course Phy 121 xEW]|غs֩assignment #007
......!!!!!!!!...................................
09:04:22 Physics video clip 09 displacement for linear v vs. t graph: common sense, formula, area If we know the initial and final velocities over some time interval, and if the rate which velocity changes is constant, then how do we calculate the displacement over a the time interval?
......!!!!!!!!...................................
RESPONSE --> vAve = 'ds / 'dt Rearrange this to solve for 'ds. Multiply both sides by 'dt to get it to cancel on the right side. 'dt * vAve = 'ds It is important to consider that vAve = (v0 + vf)/2. Since we know initial and final velocities we can find vAve, and since we know time all we have to do is plug into the above mentioned formula to find 'ds.
.................................................
......!!!!!!!!...................................
09:04:47 ** Displacement is the product of average velocity and time interval. Since acceleration is uniform average velocity is average of initial and final velocities. Displacement could therefore be calculated from the final and initial velocities => `ds =[(vf + v0)/2] * `dt. **
......!!!!!!!!...................................
RESPONSE --> I think my answer was mostly correct.
.................................................
......!!!!!!!!...................................
09:05:40 Physics video clip 10 continuing 09: calculation of area
......!!!!!!!!...................................
RESPONSE --> There is no question given here, only the clips (9 and 10) and the mention of calculation of area.
.................................................
......!!!!!!!!...................................
09:12:17 How do we use a graph of v vs. t to depict the calculation of the displacement over a time interval?
......!!!!!!!!...................................
RESPONSE --> Average velocity (or vAve) = 'ds / 'dt Therefore, when considering a graph of v vs. t to depict the calculation of the displacement over a time interval we must know the following: v is the y-axis, and t is the x-axis. As depicted in the clip, if there is a linear relationship between the two variables then you can determine from the line initial and final velocities over a particular time interval. Remember that we can rearrange the top equation to solve for 'ds, and also that vAve can be found using (vf + v0)/2. Also, multiplying altitude by width will give displacement.
.................................................
......!!!!!!!!...................................
09:12:59 ** Looking at the graph, we notice a trapezoid created by the slope line across the top, and an imaginary line drawn from the y value down to and perpendicular to the x axis. The area of this trapezoid represents the displacement or 'signed distance' the object travels. The displacement for any time interval can be found finding the average of the two 'altitudes' of the trapezoid, which represent initial and final velocities. Multiplying the average 'altitude' by the width is therefore equivalent to multiplying the approximate average velocity by the time interval, giving us the area of the trapezoid, which represents the approximate displacement. In the case where the graph is linear (which corresponds to uniform acceleration) the average of the two altitudes in fact represents the average velocity, and the result is the displacement, not the approximate displacement. **
......!!!!!!!!...................................
RESPONSE --> I think my answer was at least partially correct.
.................................................
......!!!!!!!!...................................
09:13:24 What aspect of the graph gives the displacement during the time interval?
......!!!!!!!!...................................
RESPONSE --> Altitude multiplied by width gives displacement.
.................................................
......!!!!!!!!...................................
09:13:56 ** STUDENT ANSWER: It is the average of the two sides that are as high as 'y' in each case, multiplied by the width-units of 'x'. **
......!!!!!!!!...................................
RESPONSE --> I think I may have been at least close to what the answer should have been.
.................................................
......!!!!!!!!...................................
09:18:50 Describe the flow diagram you would use for the uniform acceleration situation in which you are given v0, vf, and `dt.
......!!!!!!!!...................................
RESPONSE --> The flow diagram provides in a class notes example begins with the first level being 'dt, a, and v0. However, we don't know a so we must use other variables to find it. The next level is 'dv = a * 'dt. Since we know the initial and final velocities as well as the time interval we can solve for a. confidence assessment: 1
.................................................
......!!!!!!!!...................................
09:20:23 ** We start with v0, vf and `dt on the first line of the diagram. We use vO and vf to find Vave, indicated by lines from v0 and vf to vAve. Use Vave and 'dt to find 'ds, indicated by lines from vAve and `dt to `ds. Then use `dv and 'dt to find acceleration, indicated by lines from vAve and `dt to a. **
......!!!!!!!!...................................
