course Phy 121 I tried to do the class notes #8 and #9 but it said
......!!!!!!!!...................................
16:56:38 What do we mean by velocity?
......!!!!!!!!...................................
RESPONSE --> velocity is the rate at which something moves
.................................................
......!!!!!!!!...................................
16:59:19 How can we determine the velocity of a ball rolling down an incline?
......!!!!!!!!...................................
RESPONSE --> bymeasuring th etime it takes for the ball to travel between two distinct distances.
.................................................
......!!!!!!!!...................................
16:59:41 11-02-2005 16:59:41 ** We divide displacement by the time interval to get average velocity. If the ball happens to be starting from rest and acceleration is uniform we can also conclude that the final velocity attained on the incline is double the average velocity. **
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
17:07:18 We anticipate from our experience that a ball traveling down a greater incline, starting from rest, will experience a greater change in velocity. How can we determine whether the velocity actually changes, and whether the velocity increases in the manner that we expect?
......!!!!!!!!...................................
RESPONSE --> we can determine this by timing the ball rolling down the first incline, recording that data, and then timing the velocity of the ball on the second incline, this will show the difference
.................................................
......!!!!!!!!...................................
17:07:30 11-02-2005 17:07:30 ** We divide displacement by the time interval to get average velocity. We time the ball down one incline, then down the other and determine average velocity for each. We then infer that since both balls started from rest, the greater average velocity implies a greater change in velocity. **
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
17:08:49 How could we determine the velocity of the ball at a specific point? The specific points are measured for distance and the ball is timed when it reaches these specific points. The distance is then divided by the time.
......!!!!!!!!...................................
RESPONSE --> The specific points are measured for distance and the ball is timed when it reaches these specific points. The distance is then divided by the time.
.................................................
......!!!!!!!!...................................
17:10:23 ** Short answer: The question concerned one specific point. We can't really measure this precisely. The best we can do is use two points close together near the point we are interested in, but not so close we can't measure the time accurately enough to trust our result. More detailed answer: The question really asks how we determine the velocity at a given point, for an object in the real world. Assuming that the velocity is always changing, how can we ever know the velocity at an instant? This involves a limiting process, thinking of shorter and shorter time intervals and shorter and shorter position changes. If we know the velocity function, or if we can accurately infer the velocity function from our data, then the velocity of a ball at a specific point is obtained by finding the slope of the tangent line of the position vs. clock time graph at that point, which calculus-literate students will recognize as the derivative of the velocity function. **
......!!!!!!!!...................................
RESPONSE --> I thought that was the answer within the question (stupid me) I understand the whole process of the shorter time intervals.
.................................................
......!!!!!!!!...................................
17:13:36 How do we determine the rate at which the velocity changes? How can we understand the concept of the rate at which velocity changes?
......!!!!!!!!...................................
RESPONSE --> the average acceleration of an object between to timings is how we get the rate at which velocity changes, we can understand this concept by making a graph of velocity vs. clock time
.................................................
......!!!!!!!!...................................
17:15:00 ** We find the change in velocity then divide by the change in the clock time. Any rate consists of the change in one quantity divided by the change in another. **
......!!!!!!!!...................................
RESPONSE --> ok the rate of something is the change in one quanity divided by the change in another
.................................................
......!!!!!!!!...................................
17:27:24 ** Since the rise represents the change in velocity and the run represents the change in clock time, slope represents `dv / `dt = vAve, the average velocity over the corresponding time interval. Since the average altitude represents the average velocity and the width of the trapezoid represents the time interval the area of the trapezoid represents vAve * `dt, which is the displacement `ds. **
......!!!!!!!!...................................
RESPONSE --> I""m not sure what the rise over run is, I'd know it if I saw it, but I'm pretty sure it has something to do with slope I"" think that the ""x""is the run and the ""y"" is the rise, I'm not sure about the trapezoid
.................................................
......!!!!!!!!...................................
17:35:21 What does the graph of position vs. clock time look like for constant-acceleration motion?
