to the amount of fuel used. Be sure you see the difference. Note that in this problem we again have here an example of a rate, but unlike previous instances this rate is not calculated with respect to time. This rate is calculated with respect to the amount of fuel used. We divide the accumulated quantity, in this case miles, by the amount of fuel required to cover those miles. Note that again we call the result of this problem an average rate because there are always at least subtle differences in driving conditions that require the use of more fuel on some miles than on others. It's very important to understand the phrase 'with respect to'. Whether the calculation makes sense or not, it is defined by the order of the terms. In this case gallons / mile tells you how many gallons you are burning, on the average, per mile. This concept is not as familiar as miles / gallon, but except for familiarity it's technically no more difficult. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> ok confidence assessment: 3
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22:38:55 `q007. The word 'average' generally connotes something like adding two quantities and dividing by 2, or adding several quantities and dividing by the number of quantities we added. Why is it that we are calculating average rates but we aren't adding anything?
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RESPONSE --> It often involves values that change within the given time frame and because we have a final value for example distance and time the speed of travel is likely variable within that so it cannot be said that it was traveled at an exact rate but an average. confidence assessment: 1
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22:39:59 The word 'average' in the context of the dollars / month, miles / gallon types of questions we have been answering was used because we expect that in different months different amounts were earned, or that over different parts of the trip the gas mileage might have varied, but that if we knew all the individual quantities (e.g., the dollars earned each month, the number of gallons used with each mile) and averaged them in the usual manner, we would get the .05 gallons / mile, or the $5000 / month. In a sense we have already added up all the dollars earned in each month, or the miles traveled on each gallon, and we have obtained the total $60,000 or 1200 miles. Thus when we divide by the number of months or the number of gallons, we are in fact calculating an average rate. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> In a way the values have already been added in most average rate problems. confidence assessment: 2
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22:45:16 `q008. In a study of how lifting strength is influenced by various ways of training, a study group was divided into 2 subgroups of equally matched individuals. The first group did 10 pushups per day for a year and the second group did 50 pushups per day for year. At the end of the year to lifting strength of the first group averaged 147 pounds, while that of the second group averaged 162 pounds. At what average rate did lifting strength increase per daily pushup?
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RESPONSE --> 50 push ups - 10 push ups = 40 push ups 15 lbs/ 40 push ups = .375lbs/pushup confidence assessment: 1
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22:45:34 The second group had 15 pounds more lifting strength as a result of doing 40 more daily pushups than the first. The desired rate is therefore 15 pounds / 40 pushups = .375 pounds / pushup. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> ok confidence assessment: 3
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22:47:51 `q009. In another part of the study, participants all did 30 pushups per day, but one group did pushups with a 10-pound weight on their shoulders while the other used a 30-pound weight. At the end of the study, the first group had an average lifting strength of 171 pounds, while the second had an average lifting strength of 188 pounds. At what average rate did lifting strength increase with respect to the added shoulder weight?
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RESPONSE --> 30 lbs - 10 lbs = 20 lbs shoulder weight 188 lbs - 171 lbs = 17 lbs strength 17lbs strength / 20 lbs shoulder weight = .85 lbs strength/ lb shoulder weight confidence assessment: 3
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22:48:03 The difference in lifting strength was 17 pounds, as a result of a 20 pound difference in added weight. The average rate at which strength increases with respect added weight would therefore be 17 lifting pounds / (20 added pounds) = .85 lifting pounds / added pound. The strength advantage was .85 lifting pounds per pound of added weight, on the average. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> ok confidence assessment: 3
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22:49:16 `q010. During a race, a runner passes the 100-meter mark 12 seconds after the start and the 200-meter mark 22 seconds after the start. At what average rate was the runner covering distance between those two positions?
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RESPONSE --> 22s - 12 s = 10 s 200 m - 100 m = 100 m 100m/ 10 s = 10m/s confidence assessment: 2
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22:49:27 The runner traveled 100 meters between the two positions, and required 10 seconds to do so. The average rate at which the runner was covering distance was therefore 100 meters / (10 seconds) = 10 meters / second. Again this is an average rate; at different positions in his stride the runner would clearly be traveling at slightly different speeds. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> ok confidence assessment: 3
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22:52:38 `q011. During a race, a runner passes the 100-meter mark moving at 10 meters / second, and the 200-meter mark moving at 9 meters / second. What is your best estimate of how long it takes the runner to cover the 100 meter distance?
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RESPONSE --> (10 m/s + 9 m/s)/2 = 9.5m/s 100 m / 9.5 m/s = 10.5s confidence assessment: 2
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22:52:59 At 10 meters/sec, the runner would require 10 seconds to travel 100 meters. However the runner seems to be slowing, and will therefore require more than 10 seconds to travel the 100 meters. We don't know what the runner's average speed is, we only know that it goes from 10 m/s to 9 m/s. The simplest estimate we could make would be that the average speed is the average of 10 m/s and 9 m/s, or (10 m/s + 9 m/s ) / 2 = 9.5 m/s. Taking this approximation as the average rate, the time required to travel 100 meters will be (100 meters) / (9.5 m/s) = 10.5 sec, approx.. Note that simply averaging the 10 m/s and the 9 m/s might not be the best way to approximate the average rate--for example we if we knew enough about the situation we might expect that this runner would maintain the 10 m/s for most of the remaining 100 meters, and simply tire during the last few seconds. However we were not given this information, and we don't add extraneous assumptions without good cause. So the approximation we used here is pretty close to the best we can do with the given information. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> ok confidence assessment: 3
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22:54:14 `q012. We just averaged two quantities, adding them in dividing by 2, to find an average rate. We didn't do that before. Why we do it now?
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RESPONSE --> Because we were given a starting rate and a final rate and a distance and asked to give an estimate of time. So we must use and average rate because it was given that it was not a constant rate. confidence assessment: 2
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22:54:37 In previous examples the quantities weren't rates. We were given the amount of change of some accumulating quantity, and the change in time or in some other quantity on which the first was dependent (e.g., dollars and months, miles and gallons). Here we are given 2 rates, 10 m/s and 9 m/s, in a situation where we need an average rate in order to answer a question. Within this context, averaging the 2 rates was an appropriate tactic. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> Ok confidence assessment: 3
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