RESPONSE --> I didn't actually work through the whole diagram to find every variable. I just thought I had to find a. However, I had the right reference and example to illustrate the same point, I think. self critique assessment: 2
.................................................
......!!!!!!!!...................................
09:23:40 Describe the flow diagram you would use for the uniform acceleration situation in which you are given `dt, a, v0
......!!!!!!!!...................................
RESPONSE --> As illustrated by an example in the class notes, the first level of the diagram is 'dt, a, and v0. From this data we can complete the rest of the diagram. 'dt and a can be use to find 'dv = a * 'dt. 'dv and v0 can be used to find vf = v0 + 'dv. Vf and v0 can be used to find the average velocity (vAve = (vf + v0)/2). And, vAve and 'dt can be used to find 'ds = vAve * 'dt. confidence assessment: 2
.................................................
......!!!!!!!!...................................
09:23:56 ** Student Solution: Using 'dt and a, find 'dv. Using 'dv and v0, find vf. Using vf and vO, find vave. Using 'dt and Vave, find 'ds. **
......!!!!!!!!...................................
RESPONSE --> I was correct. self critique assessment: 3
.................................................
......!!!!!!!!...................................
09:32:04 Explain in detail how the flow diagram for the situation in which v0, vf and `dt are known gives us the two most fundamental equations of motion.
......!!!!!!!!...................................
RESPONSE --> The equations sought after here are the following: 'ds = v0 'dt + .5 a 'dt^2 'ds = vAve * 'dt Now, let me tell you how we get there: Use vf and v0 to find 'dv. Use 'dv = a * 'dt, and solve for a. Use vf and v0 to find vAve. Use vAve and 'dt to find 'ds. If you look at the diagrams and the each respective equation it takes to find each variable, these variables can be arranged into the above equations to more easily solve for 2 missing other variables. confidence assessment: 1
.................................................
......!!!!!!!!...................................
09:32:41 **Student Solution: v0 and vf give you `dv = vf - v0 and vAve = (vf + v0) / 2. `dv is divided by `dt to give accel. So we have a = (vf - v0) / `dt. Rearranging this we have a `dt = vf - v0, which rearranges again to give vf = v0 + a `dt. This is the second equation of motion. vAve is multiplied by `dt to give `ds. So we have `ds = (vf + v0) / 2 * `dt. This is the first equation of motion Acceleration is found by dividing the change in velocity by the change in time. v0 is the starting velocity, if it is from rest it is 0. Change in time is the ending beginning time subtracted by the ending time. **
......!!!!!!!!...................................
RESPONSE --> I don't think I explained it as well as is given in the solution, but I was on the right track anyway. self critique assessment: 3
.................................................
......!!!!!!!!...................................
09:36:04 qaExplain in detail how the flow diagram for the situation in which v0, a and `dt are known gives us the third fundamental equations of motion.
......!!!!!!!!...................................
RESPONSE --> Equation of Motion: 'ds = v0 * 'dt + .5 a 'dt^2 We know v0, a, and 'dt. Use a and 'dt to find 'dv. Use 'dv and v0 to find vf. Use vf and v0 to find vAve. Use vAve and 'dt to find 'ds. self critique assessment: 1
.................................................
......!!!!!!!!...................................
09:36:29 ** a and `dt give you `dv. `dv and v0 give you vf. v0 and vf give you vAve. vAve and `dt give you `ds. In symbols, `dv = a `dt. Then vf = v0 + `dv = v0 + a `dt. Then vAve = (vf + v0)/2 = (v0 + (v0 + a `dt)) / 2) = v0 + 1/2 a `dt. Then `ds = vAve * `dt = [ v0 `dt + 1/2 a `dt ] * `dt = v0 `dt + 1/2 a `dt^2. **
......!!!!!!!!...................................
RESPONSE --> I'm not quite sure where I went wrong here. self critique assessment: 2
.................................................
......!!!!!!!!...................................