......!!!!!!!!...................................
RESPONSE --> it is increasing at a steady rate
.................................................
......!!!!!!!!...................................
17:39:26 ** For constant positive acceleration velocity is increasing. The greater the velocity the steeper the position vs. clock time graph. So increasing velocity would be associated with a position vs. clock time graph which is increasing at an increasing rate. The reason velocity is the slope of the position vs. clock time graph is that the rise between two points of the position vs. clock time graph is change in position, `ds, and run is change in clock time, `dt. Slope therefore represents `ds / `dt, which is velocity. Other shapes are possible, depending on whether initial velocity and acceleration are positive, negative or zero. For example if acceleration was negative and initial velocity positive we could have a graph that's increasing at a decreasing rate. Negative initial velocity and positive acceleration could give us a graph that's decreasing at a decreasing rate. **
......!!!!!!!!...................................
RESPONSE --> ok I understand that it was increasing at an increasing rate
.................................................
......!!!!!!!!...................................
17:42:03 How can we obtain a graph of position vs. clock time from a velocity vs. clock time graph
......!!!!!!!!...................................
RESPONSE --> by being able to know how much time is required from each certian postion and then graphing these times
.................................................
......!!!!!!!!...................................
17:42:57 ** We can divide the graph of v vs. t into small strips, each forming an approximate trapezoid. The area of each strip will represent ave vel * time interval and will therefore represent the change in position during that time interval. Starting from the initial clock time and position on the position vs. clock time graph, we add each subsequent time increment to the clock time and the corresponding position change to the position to get our new position. When the graph is constructed the slopes of the position vs. clock time graph will indicate the corresponding velocities on the v vs. t graph. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
17:44:43 How can we obtain a graph of acceleration vs. clock time from a velocity vs. clock time graph?
......!!!!!!!!...................................
RESPONSE --> by measuring the rate of acceleration by the rate of velocity at different time intervals
.................................................
......!!!!!!!!...................................
17:45:09 11-02-2005 17:45:09 ** Accel is the rate of change of velocity, represented by the slope of the v vs. t graph. So we would plot the slope of the v vs. t graph vs. t, in much the same way as we plotted slopes of the position vs. clock time graph to get the v vs. t graph. }University Physics Students note: Acceleration is the derivative of the velocity. COMMON MISCONCEPTION: Take speed/ time to find the acceleration per second. The form an acceleration v. time graph and draw a straight line out from the number calculated for acceleration above. INSTRUCTOR RESPONSE: Ave acceleration is change in velocity divided by change in clock time. (note that this is different from velocity divided by time--we must use changes in velocity and clock time). (Advanced note: Velocity is always measured with respect to some reference frame, and the velocity of the reference frame itself affects a velocity/time result, but which would not affect change in velocity/time). Graphically acceleration is the slope of the velocity vs. clock time graph. If it was velocity divided by time, it would be the slope of a line from the origin to the graph point. **
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
17:52:01 How can we obtain a graph of velocity vs. clock time from an acceleration vs. clock time graph
......!!!!!!!!...................................
RESPONSE --> the same way as the previous graph but in opposite order by adding time increments in where neededand plotting them in relationto the velocity
.................................................
ҥҾx~]ʐo穭 assignment #002 ~zWJ˘FچŸ Physics I Class Notes 11-02-2005
......!!!!!!!!...................................
18:29:25 How does the shape of the corresponding position vs. clock time graph, with its upward curvature, show us that the time required to travel the first half of the incline is greater than that required to travel the second half?
......!!!!!!!!...................................
RESPONSE --> l
.................................................
......!!!!!!!!...................................
18:30:25 Given the constant rate at which velocity changes, initial velocity, and time duration, how do we reason out the corresponding change in the position of an
......!!!!!!!!...................................
RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how tm o avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.
.................................................
......!!!!!!!!...................................
