course Phy 121 ǚ}Yy|ډXkTassignment #007 007. Precalculus I 02-10-2008
......!!!!!!!!...................................
00:41:25 `q001. Note that this assignment has 8 questions Sketch a graph of the following (x, y) points: (1,2), (3, 5), (6, 6). Then sketch the straight line which appears to come as close as possible, on the average, to the four points. Your straight line should not actually pass through any of the given points. Describe how your straight line lies in relation to the points. Give the coordinates of the point at which your straight line passes through the y axes, and give the coordinates of the x = 2 and x = 7 points on your straight line. Determine the slope of the straight line between the last two points you gave.
......!!!!!!!!...................................
RESPONSE --> It passes through the y-axis at coordinate (0, 2). At x = 2 and x = 7 the coordinates would be (2, 3) and (7, 8) The slope of the line would be (8-3)/(7-2) which equals 5/5 or 1. confidence assessment: 3
.................................................
......!!!!!!!!...................................
00:48:34 `q004. The equation y = x + 1.4 gives you points (1, 2.2), (3, 3.8), and (6,6.2). How close, on the average, do these points come to the original data points (1,2), (3, 5), and (6, 6)?
......!!!!!!!!...................................
RESPONSE --> Plugging m = 1 into the first equation we get 3 = 2 * 1 + b, so 3 = 2 + b and b = 3 - 2 = 1. Now the equation y = m x + b becomes y = x + 1. confidence assessment:
.................................................
......!!!!!!!!...................................
01:02:27 `q005. Using the best-fit equation y =.76 x + 1.79, with the numbers accurate to the nearest .01, how close do the predicted points corresponding to x = 1, 3, and 6 come to the original data points (1,2), (3, 5), and (6, 6)?
......!!!!!!!!...................................
RESPONSE --> For 1, 3, and 6 using the best-fit equation y = .76x + 1.79 we get the following coordinates: (1, 2.55), (3, 4.07), (6, 6.35) These three points on average differ from the the original data points by -.01 confidence assessment: 3
.................................................
......!!!!!!!!...................................
01:03:51 Evaluating y =.76 x + 1.79 at x = 1, 3 and 6 we obtain y values 2.55, 4.07 and 6.35. This gives us the points (1,2.55), (3,4.07) and (6, 6.35). These points lie at distances of .55, .93, and .35 from the original data points. The average distance is (.55 + .93 + .35) / 3 = .58 from the points.
......!!!!!!!!...................................
RESPONSE --> I messed up the average distance because I subtracted .93. I thought we should since it was below the given data point. self critique assessment: 3
.................................................
......!!!!!!!!...................................
01:06:30 The square distances for the best-fit model are .55^2 = .30, .93^2 = .87 and .35^2 = .12. The average of these square distances is (.30 + .87 + .12) / 3 = .43. The squared distances for the first model are .2^2 = .04, 1.2^2 = 1.44 and .2^2 = .04, so the average of the square distances for this model is (.04 + 1.44 + .04) / 3 = .51. Thus the best-fit model does give the better result. We won't go into the reasons here why it is desirable to minimize the square of the distance rather than the distance. When doing eyeball estimates, you don't really need to take this subtlety into account. You can simply try to get is close is possible, on the average, to the points.
......!!!!!!!!...................................
RESPONSE --> The square distances for the best-fit model are .55^2 = .30, .93^2 = .87 and .35^2 = .12. The average of these square distances is (.30 + .87 + .12) / 3 = .43 The squared distances for the first model are .2^2 = .04, 1.2^2 = 1.44 and .2^2 = .04, so the average of the square distances for this model is (.04 + 1.44 + .04) / 3 = .51 This tells us that the best fit model is the better fit. self critique assessment: 3
.................................................
......!!!!!!!!...................................
01:13:46 `q007. If the original data points (1,2), (3, 5), and (6, 6) represent the selling price in dollars of a bag of widgets vs. the number of widgets in the bag, then how much is paid for a bag of 3 widgets? How much would you estimate a bag of 7 widgets would cost?
......!!!!!!!!...................................
RESPONSE --> 3 widgets would cost 5 dollars. 7 widgets would cost around $2.80 according to my graph and calculation because the line is decreasing at an increasing rate. confidence assessment: 1
.................................................
......!!!!!!!!...................................
01:15:28 If we use the best-fit function y =.76 x + 1.79, noting that y represents the cost and x the number of widgets, then the cost of 3 widgets is y = .76 * 3 + 1.79 = 4.05, representing cost of $4.05. The cost of 7 widgets would be y = .76 * 7 + 1.79 = 7.11. The cost of 7 widgets would be $7.11.