09:37:37 Why do we think in terms of seven fundamental quantities while we model uniformly accelerated motion in terms of five?
......!!!!!!!!...................................
RESPONSE --> The other two quantities can be found or be used to find the ones that we do use to model uniformly accelerated motion. confidence assessment: 0
.................................................
......!!!!!!!!...................................
09:38:31 ** ONE WAY OF PUTTING IT: The four equations are expressed in terms of five fundamental quantities, v0, vf, a, `dt and `ds. However to think in terms of meanings we have to be able to think not only in terms of these quantities but also in terms of average velocity vAve and change in velocity `dv, which aren't among these five quantities. Without the ideas of average velocity and change in velocity we might be able to use the equations and get some correct answers but we'll never understand motion. ANOTHER WAY: The four equations of unif accelerated motion are expressed in terms of five fundamental quantities, v0, vf, a, `dt and `ds. The idea here is that to intuitively understand uniformly accelerated motion, we must often think in terms of average velocity vAve and change in velocity `dv as well as the five quantities involved in the four fundamental equations. one important point is that we can use the five quantities without any real conceptual understanding; to reason things out rather than plugging just numbers into equations we need the concepts of average velocity and change in velocity, which also help us make sense of the equations. **
......!!!!!!!!...................................
RESPONSE --> Well, I think I at least got the part about using what we know to make sense of everything else. self critique assessment: 2
.................................................
......!!!!!!!!...................................
09:42:22 Accelerating down an incline through a given distance vs. accelerating for a given time Why does a given change in initial velocity result in the same change in final velocity when we accelerated down a constant incline for the same time, but not when we accelerated down the same incline for a constant distance?
......!!!!!!!!...................................
RESPONSE --> Distance is not a constant variable in this case. There are other factors affecting acceleration for constant distance such as friction, air resistance, etc. The length of the incline is constant and will not change, however change in velocity will not always result in the same distance regardless of the incline. confidence assessment: 0
.................................................
......!!!!!!!!...................................
09:43:23 ** If we accelerate down a constant incline our rate of change of velocity is the same whatever our initial velocity. So the change in velocity is determined only by how long we spend coasting on the incline. Greater `dt, greater `dv. If you travel the same distance but start with a greater speed there is less time for the acceleration to have its effect and therefore the change in velocity will be less. You might also think back to that introductory problem set about the car on the incline and the lamppost. Greater initial velocity results in greater average velocity and hence less time on the incline, which gives less time for the car to accelerate. **
......!!!!!!!!...................................
RESPONSE --> This solution makes sense, however I am not sure where I went wrong here. self critique assessment: 2
.................................................
r˯ƘT assignment #007 007. Acceleration of Gravity Physics I 07-01-2007
......!!!!!!!!...................................
19:35:45 `q001. We obtain an estimate of the acceleration of gravity by determining the slope of an acceleration vs. ramp slope graph for an object gliding down an incline. Sample data for an object gliding down a 50-cm incline indicate that the object glides down the incline in 5 seconds when the raised end of the incline is .5 cm higher than the lower end; the time required from rest is 3 seconds when the raised end is 1 cm higher than the lower end; and the time from rest is 2 seconds when the raised end is 1.5 cm higher than the lower end. What is the acceleration for each trial?
......!!!!!!!!...................................