18:30:31 object?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
18:31:49 ** The reasoning process is as follows: To get the change in velocity you multiply the average rate at which velocity changes (i.e., the acceleration) by the time interval. This change in velocity will be added to the initial velocity to get the final velocity. Since the rate of change of velocity is constant, we can average initial and final velocities to get average velocity. Since velocity is rate of change of position, meaning that ave velocity = change in position / change in clock time, then position change (i.e., displacement) is the product of average velocity and the time interval (i.e., the period of time `dt). In symbols using v0, vf and `dt: vAve = (v0 + vf) / `dt because accel is uniform. vAve = `ds / `dt (this is the definition of vAve and applies whether accel is uniform or not) `ds = vAve * `dt = (vf + v0) / 2 * `dt. FORMULA VS. EXPLANATION: Distance is x. x=1/2 A t ^2 + the initial velocity times time INSTRUCTOR COMMENT: That's a (correct) formula, not a reasoning process. Be sure you know the difference, which will be important on some of the tests. The formula is the end result of a reasoning process, but it is not the process. I emphasize the process at the begiinning, as you can tell from the fact that we devote over a week establishing and reasoning through these concepts. The formulas can be taught in a day; really understanding motion takes longer. **
......!!!!!!!!...................................
RESPONSE --> i
.................................................
......!!!!!!!!...................................
18:32:09 How do we determine position changes over specified time intervals from a graph of velocity vs. clock time, and how can we then construct a graph of
......!!!!!!!!...................................
RESPONSE --> l
.................................................
......!!!!!!!!...................................
18:32:20 position vs. clock time?
......!!!!!!!!...................................
RESPONSE --> k
.................................................
......!!!!!!!!...................................
18:33:15 ** The trapezoid as constructed has a single width, which represents the time interval. The average height represents the average velocity. So ave ht multiplied by width represents the product of ave vel and time interval, which gives us displacement. Good Student Response: Change in position can easily be determined from a graph of velocity versus clock time. Using the idea of a trapezoid or triangle you take the average height, which is the average velocity and multiply it by the average width, which is the time interval.. This will give us the area, which is the position change.
......!!!!!!!!...................................
RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always cri,tique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.
.................................................
......!!!!!!!!...................................
18:33:27 How can a series of velocity vs. clock time trapezoids help us to calculate and visualize position vs. clock time information?
......!!!!!!!!...................................
RESPONSE --> l
.................................................
......!!!!!!!!...................................
18:34:41 ** The position change for a given time interval corresponds to the area of the v vs. t trapezoid which covers that time interval. The greater the area of the v vs. t trapezoid, then, the greater the change in position. Starting with the first position vs. clock time point, we increase the clock time and change the position coordinate according to the dimensions of each new trapezoid, adding each new trapezoid area to the previous position. Assuming the trapezoids are constructed on equal time intervals, then, the greater the average altitude of the trapezoid the greater the position change, so that the average height of the triangle dictates the slope of the position vs. clock time graph. ANOTHER INSIGHTFUL STUDENT RESPONSE: Well the series would be trapezoids for each time interval. Using each interval and plugging it in to the equation mentioned in the previous question you would come up with different positions. The total position change up to a given clock time is easily found by adding the position changes during all the time intervals up to that clock and calculating the areas of successive trapezoids gives us a series of successive displacements, each added to the previous position, so that trapezoid by trapezoid we accumulate our change in position **
......!!!!!!!!...................................
RESPONSE --> 9
.................................................
仑Z˴D̂I{cCcvԊ assignment #002 ~zWJ˘FچŸ Physics I Class Notes 11-02-2005
......!!!!!!!!...................................
18:51:14 How does the shape of the corresponding position vs. clock time graph, with its upward curvature, show us that the time required to travel the first half of the incline is greater than that required to travel the second half?
......!!!!!!!!...................................
RESPONSE --> The shape of the graph, with its upward curve, shows us that it is increasing at an increasing rate over time.. Its speed isgreaterbecause it speeds up as time goes on it gets faster and faster
.................................................
......!!!!!!!!...................................
