#$&* course Phy 232 `q001 A straight line connects the points (3, 5) and (7, 17), while another straight line continues on from (7, 17) to the point (10, 29). Which line is steeper and on what basis to you claim your result?
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Given Solution: `aYour sketch should show that while the first trapezoid averages a little more than double the altitude of the second, the second is clearly much more than twice as wide and hence has the greater area. To justify this a little more precisely, the first trapezoid, which runs from x = 3 to x = 7, is 4 units wide while the second runs from x = 10 and to x = 50 and hence has a width of 40 units. The altitudes of the first trapezoid are 5 and 9,so the average altitude of the first is 7. The average altitude of the second is the average of the altitudes 2 and 4, or 3. So the first trapezoid is over twice as high, on the average, as the first. However the second is 10 times as wide, so the second trapezoid must have the greater area. This is all the reasoning we need to answer the question. We could of course multiply average altitude by width for each trapezoid, obtaining area 7 * 4 = 28 for the first and 3 * 40 = 120 for the second. However if all we need to know is which trapezoid has a greater area, we need not bother with this step. ********************************************* Question: `q005. Sketch graphs of y = x^2, y = 1/x and y = `sqrt(x) [note: `sqrt(x) means 'the square root of x'] for x > 0. We say that a graph increases if it gets higher as we move toward the right, and if a graph is increasing it has a positive slope. Explain which of the following descriptions is correct for each graph: As we move from left to right the graph increases as its slope increases. As we move from left to right the graph decreases as its slope increases. As we move from left to right the graph increases as its slope decreases. As we move from left to right the graph decreases as its slope decreases. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First statement:x^2 Second:x^2 Third:sqrt(x) Fourth:1/x confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aFor x = 1, 2, 3, 4: The function y = x^2 takes values 1, 4, 9 and 16, increasing more and more for each unit increase in x. This graph therefore increases, as you say, but at an increasing rate. The function y = 1/x takes values 1, 1/2, 1/3 and 1/4, with decimal equivalents 1, .5, .33..., and .25. These values are decreasing, but less and less each time. The decreasing values ensure that the slopes are negative. However, the more gradual the decrease the closer the slope is to zero. The slopes are therefore negative numbers which approach zero. Negative numbers which approach zero are increasing. So the slopes are increasing, and we say that the graph decreases as the slope increases. We could also say that the graph decreases but by less and less each time. So the graph is decreasing at a decreasing rate. For y = `sqrt(x) we get approximate values 1, 1.414, 1.732 and 2. This graph increases but at a decreasing rate. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q006. If the population of the frogs in your frog pond increased by 10% each month, starting with an initial population of 20 frogs, then how many frogs would you have at the end of each of the first three months (you can count fractional frogs, even if it doesn't appear to you to make sense)? Can you think of a strategy that would allow you to calculate the number of frogs after 300 months (according to this model, which probably wouldn't be valid for that long) without having to do at least 300 calculations? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Frist month: 2 new frogs-22 frogs Second month: 2.2 more frogs-24.2 frogs Third month: 2.42 frogs-26.62 frogs After 300 months it would be the same as 1.1*1.1 300 times. So you could figure that out by 20*1.1^300 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aAt the end of the first month, the number of frogs in the pond would be (20 * .1) + 20 = 22 frogs. At the end of the second month there would be (22 * .1) + 22 = 24.2 frogs while at the end of the third month there would be (24.2 * .1) + 24.2 = 26.62 frogs. The key to extending the strategy is to notice that multiplying a number by .1 and adding it to the number is really the same as simply multiplying the number by 1.1. We therefore get 20 * 1.1 = 22 frogs after the first month 22 * 1.1 = 24.2 after the second month etc., multiplying by for 1.1 each month. So after 300 months we will have multiplied by 1.1 a total of 300 times. This would give us 20 * 1.1^300, whatever that equals (a calculator, which is appropriate in this situation, will easily do the arithmetic). A common error is to say that 300 months at 10% per month gives 3,000 percent, so there would be 30 * 20 = 600 frogs after 30 months. That doesn't work because the 10% increase is applied to a greater number of frogs each time. 3000% would just be applied to the initial number, so it doesn't give a big enough answer. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q007. Calculate 1/x for x = 1, .1, .01 and .001. Describe the pattern you obtain. Why do we say that the values of x are approaching zero? What numbers might we use for x to continue approaching zero? What happens to the values of 1/x as we continue to approach zero? What do you think the graph of y = 1/x vs. x looks for x values between 0 and 1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: X=1->1 X=.1->10 X=.01->100 X=.001->1000 As x gets smaller the output is larger. They are approaching zero because x cannot be zero but you can get very close. You could use .0000000001. This would continue the pattern of making the values greater and greater. Between 0 and 1 the graph would almost have a vertical line. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIf x = .1, for example, 1 / x = 1 / .1 = 10 (note that .1 goes into 1 ten times, since we can count to 1 by .1, getting.1, .2, .3, .4, ... .9, 10. This makes it clear that it takes ten .1's to make 1. So if x = .01, 1/x = 100 Ithink again of counting to 1, this time by .01). If x = .001 then 1/x = 1000, etc.. Note also that we cannot find a number which is equal to 1 / 0. Deceive why this is true, try counting to 1 by 0's. You can count as long as you want and you'll ever get anywhere. The values of 1/x don't just increase, they increase without bound. If we think of x approaching 0 through the values .1, .01, .001, .0001, ..., there is no limit to how big the reciprocals 10, 100, 1000, 10000 etc. can become. The graph becomes steeper and steeper as it approaches the y axis, continuing to do so without bound but never touching the y axis. This is what it means to say that the y axis is a vertical asymptote for the graph . &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q008. At clock time t the velocity of a certain automobile is v = 3 t + 9. At velocity v its energy of motion is E = 800 v^2. What is the energy of the automobile at clock time t = 5? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: V=3(5)+9=24 800*24^2=460800 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aFor t=5, v = 3 t + 9 = (3*5) + 9 = 24. Therefore E = 800 * 24^2 = 460800. • ********************************************* Question: `q009. Continuing the preceding problem, can you give an expression for E in terms of t? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: V=3t+9 E=800v^2 (V-9)/3=t 800*((v-9)/3)^2=E confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSince v = 3 t + 9 the expression would be E = 800 v^2 = 800 ( 3t + 9) ^2. This is the only answer really required here. For further reference, though, note that this expression could also be expanded by applying the Distributive Law:. Since (3t + 9 ) ^ 2 = (3 t + 9 ) * ( 3 t + 9 ) = 3t ( 3t + 9 ) + 9 * (3 t + 9) = 9 t^2 + 27 t + 27 t + 81 = 9 t^2 + 54 t + 81, we get E = 800 ( 9 t^2 + 54 t + 81) = 7200 t^2 + 43320 t + 64800 (check my multiplication because I did that in my head, which isn't always reliable). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: For what x values is the value of the expression (2^x - 1) ( x^2 - 25 ) ( 2x + 6) zero? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Any of the factors must equal 0 2^x-1=0-> 2^x=1 x=0 X^2-25=0->sqrt(25)=x=5
#$&* course Phy 232 Question: `q001. There are 12 questions in this document.
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Given Solution: `aSlope = rise / run. Between points (7, 17) and (10, 29) we get rise / run = (29 - 17) / (10 - 7) =12 / 3 = 4. The slope between points (3, 5) and (7, 17) is 3 / 1. (17 - 5) / (7 -3) = 12 / 4 = 3. The segment with slope 4 is the steeper. The graph being a smooth curve, slopes may vary from point to point. The slope obtained over the interval is a specific type of average of the slopes of all points between the endpoints. 2. Answer without using a calculator: As x takes the values 2.1, 2.01, 2.001 and 2.0001, what values are taken by the expression 1 / (x - 2)? 1. As the process continues, with x getting closer and closer to 2, what happens to the values of 1 / (x-2)? It approaches infinity. 2. Will the value ever exceed a billion? Will it ever exceed one trillion billions? Yes with enough 0s between the decimal and the one. 3. Will it ever exceed the number of particles in the known universe? Yes the function will go to infinity. 4. Is there any number it will never exceed? Infinity itself 5. What does the graph of y = 1 / (x-2) look like in the vicinity of x = 2? On the right side there is a large spike upward and on the left side of x=2 there is a large spike downward. At x=2 there is an asymptote. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It approaches infinity. Yes with enough 0s between the decimal and the one. Yes the function will go to infinity. Infinity itself. On the right side there is a large spike upward and on the left side of x=2 there is a large spike downward. At x=2 there is an asymptote. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aFor x = 2.1, 2.01, 2.001, 2.0001 we see that x -2 = .1, .01, .001, .0001. Thus 1/(x -2) takes respective values 10, 100, 1000, 10,000. It is important to note that x is changing by smaller and smaller increments as it approaches 2, while the value of the function is changing by greater and greater amounts. As x gets closer in closer to 2, it will reach the values 2.00001, 2.0000001, etc.. Since we can put as many zeros as we want in .000...001 the reciprocal 100...000 can be as large as we desire. Given any number, we can exceed it. Note that the function is simply not defined for x = 2. We cannot divide 1 by 0 (try counting to 1 by 0's..You never get anywhere. It can't be done. You can count to 1 by .1's--.1, .2, .3, ..., .9, 1. You get 10. You can do similar thing for .01, .001, etc., but you just can't do it for 0). As x approaches 2 the graph approaches the vertical line x = 2; the graph itself is never vertical. That is, the graph will have a vertical asymptote at the line x = 2. As x approaches 2, therefore, 1 / (x-2) will exceed all bounds. Note that if x approaches 2 through the values 1.9, 1.99, ..., the function gives us -10, -100, etc.. So we can see that on one side of x = 2 the graph will approach +infinity, on the other it will be negative and approach -infinity. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q003. One straight line segment connects the points (3,5) and (7,9) while another connects the points (10,2) and (50,4). From each of the four points a line segment is drawn directly down to the x axis, forming two trapezoids. Which trapezoid has the greater area? Try to justify your answer with something more precise than, for example, 'from a sketch I can see that this one is much bigger so it must have the greater area'. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Cut the trapezoids into a triangle and a rectangle. The first trapezoid has an area of 28 because of a 4x4 square and a 4x4 (area 20 due to A=bh) triangle (area 8 due to A=1/2bh). The second trapezoid is much larger with an area of 120 because of a 40x2 rectangle (area 80 because of A=bh) and a 40x2 triangle (area 40). confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aYour sketch should show that while the first trapezoid averages a little more than double the altitude of the second, the second is clearly much more than twice as wide and hence has the greater area. To justify this a little more precisely, the first trapezoid, which runs from x = 3 to x = 7, is 4 units wide while the second runs from x = 10 and to x = 50 and hence has a width of 40 units. The altitudes of the first trapezoid are 5 and 9,so the average altitude of the first is 7. The average altitude of the second is the average of the altitudes 2 and 4, or 3. So the first trapezoid is over twice as high, on the average, as the first. However the second is 10 times as wide, so the second trapezoid must have the greater area. This is all the reasoning we need to answer the question. We could of course multiply average altitude by width for each trapezoid, obtaining area 7 * 4 = 28 for the first and 3 * 40 = 120 for the second. However if all we need to know is which trapezoid has a greater area, we need not bother with this step. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. If f(x) = x^2 (meaning 'x raised to the power 2') then which is steeper, the line segment connecting the x = 2 and x = 5 points on the graph of f(x), or the line segment connecting the x = -1 and x = 7 points on the same graph? Explain the basis of your reasoning. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Slope of x=2 to x=5-> (25-4)/(5-2)=7 Slope of x=-1 to x=7-> (49-1)/(7—1)=6 There for it is steeper from x=2 to x=5 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe line segment connecting x = 2 and the x = 5 points is steeper: Since f(x) = x^2, x = 2 gives y = 4 and x = 5 gives y = 25. The slope between the points is rise / run = (25 - 4) / (5 - 2) = 21 / 3 = 7. The line segment connecting the x = -1 point (-1,1) and the x = 7 point (7,49) has a slope of (49 - 1) / (7 - -1) = 48 / 8 = 6. The slope of the first segment is greater. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q005. Suppose that every week of the current millennium you go to the jeweler and obtain a certain number of grams of pure gold, which you then place in an old sock and bury in your backyard. Assume that buried gold lasts a long, long time ( this is so), that the the gold remains undisturbed (maybe, maybe not so), that no other source adds gold to your backyard (probably so), and that there was no gold in your yard before.. 1. If you construct a graph of y = the number of grams of gold in your backyard vs. t = the number of weeks since Jan. 1, 2000, with the y axis pointing up and the t axis pointing to the right, will the points on your graph lie on a level straight line, a rising straight line, a falling straight line, a line which rises faster and faster, a line which rises but more and more slowly, a line which falls faster and faster, or a line which falls but more and more slowly? 2. Answer the same question assuming that every week you bury 1 more gram than you did the previous week. 3. Answer the same question assuming that every week you bury half the amount you did the previous week. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The first question is undetermined. However assuming the certain amount of gold you collect each week is constant then the slope will be constant because the after one week you could have 1 gram then two weeks with two grams and it would go on and on. If you buy an extra gram each week then the line would getting steeper at a constant rate. If you bought half than the week before then your line would be getting less steep and falls more and more slowly. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a1. If it's the same amount each week it would be a straight line. 2. Buying gold every week, the amount of gold will always increase. Since you buy more each week the rate of increase will keep increasing. So the graph will increase, and at an increasing rate. 3. Buying gold every week, the amount of gold won't ever decrease. Since you buy less each week the rate of increase will just keep falling. So the graph will increase, but at a decreasing rate. This graph will in fact approach a horizontal asymptote, since we have a geometric progression which implies an exponential function. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q006. Suppose that every week you go to the jeweler and obtain a certain number of grams of pure gold, which you then place in an old sock and bury in your backyard. Assume that buried gold lasts a long, long time, that the the gold remains undisturbed, and that no other source adds gold to your backyard. 1. If you graph the rate at which gold is accumulating from week to week vs. the number of weeks since Jan 1, 2000, will the points on your graph lie on a level straight line, a rising straight line, a falling straight line, a line which rises faster and faster, a line which rises but more and more slowly, a line which falls faster and faster, or a line which falls but more and more slowly? 2. Answer the same question assuming that every week you bury 1 more gram than you did the previous week. 3. Answer the same question assuming that every week you bury half the amount you did the previous week. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A flat straight line. A increasing straight line. A decreasing slope but the value will never equal 0. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThis set of questions is different from the preceding set. This question now asks about a graph of rate vs. time, whereas the last was about the graph of quantity vs. time. Question 1: This question concerns the graph of the rate at which gold accumulates, which in this case, since you buy the same amount eact week, is constant. The graph would be a horizontal straight line. Question 2: Each week you buy one more gram than the week before, so the rate goes up each week by 1 gram per week. You thus get a risingstraight line because the increase in the rate is the same from one week to the next. Question 3. Since half the previous amount will be half of a declining amount, the rate will decrease while remaining positive, so the graph remains positive as it decreases more and more slowly. The rate approaches but never reaches zero. STUDENT COMMENT: I feel like I am having trouble visualizing these graphs because every time for the first one I picture an increasing straight line INSTRUCTOR RESPONSE: The first graph depicts the amount of gold you have in your back yard. The second depicts the rate at which the gold is accumulating, which is related to, but certainly not the same as, the amount of gold. For example, as long as gold is being added to the back yard, the amount will be increasing (though not necessarily on a straight line). However if less and less gold is being added every year, the rate will be decreasing (perhaps along a straight line, perhaps not). FREQUENT STUDENT RESPONSE This is the same as the problem before it. No self-critique is required. INSTRUCTOR RESPONSE This question is very different that the preceding, and in a very significant and important way. You should have self-critiqued; you should go back and insert a self-critique on this very important question and indicate your insertion by preceding it with ####. The extra effort will be more than worth your trouble. These two problems go to the heart of the Fundamental Theorem of Calculus, which is the heart of this course, and the extra effort will be well worth it in the long run. The same is true of the last question in this document. STUDENT COMMENT Aha! Well you had me tricked. I apparently misread the question. Please don’t do this on a test! INSTRUCTOR RESPONSE I don't usually try to trick people, and wasn't really trying to do so here, but I was aware when writing these two problems that most students would be tricked. My real goal: The distinction between these two problems is key to understanding what calculus is all about. I want to at least draw your attention to it early in the course. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:
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Given Solution: `aAt t = 30 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 30 + .01 * 30^2 = 49. At t = 40 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 40 + .01 * 40^2 = 36. At t = 60 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 60 + .01 * 60^2 = 16. 49 cm - 36 cm = 13 cm change in 10 sec or 1.3 cm/s on the average. 36 cm - 16 cm = 20 cm change in 20 sec or 1.0 cm/s on the average. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q008. If the rate at which water descends in a container is given, in cm/s, by 10 - .1 t, where t is clock time in seconds, then at what rate is water descending when t = 10, and at what rate is it descending when t = 20? How much would you therefore expect the water level to change during this 10-second interval? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: T=10->10-.1(10)=9 T=20->10-.1(20)=8 Average change=8.5cm/s 8.5cm/s*10s=85cm
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Given Solution: `aAt t = 10 sec the rate function gives us 10 - .1 * 10 = 10 - 1 = 9, meaning a rate of 9 cm / sec. At t = 20 sec the rate function gives us 10 - .1 * 20 = 10 - 2 = 8, meaning a rate of 8 cm / sec. The rate never goes below 8 cm/s, so in 10 sec the change wouldn't be less than 80 cm. The rate never goes above 9 cm/s, so in 10 sec the change wouldn't be greater than 90 cm. Any answer that isn't between 80 cm and 90 cm doesn't fit the given conditions.. The rate change is a linear function of t. Therefore the average rate is the average of the two rates, or 8.5 cm/s. The average of the rates is 8.5 cm/sec. In 10 sec that would imply a change of 85 cm. STUDENT RESPONSES The following, or some variation on them, are very common in student comments. They are both very good questions. Because of the importance of the required to answer this question correctly, the instructor will typically request for a revision in response to either student response: I don't understand how the answer isn't 1 cm/s. That's the difference between 8 cm/s and 9 cm/s. I don't understand how the answer isn't 8.5 cm/s. That's the average of the 8 cm/s and the 9 cm/s. INSTRUCTOR RESPONSE A self-critique should include a full statement of what you do and do not understand about the given solution. A phrase-by-phrase analysis of the solution is not unreasonable (and would be a good idea on this very important question), though it wouldn't be necessary in most situations. An important part of any self-critique is a good question, and you have asked one. However a self-critique should if possible go further. I'm asking that you go back and insert a self-critique on this very important question and indicate your insertion by preceding it with ####, before submitting it. The extra effort will be more than worth your trouble. This problem, along with questions 5 and 6 of this document, go to the heart of the Fundamental Theorem of Calculus, which is the heart of this course, and the extra effort will be well worth it in the long run. You should review the instructions for self-critique, provided at the link given at the beginning of this document. STUDENT COMMENT The question is worded very confusingly. I took a stab and answered correctly. When answering, """"How much would you therefore expect the water level to change during this 10-second interval?"""" It is hard to tell whether you are asking for what is the expected change in rate during this interval and what is the changing """"water level."""" But now, after looking at it, with your comments, it is clearer that I should be looking for the later. Thanks! INSTRUCTOR RESPONSE 'Water level' is clearly not a rate. I don't think there's any ambiguity in what's being asked in the stated question. The intent is to draw the very important distinction between the rate at which a quantity changes, and the change in the quantity. It seems clear that as a result of this question you understand this and will be more likely to make such distinctions in your subsequent work. This distinction is at the heart of the calculus and its applications. It is in fact the distinction between a derivative and an integral. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q009. Sketch the line segment connecting the points (2, -4) and (6, 4), and the line segment connecting the points (2, 4) and (6, 1). The first of these lines if the graph of the function f(x), the second is the graph of the function g(x). Both functions are defined on the interval 2 <= x <= 6. Let h(x) be the function whose value at x is the product of the values of these two functions. For example, when x = 2 the value of the first function is -4 and the value of the second is 4, so when x = 2 the value of h(x) is -4 * 4 = -16. Answer the following based just on the characteristics of the graphs you have sketched. (e.g., you could answer the following questions by first finding the formulas for f(x) and g(x), then combining them to get a formula for h(x); that's a good skill but that is not the intent of the present set of questions). What is the value of h(x) when x = 6? Is the value of h(x) ever greater than its value at x = 6? What is your best description of the graph of h(x)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4*1=4 The slope of the first line is 2 and the slope of the second line is -3/4. At x=4 the vales are 0 and 2.5. This comes to zero. After a few trials there is no answer greater than 4. The graph is fluctuating line that has a range from -16 to 4
#$&* course Phy 232 Question: `q001. There are two parts to this problem. Reason them out using common sense.