RESPONSE --> Definition and formulas for acceleration: aAve = 'dv/'dt Average acceleration is defined as the change in velocity divided by the time taken to make this change. Instantaneous acceleration, a, can be defined in analogy to instantaneous velocity, for any specific instant. Therefore, (concerning the first trial) we know that the object covers a distance of 50 cm over a period of 5 seconds. We need to know final velocity. vAve = 'ds/'dt vAve = 50cm / 5 sec vAve = 10 cm/sec vAve = (v0 + vf) / 2 10 cm/sec = (0 + vf)/2 Multiply both sides by 2 to get it to cancel on the right side of the equation. 20 = 0 + vf 20 cm/sec = vf aAve = 'dv/'dt aAve = (20 - 0)/(5 - 0) aAve = 20/5 aAve = 4 cm/sec for trial 1 vAve = 'ds/'dt vAve = 50cm / 3sec vAve = 17 cm/sec vAve = (v0 + vf) / 2 17 cm/sec = (0 + vf)/2 Multiply both sides by 2 to get it to cancel on the right side of the equation. 34 = 0 + vf 34 cm/sec = vf aAve = 'dv/'dt aAve = (34 - 0)/(3 - 0) aAve = 34/3 aAve = 11.3 cm/sec for trial 2 vAve = 'ds/'dt vAve = 50cm / 2 sec vAve = 25 cm/sec vAve = (v0 + vf) / 2 25 cm/sec = (0 + vf)/2 Multiply both sides by 2 to get it to cancel on the right side of the equation. 50 = 0 + vf 50 cm/sec = vf aAve = 'dv/'dt aAve = (50 - 0)/(2 - 0) aAve = 50/2 aAve = 25 cm/sec for trial 3 confidence assessment: 1
.................................................
......!!!!!!!!...................................
19:36:19 We can find the accelerations either using equations or direct reasoning. To directly reason the acceleration for the five-second case, we note that the average velocity in this case must be 50 cm/(5 seconds) = 10 cm/s. Since the initial velocity was 0, assuming uniform acceleration we see that the final velocity must be 20 cm/second, since 0 cm/s and 20 cm/s average out to 10 cm/s. This implies a velocity change of 20 cm/second a time interval of 5 seconds, or a uniform acceleration of 20 cm/s / (5 s) = 4 cm/s^2. The acceleration in the 3-second case could also be directly reasoned, but instead we will note that in this case we have the initial velocity v0 = 0, the time interval `dt = 3 sec, and the displacement `ds = 50 cm. We can therefore find the acceleration from the equation `ds = v0 `dt + .5 a `dt^2. Noting first that since v0 = 0 the term v0 `dt must also be 0,we see that in this case the equation reduces to `ds = .5 a `dt^2. We easily solve for the acceleration, obtaining a = 2 `ds / `dt^2. In this case we have a = 2 * (50 cm) / (3 sec)^2 = 11 cm/s^2 (rounded to nearest cm/s^2). For the 2-second case we can use the same formula, obtaining a = 2 * (50 cm) / (2 sec)^2 = 25 cm/s^2.
......!!!!!!!!...................................
RESPONSE --> I got the correct numbers, but I messed up the units. self critique assessment: 3
.................................................
......!!!!!!!!...................................
19:38:19 `q002. What are the ramp slopes associated with these accelerations?
......!!!!!!!!...................................
RESPONSE --> The slopes correspond to how much each ramp was raised which was mentioned in the first problem. Because those numbers were not pertinent to the calculations I was doing at the time, I didn't write them down. Slope is determined by rise / run. Those numbers correspond to the rise and the time to the run. confidence assessment: 1
.................................................
......!!!!!!!!...................................
19:38:57 For the 5-second trial, where acceleration was 4 cm/s^2, the 'rise' of the ramp was .5 cm and the 'run' was nearly equal to the 50-cm length of the ramp so the slope was very close to .5 cm / (50 cm) = .01. For the 3-second trial, where acceleration was 11 cm/s^2, the 'rise' of the ramp was 1 cm and the 'run' was very close to the 50-cm length, so the slope was very close to 1 cm / (50 cm) = .02. For the 2-second trial, where the acceleration was 25 cm/s^2, the slope is similarly found to be very close to 1.5 cm / (50 cm) = .03.
......!!!!!!!!...................................
RESPONSE --> Had I written down how much each ramp was raised I would have gotten the correct answer. self critique assessment: 3
.................................................
......!!!!!!!!...................................
19:42:08 `q003. Sketch a reasonably accurate graph of acceleration vs. ramp slope and give a good description and interpretation of the graph. Be sure to include in your description how the graph points seem to lie with respect to the straight line that comes as close as possible, on the average, to the three points.
......!!!!!!!!...................................