18:51:31 Given the constant rate at which velocity changes, initial velocity, and time duration, how do we reason out the corresponding change in the position of an
......!!!!!!!!...................................
RESPONSE --> ok.
.................................................
......!!!!!!!!...................................
18:53:26 object?
......!!!!!!!!...................................
RESPONSE --> : It is determined and multipled it by half the accelleration. You could construct a graph by determining position changes over specified time intervals by subtracting the initial time from the final time for that specified time period. Then take the square of the time that we just simply finding all the position points and plotting those on the graph instead of the velocity.
.................................................
......!!!!!!!!...................................
18:54:21 ** The reasoning process is as follows: To get the change in velocity you multiply the average rate at which velocity changes (i.e., the acceleration) by the time interval. This change in velocity will be added to the initial velocity to get the final velocity. Since the rate of change of velocity is constant, we can average initial and final velocities to get average velocity. Since velocity is rate of change of position, meaning that ave velocity = change in position / change in clock time, then position change (i.e., displacement) is the product of average velocity and the time interval (i.e., the period of time `dt). In symbols using v0, vf and `dt: vAve = (v0 + vf) / `dt because accel is uniform. vAve = `ds / `dt (this is the definition of vAve and applies whether accel is uniform or not) `ds = vAve * `dt = (vf + v0) / 2 * `dt. FORMULA VS. EXPLANATION: Distance is x. x=1/2 A t ^2 + the initial velocity times time INSTRUCTOR COMMENT: That's a (correct) formula, not a reasoning process. Be sure you know the difference, which will be important on some of the tests. The formula is the end result of a reasoning process, but it is not the process. I emphasize the process at the begiinning, as you can tell from the fact that we devote over a week establishing and reasoning through these concepts. The formulas can be taught in a day; really understanding motion takes longer. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
18:56:15 position vs. clock time?
......!!!!!!!!...................................
RESPONSE --> the series would be trapezoids for each time interval. Using each interval and plugging it in to the equation mentioned in the previous question you would come up with all the different positions. The total position change up to a given clock time is easily found by adding the position changes during all new time intervals
.................................................
......!!!!!!!!...................................
18:56:41 ** STUDENT RESPONSE: We determine position changes over specified time intervals by subtracting the initial time from the final time for that specified time period. Then take the square of the time that we just determined and multiple it by half the acceleration. You could construct a graph by simply finding all the position points and plotting those on the graph instead of the velocity. INSTRUCTOR COMMENTS: We can of course do this, except that the graph doesn't directly tell us the acceleration. To find that we would first have to find the rate of velocity change, represented by the slope of the graph. The geometrical picture is very important. What we do is find the area of the trapezoid formed by the graph between the two clock times. The altitudes represent velocities, so the average altitude of a trapezoid represents the average velocity during the corresponding time interval. Since the width of the trapezoid represents the time interval, multiplying the altitude by the width to get the area represents the product of average velocity and time interval, which is the displacement corresponding to the time interval. Typically we will do this for several consecutive trapezoids, in order to get the change in position from the start of the first trapezoid to the end of the last. If this is done over a series of adjacent trapezoids, we get a series of areas, which are position changes. We then construct a graph of position vs. clock time for the given clock times by adding each successive position change to the previous position. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
19:00:13 ** The trapezoid as constructed has a single width, which represents the time interval. The average height represents the average velocity. So ave ht multiplied by width represents the product of ave vel and time interval, which gives us displacement. Good Student Response: Change in position can easily be determined from a graph of velocity versus clock time. Using the idea of a trapezoid or triangle you take the average height, which is the average velocity and multiply it by the average width, which is the time interval.. This will give us the area, which is the position change.
......!!!!!!!!...................................
RESPONSE --> Change in position is determined from a graph of velocity vs. clock time. Using the idea of a trapezoid or triangle you take the average height,(aka the average velocity) and multiply it by the average width, (aka the time interval).. This will give us the area, or the the position change.
.................................................
......!!!!!!!!...................................