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Given Solution: `aIt will take 5 seconds to complete the change. 30 mph - 20 mph = 10 mph change at 2 mph per second (i.e., 2 mph every second) implies 5 seconds to go from 20 mph to 30 mph Change in speed is 2 mph/second * 7 seconds = 14 mph Add this to the initial 10 mph and the speedometer now reads 24 mph. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q002. An automobile traveling down a hill passes a certain milepost traveling at a speed of 10 mph, and proceeds to coast to a certain lamppost further down the hill, with its speed increasing by 2 mph every second. The time required to reach the lamppost is 10 seconds. It then repeats the process, this time passing the milepost at a speed of 20 mph. This time: Will the vehicle require more or less than 10 seconds to reach the lamppost? Since its initial speed was 10 mph greater than before, does it follow that its speed at the lamppost will be 10 mph greater than before? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It will take less time and therefore it will not be 10 mph greater because it has less time to accelerate. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIf 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q003. The following example shows how we can measure the rate at which an automobile speeds up: If an automobile speeds up from 30 mph to 50 mph as the second hand of a watch moves from the 12-second position to the 16-second position, and its speed changes by 20 mph in 4 seconds. This gives us an average rate of velocity change equal to 20 mph / 4 seconds = 5 mph / second. We wish to compare the rates at which two different automobiles increase their speed: Which automobile speeds up at the greater rate, one which speeds up from 20 mph to 30 mph in five seconds or one which speeds up from 40 mph to 90 mph in 20 seconds? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The one going from 40-90 10/5=.5mph per s 50/20=2.5 mph per s confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The first automobile's speed changes from 20 mph to 30mph, a 10 mph difference, which occurs in 5 seconds. So the rate of chage in 10 mph / (5 sec) = 2 mph / sec. = rate of change of 2 mph per second. The second automobile's speed changes from 40 mph to 90 mph, a 50 mph difference in 20 seconds so the rate of change is 50 mph / (20 sec) = 2.5 mph per second. Therefore, the second auto is increasing its velocity ar a rate which is .5 mph / second greater than that of the first. Self-critique: ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. If an automobile of mass 1200 kg is pulled by a net force of 1800 Newtons, then the number of Newtons per kg is 1800 / 1200 = 1.5. The rate at which an automobile speeds up is determined by the net number of Newtons per kg. Two teams pulling on ropes are competing to see which can most quickly accelerate their initially stationary automobile to 5 mph. One team exerts a net force of 3000 Newtons on a 1500 kg automobile while another exerts a net force of 5000 Newtons on a 2000 kg automobile. Which team will win and why? If someone pulled with a force of 500 Newtons in the opposite direction on the automobile predicted to win, would the other team then win? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 3000/1500=2 N per kg 5000/2000=2.5 N per kg. The second team would win After opposing force the resultant would be 4500/2000-2.25 N per kg. The team would still win. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe first team's rate is 3000 Newtons divided by 1500 kg or 2 Newtons per kg, while the second team's rate is 5000 Newtons divided by 2000 kg or 2.5 Newtons per kg. The second team therefore increases velocity more quickly. Since both start at the same velocity, zero, the second team will immediately go ahead and will stay ahead. The second team would still win even if the first team was hampered by the 500 Newton resistance, because 5000 Newtons - 500 Newtons = 4500 Newtons of force divided by 2000 kg of car gives 2.25 Newtons per kg, still more than the 2 Newtons / kg of the first team &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q005. Both the mass and velocity of an object contribute to its effectiveness in a collision. If a 250-lb football player moving at 10 feet per second collides head-on with a 200-lb player moving at 20 feet per second in the opposite direction, which player do you precidt will be moving backward immediately after the collision, and why? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Momentum- 250*10=2500 Vs. 200*20=4000 The 250 lb player would move backward because he has less momentum. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aGreater speed and greater mass both provide advantages. In this case the player with the greater mass has less speed, so we have to use some combination of speed and mass to arrive at a conclusion. It turns out that if we multiply speed by mass we get the determining quantity, which is called momentum. 250 lb * 10 ft/sec = 2500 lb ft / sec and 200 lb * 20 ft/sec = 4000 lb ft / sec, so the second player will dominate the collision. In this course we won't use pounds as units, and in a sense that will become apparent later on pounds aren't even valid units to use here. However that's a distinction we'll worry about when we come to it. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q006. Two climbers eat Cheerios for breakfast and then climb up a steep mountain as far as they can until they use up all their energy from the meal. All other things being equal, who should be able to climb further up the mountain, the 200-lb climber who has eaten 12 ounces of Cheerios or the 150-lb climber who has eaten 10 ounces of Cheerios? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 12/200=.06 10/150=.067 The 150 lb climber has more energy per lb. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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. STUDENT COMMENT I am satisfied with how I worked out the problem, though it would be nice to know what formulas to use in case my instinct is wrong. I should have got the energy used per pound by rereading the question. INSTRUCTOR RESPONSE There are two points to these problems: 1. You can go a long ways with common sense, intuition or instinct, and you often don't need formulas. 