RESPONSE --> Visually on paper the graph looks like a foward slash ( \ ). As the incline of the ramp is increased the amount of time it take the object to cover the distance decreases. Therefore the relationship of these values is linear, and the line is decreasing at an increasing rate ( I think?). A best fit line (based on my own estimation) seems to come quite close to all three points. confidence assessment: 0
.................................................
......!!!!!!!!...................................
19:42:58 The graph will have acceleration in cm/s^2 on the vertical axis (the traditional y-axis) and ramp slope on the horizontal axis (the traditional x-axis). The graph points will be (.01, 4 cm/s^2), (.02, 11.1 cm/s^2), (.03, 25 cm/s^2). The second point lies somewhat lower than a line connecting the first and third points, so the best possible line will probably be lower than the first and third points but higher than the second. The graph indicates that acceleration increases with increasing slope, which should be no surprise. It is not clear from the graph whether a straight line is in fact the most appropriate model for the data. If timing wasn't particularly accurate, these lines could easily be interpreted as being scattered from the actual linear behavior due to experimental errors. Or the graph could indicate acceleration vs. ramp slope behavior that is increasing at an increasing rate.
......!!!!!!!!...................................
RESPONSE --> Okay, I don't think I explained this quite as well as I should have, but the solution makes sense. self critique assessment: 2
.................................................
......!!!!!!!!...................................
20:06:20 `q004. Carefully done experiments show that for small slopes (up to a slope of about .1) the graph appears to be linear or very nearly so. This agrees with theoretical predictions of what should happen. Sketch a vertical line at x = .05. Then extend the straight line you sketched previously until it intersects the y axis and until it reaches past the vertical line at x = .05. What are the coordinates of the points where this line intersects the y-axis, and where it intersects the x =.05 line? What are the rise and the run between these points, and what therefore is the slope of your straight line?
......!!!!!!!!...................................
RESPONSE --> Intersection of the vertical line coordinates: (.5, -3) Intersection of the y-axis coordinates: (0, -4) Slope of the straight line: (rise/run) = -1/.5 = -2 (slope may not be exactly right?) confidence assessment: 1
.................................................
......!!!!!!!!...................................
20:08:28 A pretty good straight line goes through the points (0, -6 cm/s^2) and (.05, 42 cm/s^2). Your y coordinates might differ by a few cm/s^2 either way. For the coordinates given here, the rise is from -6 cm/s^2 to 42 cm/s^2, a rise of 48 cm/s^2. The run is from 0 to .05, a run of .05. The slope of the straight line is approximately 48 cm/s^2 / .05 = 960 cm/s^2. Note that this is pretty close to the accepted value, 980 cm/second^2, of gravity. Carefully done, this experiment will give us a very good estimate of the acceleration of gravity.
......!!!!!!!!...................................
RESPONSE --> I think my answer was pretty close, or I think I was at least on the right track. self critique assessment: 1
.................................................
......!!!!!!!!...................................
20:15:25 `q005. The most accurate way to measure the acceleration of gravity is to use the relationship T = 2 `pi / `sqrt(g) * `sqrt(L) for the period of a pendulum. Use your washer pendulum and time 100 complete back-and-forth cycles of a pendulum of length 30 cm. Be sure to count carefully and don't let the pendulum swing out to a position more than 10 degrees from vertical. How long did it take, and how long did each cycle therefore last?
......!!!!!!!!...................................
RESPONSE --> It took 130 seconds for the pendulum to complete 100 cycles. Therefore: 130 sec / 100 cycles = 1.3 seconds per cycle. confidence assessment: 1
.................................................
......!!!!!!!!...................................
20:16:09 100 cycles of a pendulum of this length should require approximately 108 seconds. This would be 108 seconds per 100 cycles, or 108 sec / (100 cycles) = 1.08 sec / cycle. If you didn't count very carefully or didn't time very accurately, you might differ significantly from this result; differences of up to a couple of cycles are to be expected.
......!!!!!!!!...................................
RESPONSE --> My answer varied from this result, but I used the correct method of achieving my result. self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:23:13 `q006. You now have values for the period T and the length L, so you can use the relationship T = 2 `pi / `sqrt(g) * `sqrt(L) to find the acceleration g of gravity. Solve the equation for g and then use your values for T and L to determine the acceleration of gravity.