......!!!!!!!!...................................
RESPONSE --> I didn't use the y = .76 + 1.79 function. I must have misread it because I came up with a whole new funtion using the three coordinates given, but I do fully understand now self critique assessment: 3
.................................................
......!!!!!!!!...................................
01:20:05 `q008. According to the function y = .8 x + 1.4, how much will a bag of 7 widgets cost? How many widgets would you expect to get for $10?
......!!!!!!!!...................................
RESPONSE --> 7 widgets cost would be calculated as follows: y = .8(7) + 1.4 y = $7.00 If I paid $10 I would calculate the number purchased as follows: 10 = .8x + 1.4 10 - 1.4 = .8x 8.6 = .8x 8.6 / .8 = x = 10.75 widgets confidence assessment: 3
.................................................
......!!!!!!!!...................................
01:20:41 end program
......!!!!!!!!...................................
RESPONSE --> OK self critique assessment: 3
.................................................
}MwLV瑝|³ assignment #008 008. Using the Acceleration of Gravity Physics I 02-10-2008 ޜfR{|{ assignment #008 008. Using the Acceleration of Gravity Physics I 02-10-2008
......!!!!!!!!...................................
17:18:48 You would have an initial vertical velocity of 0, and would accelerate at 9.8 m/s^2 in the same direction as your 1 meter vertical displacement. You would also have a slight horizontal velocity (you don't step off of a table without moving a bit in the horizontal direction, and you would very likely maintain a small horizontal velocity as you fell), but this would have no effect on your vertical motion. So your vertical velocity is a uniform acceleration with v0 = 0, `ds = 1 meter and a = 9.8 m/s^2. The equation vf^2 = v0^2 + 2 a `ds contains the three known variables and can therefore be used to find the desired final velocity. We obtain vf = +- `sqrt( v0^2 + 2 a `ds) = +- `sqrt ( 0^2 + 2 * 9.8 m/s^2 * 1 m)= +- `sqrt ( 19.6 m^2 / s^2) = +- 4.4 m/s, approx. Since the acceleration and displacement were in the direction chosen as positive, we conclude that the final velocity will be in the same direction and we choose the solution vf = +4.4 m/s.
......!!!!!!!!...................................
RESPONSE --> vf^2 = v0^2 + 2 a `ds vf = +- `sqrt( v0^2 + 2 a `ds) which equals vf = +- `sqrt ( 0^2 + 2 * 9.8 m/s^2 * 1 m) which equals +- `sqrt ( 19.6 m^2 / s^2) = +- 4.4 m/s self critique assessment: 3
.................................................
......!!!!!!!!...................................
17:23:59 From the instant the leave the ground until the instant you reach your highest point, you have an acceleration of 9.8 m/s^2 in the downward direction. Since you are jumping upward, and since we can take our choice of whether upward or downward is the positive direction, we choose the upward direction as positive. You might have chosen the downward direction, and we will see in a moment how you should have proceeded after doing so. For now, using the upward direction as positive, we see that you have an initial velocity of v0 = + 3 m/s and an acceleration of a = -9.8 m/s^2. In order to use any of the equations of motion, each of which involves four variables, you should have the values of three variables. So far you only have two, v0 and a. {}What other variable might you know? If you think about it, you will notice that when objects tossed in the air reach their highest point they stop for an instant before falling back down. That is precisely what will happen to you. At the highest point your velocity will be 0. Since the highest point is the last point we are considering, we see that for your motion from the ground to the highest point, vf = 0. Therefore we are modeling a uniform acceleration situation with v0 = +3 m/s, a = -9.8 m/s^2 and vf = 0. We wish to find the displacement `ds. Unfortunately none of the equations of uniformly accelerated motion contain the four variables v0, a, vf and `ds. This situation can be easily reasoned out from an understanding of the basic quantities. We can find the change in velocity to be -3 meters/second; since the acceleration is equal to the change in velocity divided by the time interval we quickly determine that the time interval is equal to the change in velocity divided by the acceleration, which is `dt = -3 m/s / (-9.8 m/s^2) = .3 sec, approx.; then we multiply the .3 second time interval by the 1.5 m/s average velocity to obtain `ds = .45 meters. However if we wish to use the equations, we can begin with the equation vf = v0 + a `dt and solve to find `dt = (vf - v0) / a = (0 - 3 m/s) / (-9.8 m/s^2) = .3 sec. We can then use the equation `ds = (vf + v0) / 2 * `dt = (3 m/s + 0 m/s) / 2 * .3 sec = .45 m. This solution closely parallels and is completely equivalent to the direct reasoning process, and shows that and initial velocity of 3 meters/second should carry a jumper to a vertical height of .45 meters, approximately 18 inches. This is a fairly average vertical jump. If the negative direction had been chosen as positive then we would have a = +9.8 m/s^2, v0 = -3 m/s^2 (v0 is be in the direction opposite the acceleration so if acceleration is positive then initial velocity is negative) and again vf = 0 m/s (0 m/s is the same whether going up or down). The steps of the solution will be the same and the same result will be obtained, except that `ds will be -.45 m--a negative displacement, but where the positive direction is down. That is we move .45 m in the direction opposite to positive, meaning we move .45 meters upward.