19:04:46 ** The position change for a given time interval corresponds to the area of the v vs. t trapezoid which covers that time interval. The greater the area of the v vs. t trapezoid, then, the greater the change in position. Starting with the first position vs. clock time point, we increase the clock time and change the position coordinate according to the dimensions of each new trapezoid, adding each new trapezoid area to the previous position. Assuming the trapezoids are constructed on equal time intervals, then, the greater the average altitude of the trapezoid the greater the position change, so that the average height of the triangle dictates the slope of the position vs. clock time graph. ANOTHER INSIGHTFUL STUDENT RESPONSE: Well the series would be trapezoids for each time interval. Using each interval and plugging it in to the equation mentioned in the previous question you would come up with different positions. The total position change up to a given clock time is easily found by adding the position changes during all the time intervals up to that clock and calculating the areas of successive trapezoids gives us a series of successive displacements, each added to the previous position, so that trapezoid by trapezoid we accumulate our change in position **
......!!!!!!!!...................................
RESPONSE --> The series would be trapezoids for each time interval. I used each interval and put it in to the equation mentioned in the previous question and came up with different positions. I found total position change up to a given clock time by adding the position changes during all the time intervals up to that clock and calculating the areas of the folllowing trapezoids gave successive displacements, each added to the previous position, so that by each trapezoid I got my changed position
.................................................
汋 assignment #003 ~zWJ˘FچŸ Physics I Class Notes 11-02-2005
......!!!!!!!!...................................
19:08:11 ** COMMON ERRONEOUS STUDENT RESPONSE: To find the average acceleration, we divide the change in veolocity by the time interval. To find the change in position or displacement of the object over any time interval multiply the average velocity over that interval by the duration of the interval. You are not given the average velocity or the change in velocity. You have to first determine the average velocity; then your strategy will work. Since acceleration is constant you can say that average velocity is the average of initial and final velocities: vAve = (v0 + vf) / 2. Change in velocity is `dv = vf - v0. Now we can do as you say: To find the change in position or displacement of the object over any time interval multiply the average velocity over that interval by the duration of the interval. **
......!!!!!!!!...................................
RESPONSE --> To find the change in position or displacement of the object over any time interval multiply the average velocity over that interval by the duration of the interval.
.................................................
......!!!!!!!!...................................
19:10:44 ** If you multiply the average altitude by the width (finding the area) of the trapezoid you are multiplying the average velocity by the time interval. This gives you the displacement during the time interval. The rise of the triangle represents the change in velocity and the run represents the time interval, so slope = rise / run represents change in velocity / time interval = acceleration. **
......!!!!!!!!...................................
RESPONSE --> If you multiply the average velocity by the time interval. This gives you the displacement during the time interval. The rise of the triangle represents the change in velocity and the run represents the time interval.
.................................................
xJ{v assignment #004 ~zWJ˘FچŸ Physics I Class Notes 11-02-2005
......!!!!!!!!...................................
19:13:24 ** COMMON ERROR: By first finding the average velocity. Also we must understand this gives us the change in velocity which gives us the acceleration. INSTRUCTOR COMMENT: Average acceleration is not related to average velocity. It is related to the change in velocity, but change in velocity is not the same thing as acceleration. Acceleration is the rate of change of velocity. ANSWER TO QUESTION: Average acceleration is defined to be the average rate at which velocity changes, which is change in velocity/change in clock time, or equivalently change in velocity/time duration. However we do not know the change in velocity, nor from the given information can we directly determine the change in velocity. So we have to look at what we can reason out from what we know. Given displacement from rest and time duration we calculate average velocity: vAve = `ds / `dt = displacement/time duration. Since the initial velocity is 0, the final velocity is double the average so we find the final velocity by doubling the average velocity. Now that we know the initial velocity 0 and the final velocity we can find change in velocity by subtracting initial vel from final vel. Dividing change in velocity by a time duration we finally obtain the average acceleration. **
......!!!!!!!!...................................
RESPONSE --> Dividing change in velocity by a time duration we finally obtain the average acceleration.