......!!!!!!!!...................................
RESPONSE --> T = 2 'pi / 'sqrt(g) * 'sqrt(L) 130 seconds = (2 'pi) / 'sqrt(g) * 'sqrt(30 cm) 130 = 6.28 / 'sqrt(g) * 5.48 130 = 6.28 / 5.48 'sqrt(g)
.................................................
......!!!!!!!!...................................
20:27:23 Solving T = 2 `pi / `sqrt(g) * `sqrt(L) for g, we can first multiply both sides by `sqrt(g) to obtain T * `sqrt(g) = 2 `pi `sqrt(L). Then dividing both sides by T we obtain `sqrt(g) = 2 `pi `sqrt(L) / T. Squaring both sides we finally obtain {}g = 4 `pi^2 L / T^2. Plugging in the values given here, L = 30 cm and T = 1.08 sec, we obtain g = 4 `pi^2 * 30 cm / (1.08 sec)^2 = 1030 cm/s^2. You should check these calculations for accuracy, since they were mentally approximated.
......!!!!!!!!...................................
RESPONSE --> I not exactly sure where I went wrong here. Any suggestions or comments here would be appreciated. self critique assessment: 2
.................................................
xEW]|غs֩ assignment #007 007. `query 7 Physics I 06-26-2007 呦ުѢfweሯ assignment #004 }jhF}}Z٨ Physics I Vid Clips 06-26-2007
......!!!!!!!!...................................
09:04:22 Physics video clip 09 displacement for linear v vs. t graph: common sense, formula, area If we know the initial and final velocities over some time interval, and if the rate which velocity changes is constant, then how do we calculate the displacement over a the time interval?
......!!!!!!!!...................................
RESPONSE --> vAve = 'ds / 'dt Rearrange this to solve for 'ds. Multiply both sides by 'dt to get it to cancel on the right side. 'dt * vAve = 'ds It is important to consider that vAve = (v0 + vf)/2. Since we know initial and final velocities we can find vAve, and since we know time all we have to do is plug into the above mentioned formula to find 'ds.
.................................................
......!!!!!!!!...................................
09:04:47 ** Displacement is the product of average velocity and time interval. Since acceleration is uniform average velocity is average of initial and final velocities. Displacement could therefore be calculated from the final and initial velocities => `ds =[(vf + v0)/2] * `dt. **
......!!!!!!!!...................................
RESPONSE --> I think my answer was mostly correct.
.................................................
......!!!!!!!!...................................
09:05:40 Physics video clip 10 continuing 09: calculation of area
......!!!!!!!!...................................
RESPONSE --> There is no question given here, only the clips (9 and 10) and the mention of calculation of area.
.................................................
......!!!!!!!!...................................
09:12:17 How do we use a graph of v vs. t to depict the calculation of the displacement over a time interval?
......!!!!!!!!...................................
RESPONSE --> Average velocity (or vAve) = 'ds / 'dt Therefore, when considering a graph of v vs. t to depict the calculation of the displacement over a time interval we must know the following: v is the y-axis, and t is the x-axis. As depicted in the clip, if there is a linear relationship between the two variables then you can determine from the line initial and final velocities over a particular time interval. Remember that we can rearrange the top equation to solve for 'ds, and also that vAve can be found using (vf + v0)/2. Also, multiplying altitude by width will give displacement.
.................................................
......!!!!!!!!...................................
09:12:59 ** Looking at the graph, we notice a trapezoid created by the slope line across the top, and an imaginary line drawn from the y value down to and perpendicular to the x axis. The area of this trapezoid represents the displacement or 'signed distance' the object travels. The displacement for any time interval can be found finding the average of the two 'altitudes' of the trapezoid, which represent initial and final velocities. Multiplying the average 'altitude' by the width is therefore equivalent to multiplying the approximate average velocity by the time interval, giving us the area of the trapezoid, which represents the approximate displacement. In the case where the graph is linear (which corresponds to uniform acceleration) the average of the two altitudes in fact represents the average velocity, and the result is the displacement, not the approximate displacement. **
......!!!!!!!!...................................