2. Common sense, intuition and instinct aren't the easiest things to apply correctly, and it's really easy to get things turned around. A corollary: When we do use formulas it will be important to understand them, as best we can, in terms of common sense and experience. Either way, practice makes the process easier, and one of the great benefits of studying physics is that we get the opportunity to apply common sense in situations where we can get feedback by experimentally testing our thinking. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q007. Two automobiles are traveling up a long hill with an steepness that doesn't change until the top, which is very far away, is reached. One automobile is moving twice as fast as the other. At the instant the faster automobile overtakes the slower their drivers both take them out of gear and they coast until they stop. Which automobile will take longer to come to a stop? Will that automobile require about twice as long to stop, more than twice as long or less than twice as long? Which automobile will have the greater average coasting velocity? Will its average coasting velocity by twice as great as the other, more than twice as great or less than twice as great? Will the distance traveled by the faster automobile be equal to that of the slower, twice that of the slower or more than twice that of the slower? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The faster car will take longer to stop, take twice as long assuming all things constant. It will also have a greater average coasting velocity of twice as great and the distance will be 4 times that of the slower car. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIt 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. STUDENT COMMENT: I do not understand why the car would go four times as far as the slower car. INSTRUCTOR RESPONSE: The faster car takes twice as long to come to rest, and have twice the average velocity. If the car traveled at the same average velocity for twice as long it would go twice as far. If it traveled at twice the average velocity for the same length of time it would go twice as far. However it travels at twice the average velocity for twice as long, so it goes four times as far. STUDENT COMMENT: it’s hard to know this stuff without having first discussed it in notes or read it in the book, or have an equation handy. I guess this will all come with the class. INSTRUCTOR RESPONSE One purpose of this and similar exercises is to get students into the habit of thinking for themselves, as opposed to imitating what they see done in a textbook. You're doing some good thinking. When you get to the text and other materials, ideally you'll be better prepared to understand them as a result of this process. This works better for some students than others, but it's beneficial to just about everyone. STUDENT COMMENT I understand, it seems as though it would be easier if there were formulas to apply. I used a little common sense on all but the last one. Reading the responses I somewhat understand the last one. ?????The problem doesn’t indicate the vehicle travels twice the average velocity for twice as long. Should I have known that by reading the problem or should that have become clear to me after working it some????? INSTRUCTOR RESPONSE You did know these things when you thought about the problem. You concluded that the automobile would take twice as long to come to rest, and that it would have twice the coasting velocity. You just didn't put the two conclusions together (don't feel badly; very few students do, and most don't get as close as you did). You should now see how your two correct conclusions, when put together using common sense, lead to the final conclusion that the second automobile travels four times as far. No formula is necessary to do this. In fact if students are given a formula, nearly all will go ahead and use it without ever thinking about or understanding what is going on. In this course we tend to develop an idea first, and then summarize the idea with one or more formulas. Once we've formulated a concept, the formula gives us a condensed expression of our understanding. The formula then becomes a means of remembering the ideas it represents, and gives us a tool to probe even more deeply into the relationships it embodies. There are exceptions in which we start with a formula, but usually by the time we get to the formula we will understand, at least to some extent, what it's about. I suppose this could be put succinctly as 'think before formulating'. STUDENT COMMENT I feel that I did decent on the problem, but I am the student that likes to have formulas. Your insight has opened my eyes to a different way of looking at this problem. I like the comment “Think before Formulate” INSTRUCTOR RESPONSE Your solution was indeed well thought out. I should probably add another comment: 'Think after formulating.' Formulas are essential, but can't be applied reliably without the thinking, which should come first and last. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 100 lbs to 150 lbs stretches 4 more feet. 150 lbs to 200 lbs stretches 3 more feet. 125 lb stretch is halfway but the other half of weight requires less than 2 ft of stretch. Therefore it wuld stretch more than 7 ft. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aFrom 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). STUDENT COMMENT I feel like I nailed this one. Probably just didn’t state things very clearly. INSTRUCTOR RESPONSE You explanation was very good. Remember that I get to refine my statements, semester after semester, year after year. You get one shot and you don't have time to hone it to perfection (not to say that my explanations ever achieve that level). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q009. When given a push of 10 pounds, with the push maintained through a distance of 4 feet, a certain ice skater can coast without further effort across level ice for a distance of 30 feet. When given a push of 20 pounds (double the previous push) through the same distance, the skater will be able to coast twice as far, a distance of 60 feet. When given a push of 10 pounds for a distance of 8 feet (twice the previous distance) the skater will again coast a distance of 60 feet. The same skater is now accelerated by a sort of a slingshot consisting of a bungee-type cord slung between two posts in the ice. The cord, as one might expect, exerts greater and greater force as it is pulled back further and further. Assume that the force increases in direct proportion to pullback (ie.g., twice the pullback implies twice the force). When the skater is pulled back 4 feet and released, she travels 20 feet. When she is pulled back 8 feet and released, will she be expected to travel twice as far, more than twice as far or less than twice as far as when she was pulled back 4 feet? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4 times as far because you have 2 doubling effects. One with the doubled distance and the other being the doubled pullback. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe distance through which the force acts will be twice as great, which alone would double the distance; because of the doubled pullback and the linear proportionality relationship for the force the average force is also twice as great, which alone would double the distance. So we have to double the doubling; she will go 4 times as far STUDENT COMMENT: I do not understand the linear proportionality relationship for the force. If the skater is pulled back an extra four feet, does that mean that the amount of pounds propelling her is also doubled? INSTRUCTOR COMMENT: That is so. However the force propelling her isn't the only thing that influences how far she slides. The distance through which the force is applied is also a factor. Doubling the force alone would double the sliding distance. Doubling the distance through which the force is applied would double the sliding distane. Doubling both the applied force and the distance through which it is applied quadruples the sliding distance. STUDENT SOLUTION AND QUESTION She should travel three times as far. The first four feet pulled back yield 20 feet of travel. The second four feet (i.e., feet 5 through 8) will propel her with twice the force as the first four feet. So this interval, by itself, would propel her 40 feet. The 20 feet of the first four-foot interval plus the 40 feet of the second four-foot interval is 60 feet total. But wouldn’t it be the case that by the time the slingshot reaches the four-foot position, the force exerted on the skater would only be half of that exerted when she was eight feet out? I understand why it would be a multiplier of four if the force were the same throughout, but I’m assuming that the force will decrease as the slingshot is contracts. I would appreciate help with this question. Thanks. INSTRUCTOR RESPONSE The average force for the entire 8-foot pull would be double the average force for the 4-foot pull. At this point we don't want to get too mathematical so we'll stick to a numerical plausibility argument. This argument could be made rigorous using calculus (just integrate the force function with respect to position), but the numerical argument should be compelling: Compare the two pulls at the halfway point of each. For a convenient number assume that the 4-foot pull results in a force of 100 lb. Then the 8-foot pull will therefore exert a force of 200 lb. When released at the 4-foot mark, the skater will be halfway back at the 2-foot mark, where she will experience a 50-lb force. When released at the 8-foot mark, the skater will be halfway back at the 4-foot mark, where she will experience a 100-lb force. Since the force is proportional to pullback, the halfway force is in fact the average force. Note that during the second 4 ft of the 8 ft pull the force goes from 100 lb to 200 lb, so the average force for the second 4 ft is 150 lb, three times as great as the average force for the first 4 ft. The max force for the second 4 ft is double that of the first 4 ft, but the second 4 ft starts out with 100 lbs of force, while the first 4 ft starts out with 0 lbs. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The first moth will not tell a difference because from a distance they will appear identical. The second moth will not notice the dimmer larger bulb because it is on the surface of the brighter bulb confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aBoth bulbs send out the same energy per second. The surface of the second bulb will indeed be dimmer than the first, as we will see below. However the same total energy per second reaches the eye (identically frosted bulbs will dissipate the same percent of the bulb energy) and from a great distance you can't tell the difference in size, so both will appear the same. The second sphere, while not as bright at its surface because it has proportionally more area, does have the extra area, and that exactly compensates for the difference in brightness. Specifically the brightness at the surface will be 1/4 as great (twice the radius implies 4 times the area which results in 1/4 the illumination at the surface) but there will be 4 times the surface area. Just as a 2' x 2' square has four times the area of a 1' x 1' square, a sphere with twice the diameter will have four times the surface area and will appear 1 / 4 as bright at its surface. Putting it another way, the second sphere distributes the intensity over four times the area, so the light on 1 square inch has only 1 / 4 the illumination. STUDENT COMMENT: I understand the first part of the problem about the distances. But the second part really confuses me. Looking straight down from the top of the spheres, the bulb is the same intensity and the frosted glass is exactly the same, so why would it seem dimmer? I would think that if a person was standing in front of the spheres, that person would be able to tell a difference, but not extremely close. INSTRUCTOR RESPONSE: Imagine a light bulb inside a frosted glass lamp of typical size. Imagine it outside on a dark night. If you put your eye next to the glass, the