......!!!!!!!!...................................
RESPONSE --> `dt = (vf - v0) / a = (0 - 3 m/s) / (-9.8 m/s^2) = .3 sec. `ds = (vf + v0) / 2 * `dt = (3 m/s + 0 m/s) / 2 * .3 sec = .45 m. Which means we have jumped .45 meters up. self critique assessment: 2
.................................................
......!!!!!!!!...................................
17:27:20 A ball dropped from rest at a height of .9 meters will fall to the ground with a uniform vertical acceleration of 9.8 m/s^2 downward. Selecting the downward direction as positive we have `ds = .9 meters, a = 9.8 m/s^2 and v0 = 0. Using the equation `ds = v0 `dt + .5 a `dt^2 we see that v0 = 0 simplifies the equation to `ds = .5 a `dt^2, so `dt = `sqrt( 2 `ds / a) = `sqrt(2 * .9 m / (9.8 m/s^2) ) = .42 sec, approx.. Since the ball rolls off the edge of the table with only a horizontal velocity, its initial vertical velocity is still zero and it still falls to the floor in .42 seconds. Since its horizontal velocity remains at 3 m/s, it travels through a displacement of 3 m/s * .42 sec = 1.26 meters in this time.
......!!!!!!!!...................................
RESPONSE --> `ds = v0 `dt + .5 a `dt^2 `sqrt(2 * .9 m / (9.8 m/s^2) ) = .42 sec So it travels through a displacement of 3 m/s * .42 sec = 1.26 meters in this time. self critique assessment: 2
.................................................
÷UXѮⲌ| assignment #008 008. Using the Acceleration of Gravity Physics I 02-10-2008
course Phy 121 Physics I Class Notes02-10-2008
......!!!!!!!!...................................
19:31:10 QUESTION FROM STUDENT--Please define the differnece between Fnet and Force. See if you can answer this question.
......!!!!!!!!...................................
RESPONSE --> Fnet is the sum of a group of different force readings while force is the force exerted by one object. confidence assessment: 0
.................................................
......!!!!!!!!...................................
19:35:15 ** It's very important in physics to be able to think in terms of proportionality. To say that y is proportional to x is to say that for some k, y = k x. That is, y is a constant multiple of x. To say that a * `ds is proportional to the change in v^2 is to say that for some k, a * `ds = k * ( change in v^2)--i.e., that a * `ds is a constant multiple of the change in v^2. In terms of the equations of motion, we know that vf^2 = v0^2 + 2 a `ds so a `ds = 1/2 (vf^2 - v0^2), which is 1/2 the change in v^2. So a `ds is a constant multiple (1/2) of the change in v^2. Formally we have a `ds = k ( change in v^2) for k = 1/2. Now since Fnet = m a we see that Fnet is proportional to a for a given mass m, and it follows that Fnet `ds = k * change in v^2, for the appropriate k (specifically for k = mass / 2. **
......!!!!!!!!...................................
RESPONSE --> OK self critique assessment: 3
.................................................
......!!!!!!!!...................................
19:37:13 ** We didn't actually do this part of the experiment, but on a ramp with fixed slope a `ds is simply proportional to `ds. When we measured `dt for different distances from rest down the same ramp, we were then able to determine the average and final velocities. The change in v^2 for each timing would be from 0 to vf^2. The change would therefore be just vf^2. If a `ds is proportional to the change in vf^2 then a graph of vf^2 vs. a `ds should be linear. Since a is constant we don't even need to determine it--a graph of vf^2 vs. `ds would be linear. This would confirm the hypothesis. **
......!!!!!!!!...................................
RESPONSE --> I don't remember doing an experiment using Force. self critique assessment: 3
.................................................