.................................................
......!!!!!!!!...................................
19:16:52 ** If we start with a position vs. clock time graph we do divide the graph into intervals. On each interval velocity, which is rate of change of position, is represented by the slope of the graph. So we have to calculate the slope of the graph on each interval. If we then graph the slopes vs. the midpoint times of the intervals we get a good approximation of the velocity vs. clock time graph. The velocity at a given instant is the slope of the position vs. clock time graph. STUDENT ANSWER: We first calculate the time interval and displacement between each pair of data points. We use these calculations to calculate the average velocity. To obtain an approximate accelerations. Clock timetable and graph by associating the average velocity on a time interval with the midpoint clock time on that interval. INSTRUCTOR CRITIQUE: Using your words and amplifying a bit: We first calculate the time interval and displacement between each pair of data points. We use these results to calculate the average velocity, dividing displacement by time interval for each interval. Then we make a table, showing the average velocity vs. the midpoint time for each time interval. To obtain approximate accelerations we use the table and graph obtained by associating the average velocity on a time interval with the midpoint clock time on that interval. We find the time interval between each pair of midpoint times, and the change in average velocities between those two midpoint times. Dividing velocity change by time interval we get the rate of velocity change, or acceleration. **
......!!!!!!!!...................................
RESPONSE --> First, calculate the time interval and displacement between each pair of points. We use these calculations to get the average velocity. To get the approximate accelerations, use the clock timetable and graph by putting the average vel. on a time interval with the midpoint clock time on that interval.
.................................................
......!!!!!!!!...................................
19:18:55 ** If we first calculate velocities from the position vs. clock time data we get decreasing velocities. If we graph these velocities vs. midpoint clock times we get a graph which appears to be well-fit by a straight line. This is evidence that the acceleration of the water surface is uniform. If we calculate average accelerations based on average velocities and midpoint clock times we get a lot of variation in our results. However since acceleration results depend on velocities and changes in clock times, and since the velocities themselves were calculated based on changes in clock times, our results are doubly dependent on the accuracy of our clock times. So these fluctuating results don't contradict the linearity of the v vs. t graph. We also find that the position vs. clock time data are very well-fit by a quadratic function of clock time. If the position vs. clock time graph is quadratic then the velocity is a linear function of clock time (University Physics students note that the derivative of a quadratic function is a linear function) and acceleration is constant (University Physics students note that the second derivative of a quadratic function is constant). **
......!!!!!!!!...................................
RESPONSE --> If we graph the velocities vs. midpoint clock times we get a graph that seems to be a straight line. This shows that the acceleration of the water surface is uniform.
.................................................
......!!!!!!!!...................................
19:21:47 ** If the graph is linear then the average velocity occurs at the midpoint clock time, and is halfway between the initial and final velocities. In this case 4 cm/s would be halfway between 0 and the final velocity, so the final velocity would have to be 8 cm/s (4 is halfway between 0 and 8). **
......!!!!!!!!...................................
RESPONSE --> The graph is linear and the average velocity occurs at the midpoint clock time, and is halfway between the initial and final velocities. So 4 would be halfway between 0 and the final velocity, so the final velocity would have to be 8
.................................................
......!!!!!!!!...................................
22:42:52 ** Assuming uniform time intervals, a linear v vs. t graph implies that over every time interval the average velocity will be different that over the previous time intervals, and that it will be changing by the same amount from one time interval to the next. The result is that the distance moved changes by the same amount from one time interval to the next. The distance moved is the rise of the position vs. clock time graph. If over uniform intervals the rise keeps changing, by the same amount with each new interval, the graph has to curve. **
......!!!!!!!!...................................
RESPONSE --> If the time intervals are the same steady rate, a linear velocity vs. time graph shows that over each time interval the average velocity will be different than over the time intervals that come beforeit, and if over these intervals the rise keeps changing by the same amount with each new interval the graph will curve.
.................................................
......!!!!!!!!...................................