light will be bright. Not as bright as if you put your eye right next to the bulb, but certainly bright. The power of the bulb is spread out over the lamp, but the lamp doesn't have that large an area so you detect quite a bit of light. If you put the same bulb inside a stadium with a frosted glass dome over it, and put your eye next to the glass on a dark night, with just the bulb lit, you won't detect much illumination. The power of the bulb is distributed over a much greater area than that of the lamp, and you detect much less light. STUDENT COMMENT: I also didn’t get the second part of the question. I still don’t really see where the ¼ comes from. INSTRUCTOR RESPONSE: First you should address the explanation given in the problem: 'Just as a 2' x 2' square has four times the area of a 1' x 1' square, a sphere with twice the diameter will have four times the surface area and will appear 1 / 4 as bright at its surface. Putting it another way, the second sphere distributes the intensity over four times the area, so the light on 1 square inch has only 1 / 4 the illumination. ' Do you understand this explanation? If not, what do you understand about it and what don't you understand? This simple image of a 2x2 square being covered by four 1x1 squares is the most basic reason the larger sphere has four time the area of the smaller. There is, however, an alternative explanation in terms of formulas: The surface area of a sphere is 4 pi r^2. If r is doubled, r^2 increases by factor 2^2 = 4. So a sphere with double the radius has four time the area. If the same quantity is spread out over the larger sphere, it will be 1/4 as dense on the surface. STUDENT COMMENT: I also have no clue why the extra area doesn’t take away some brightness. INSTRUCTOR RESPONSE: All the light produced by the bulb is passing through either of the spheres. From a distance you see all the light, whichever sphere you're looking at; you see just as much light when looking at one as when looking at the other. From a distance you can't tell whether you're looking at the sphere with larger area but less intensity at its surface, or the sphere with lesser area and greater intensity at its surface. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Melting the ice at its melting point. Increasing the temperature of the water by 20 degrees after all the ice melted. Increasing the temperature of the ice by 20 degrees to reach its melting point. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q012. Suppose you are in the center of a long, narrow swimming pool (e.g., a lap pool). Two friends with kickboards, one at either end of the pool, are using them to push waves in your direction. Their pushes are synchronized, and the crests of the waves are six feet apart as they travel toward you, with a 'valley' between each pair of crests. Since your friends are at equal distances from you the crests from both directions always reach you at the same instant, so every time the crests reach you the waves combine to create a larger crest. Similarly when the valleys meet you experience a larger valley, and as a result you bob up and down further than you would if just one person was pushing waves at you. Now if you move a bit closer to one end of the pool the peak from that end will reach you a bit earlier, and the peak from the other end will reach you a little later. So the peaks won't quite be reaching you simultaneously, nor will the valleys, and you won't bob up and down as much. If you move far enough, in fact, the peak from one end will reach you at the same time as the valley from the other end and the peak will 'fll in' the valley, with the result that you won't bob up and down very much. If the peaks of the approaching waves are each 6 inches high, how far would you expect to bob up and down when you are at the center point? How far would you have to move toward one end or the other in order for peaks to meet valleys, placing you in relatively calm water? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You should bob 12 inches up and then 12 inches down from equilibrium. If you move 3 feet the crests and valleys will be matched up and would cancel each other out. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIf 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q013. This problem includes some questions that are fairly straightfoward, some that involve more complicated considerations, and possibly some that can't be answered without additional information. We're hoping for some correct answers, but we expect that few students coming into this course will be able to think correctly through every nuance of the more complex situations. On these questions we are hoping for your best thinking without being particularly concerned with the final answer. A steel ball and a wood ball are both thrown upward and, between release and coming to rest at maximum height, both rise with the same average speed. If not for air resistance they would both come to rest at the same time, at the same height. However air resistance causes the wood ball to stop rising more quickly than the steel ball. Each ball, having risen to its maximum height, then falls back to the ground. Which ball would you expect to have the greater average velocity as it falls? Which ball would you expect to spend the greater time falling? Which ball would you expect to hit the ground first? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The steel ball because it rises higher since the air resistance slows the wood ball faster. The average velocity should be the same for both balls because of the gravitational constant of 9.8m/s^2 The steel ball would fall for longer because it goes higher. The balls should hit the ground at the same time. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q014. If you double the voltage across a certain circuit you double the current passing through it. The power required to maintain the circuit is equal to the product of the current and the voltage. How many times as much power is required if the voltage is doubled? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: P=current*V 1/4 the power because you would have 1/4P=2current*2V