RESPONSE --> I think my answer was at least partially correct.
.................................................
......!!!!!!!!...................................
09:13:24 What aspect of the graph gives the displacement during the time interval?
......!!!!!!!!...................................
RESPONSE --> Altitude multiplied by width gives displacement.
.................................................
......!!!!!!!!...................................
09:13:56 ** STUDENT ANSWER: It is the average of the two sides that are as high as 'y' in each case, multiplied by the width-units of 'x'. **
......!!!!!!!!...................................
RESPONSE --> I think I may have been at least close to what the answer should have been.
.................................................
......!!!!!!!!...................................
09:18:50 Describe the flow diagram you would use for the uniform acceleration situation in which you are given v0, vf, and `dt.
......!!!!!!!!...................................
RESPONSE --> The flow diagram provides in a class notes example begins with the first level being 'dt, a, and v0. However, we don't know a so we must use other variables to find it. The next level is 'dv = a * 'dt. Since we know the initial and final velocities as well as the time interval we can solve for a. confidence assessment: 1
.................................................
......!!!!!!!!...................................
09:20:23 ** We start with v0, vf and `dt on the first line of the diagram. We use vO and vf to find Vave, indicated by lines from v0 and vf to vAve. Use Vave and 'dt to find 'ds, indicated by lines from vAve and `dt to `ds. Then use `dv and 'dt to find acceleration, indicated by lines from vAve and `dt to a. **
......!!!!!!!!...................................
RESPONSE --> I didn't actually work through the whole diagram to find every variable. I just thought I had to find a. However, I had the right reference and example to illustrate the same point, I think. self critique assessment: 2
.................................................
......!!!!!!!!...................................
09:23:40 Describe the flow diagram you would use for the uniform acceleration situation in which you are given `dt, a, v0
......!!!!!!!!...................................
RESPONSE --> As illustrated by an example in the class notes, the first level of the diagram is 'dt, a, and v0. From this data we can complete the rest of the diagram. 'dt and a can be use to find 'dv = a * 'dt. 'dv and v0 can be used to find vf = v0 + 'dv. Vf and v0 can be used to find the average velocity (vAve = (vf + v0)/2). And, vAve and 'dt can be used to find 'ds = vAve * 'dt. confidence assessment: 2
.................................................
......!!!!!!!!...................................
09:23:56 ** Student Solution: Using 'dt and a, find 'dv. Using 'dv and v0, find vf. Using vf and vO, find vave. Using 'dt and Vave, find 'ds. **
......!!!!!!!!...................................
RESPONSE --> I was correct. self critique assessment: 3
.................................................
......!!!!!!!!...................................
09:32:04 Explain in detail how the flow diagram for the situation in which v0, vf and `dt are known gives us the two most fundamental equations of motion.
......!!!!!!!!...................................
RESPONSE --> The equations sought after here are the following: 'ds = v0 'dt + .5 a 'dt^2 'ds = vAve * 'dt Now, let me tell you how we get there: Use vf and v0 to find 'dv. Use 'dv = a * 'dt, and solve for a. Use vf and v0 to find vAve. Use vAve and 'dt to find 'ds. If you look at the diagrams and the each respective equation it takes to find each variable, these variables can be arranged into the above equations to more easily solve for 2 missing other variables. confidence assessment: 1
.................................................
......!!!!!!!!...................................
09:32:41 **Student Solution: v0 and vf give you `dv = vf - v0 and vAve = (vf + v0) / 2. `dv is divided by `dt to give accel. So we have a = (vf - v0) / `dt. Rearranging this we have a `dt = vf - v0, which rearranges again to give vf = v0 + a `dt. This is the second equation of motion. vAve is multiplied by `dt to give `ds. So we have `ds = (vf + v0) / 2 * `dt. This is the first equation of motion Acceleration is found by dividing the change in velocity by the change in time. v0 is the starting velocity, if it is from rest it is 0. Change in time is the ending beginning time subtracted by the ending time. **
......!!!!!!!!...................................
RESPONSE --> I don't think I explained it as well as is given in the solution, but I was on the right track anyway. self critique assessment: 3
.................................................
......!!!!!!!!...................................
09:36:04 qaExplain in detail how the flow diagram for the situation in which v0, a and `dt are known gives us the third fundamental equations of motion.
......!!!!!!!!...................................
RESPONSE --> Equation of Motion: 'ds = v0 * 'dt + .5 a 'dt^2 We know v0, a, and 'dt. Use a and 'dt to find 'dv. Use 'dv and v0 to find vf. Use vf and v0 to find vAve. Use vAve and 'dt to find 'ds. self critique assessment: 1
.................................................
......!!!!!!!!...................................
09:36:29 ** a and `dt give you `dv. `dv and v0 give you vf. v0 and vf give you vAve. vAve and `dt give you `ds. In symbols, `dv = a `dt. Then vf = v0 + `dv = v0 + a `dt. Then vAve = (vf + v0)/2 = (v0 + (v0 + a `dt)) / 2) = v0 + 1/2 a `dt. Then `ds = vAve * `dt = [ v0 `dt + 1/2 a `dt ] * `dt = v0 `dt + 1/2 a `dt^2. **
......!!!!!!!!...................................
RESPONSE --> I'm not quite sure where I went wrong here. self critique assessment: 2
.................................................
......!!!!!!!!...................................
09:37:37 Why do we think in terms of seven fundamental quantities while we model uniformly accelerated motion in terms of five?
......!!!!!!!!...................................
RESPONSE --> The other two quantities can be found or be used to find the ones that we do use to model uniformly accelerated motion. confidence assessment: 0
.................................................
......!!!!!!!!...................................
09:38:31 ** ONE WAY OF PUTTING IT: The four equations are expressed in terms of five fundamental quantities, v0, vf, a, `dt and `ds. However to think in terms of meanings we have to be able to think not only in terms of these quantities but also in terms of average velocity vAve and change in velocity `dv, which aren't among these five quantities. Without the ideas of average velocity and change in velocity we might be able to use the equations and get some correct answers but we'll never understand motion. ANOTHER WAY: The four equations of unif accelerated motion are expressed in terms of five fundamental quantities, v0, vf, a, `dt and `ds. The idea here is that to intuitively understand uniformly accelerated motion, we must often think in terms of average velocity vAve and change in velocity `dv as well as the five quantities involved in the four fundamental equations. one important point is that we can use the five quantities without any real conceptual understanding; to reason things out rather than plugging just numbers into equations we need the concepts of average velocity and change in velocity, which also help us make sense of the equations. **
......!!!!!!!!...................................
RESPONSE --> Well, I think I at least got the part about using what we know to make sense of everything else. self critique assessment: 2
.................................................
......!!!!!!!!...................................
09:42:22 Accelerating down an incline through a given distance vs. accelerating for a given time Why does a given change in initial velocity result in the same change in final velocity when we accelerated down a constant incline for the same time, but not when we accelerated down the same incline for a constant distance?
......!!!!!!!!...................................
RESPONSE --> Distance is not a constant variable in this case. There are other factors affecting acceleration for constant distance such as friction, air resistance, etc. The length of the incline is constant and will not change, however change in velocity will not always result in the same distance regardless of the incline. confidence assessment: 0
.................................................
......!!!!!!!!...................................
09:43:23 ** If we accelerate down a constant incline our rate of change of velocity is the same whatever our initial velocity. So the change in velocity is determined only by how long we spend coasting on the incline. Greater `dt, greater `dv. If you travel the same distance but start with a greater speed there is less time for the acceleration to have its effect and therefore the change in velocity will be less. You might also think back to that introductory problem set about the car on the incline and the lamppost. Greater initial velocity results in greater average velocity and hence less time on the incline, which gives less time for the car to accelerate. **
......!!!!!!!!...................................
RESPONSE --> This solution makes sense, however I am not sure where I went wrong here. self critique assessment: 2
.................................................
"