23:00:08 ** For an object with positive acceleration, at the beginning of a time interval the velocity is less than at the end of the interval. We expect that the average velocity is between the beginning and ending velocities; if acceleration is uniform, in fact the average velocity is equal to the average of initial and final velocities on that interval, which is halfway between initial and final velocities. If the interval is short, even if acceleration is not uniform, then we expect the average velocity occur near the midpoint clock time. **
......!!!!!!!!...................................
RESPONSE --> because that is when the object is at its midpoint in velocity itwas slower before it got there and faster after it passed that point so the average and the midpoint go hand in hand
.................................................
ѝbʏݽڙ assignment #005 ~zWJ˘FچŸ Physics I Class Notes 11-02-2005
......!!!!!!!!...................................
23:25:48 ** For small slopes, the data indicate that the change in acceleration would be nearly proportional to change in slope. **
......!!!!!!!!...................................
RESPONSE --> In small slopes,the chage in acceleration isnearly proportional to the change in slope
.................................................
......!!!!!!!!...................................
23:28:23 ** Large slope means slope of large magnitude. For the accel vs. ramp slope experiment the slopes were all small, less than .1. The small-slope behavior appears to be linear, in the sense that acceleration appears to be proportional to ramp slope. As the ramp approaches vertical the slope approaches infinity, while acceleration approaches simply the acceleration of gravity. If the graph was linear then as slope approaches infinity we would have acceleration approaching infinity. We conclude that the linear acceleration vs. slope relationship will not continue. ** As we approach vertical slope approaches infinity, but acceleration simply approaches the acceleration of gravity. So the graph has to start leveling off at some point. **
......!!!!!!!!...................................
RESPONSE --> It continues until the slope becomesvertical and then it reaches infinity where acceleration is no more
.................................................
}xƝYΐk` assignment #006 ~zWJ˘FچŸ Physics I Class Notes 11-02-2005
......!!!!!!!!...................................
23:30:07 ** They help us to visualize how all the variables are related. Flow diagrams can also help us to obtain formulas relating the basic kinematic quantities in terms of which we have been analyzing uniformly accelerated motion. **
......!!!!!!!!...................................
RESPONSE --> It helps to see how the variables are related to each other
.................................................
......!!!!!!!!...................................
23:35:48 ** Velocity tells us the rate at which the position changes whereas the acceleration tells us the rate at which the velocity is changing. If acceleration is uniform ave velocity is the average of initial and final velocities. The change in position is found by taking the average velocity vAve = (vf+ v0) / 2 and multiplying by the'dt to get the first fundamental equation `ds = (v0 + vf)/2 * `dt. The acceleration is accel = rate of change of velocity = change in velocity / `dt = (vf - v0) / `dt. In symbols this equation is a = (vf + v0) / `dt. Algebraic rearrangement gives us this equation in the form vf = v0 + a `dt. This form also has an obvious interpretation: a `dt is the change in velocity, which when added to the initial velocity gives us the final velocity. **
......!!!!!!!!...................................
RESPONSE --> Velocity tells us the rate at the position changes and the acceleration tells us the rate the velocity is changing. .The average velocity vAve = (vf+ v0) / 2 The acceleration is accel = rate of change of velocity = change in velocity / `dt = (vf - v0) / `dt.
.................................................
......!!!!!!!!...................................
23:47:00 ** The third equation says that `ds = v0 `dt + .5 a `dt^2. This means that the displacement `ds arises independently from initial velocity v0 and acceleration a: v0 `dt is the displacement of an object with uniform velocity moving at velocity v0, and 1/2 a `dt^2 the distance moved from rest by a uniformly accelerating object. The two contributions are added to get the total `ds. **
......!!!!!!!!...................................
RESPONSE --> The third equation is `ds = v0 `dt + .5 a `dt^2. This means that the displacement `ds arises independently from initial velocity v0 and acceleration a: v0 `dt is the displacement of an object with the same velocity moving at velocity v0, and 1/2 a `dt^2 the distance moved from rest by a uniformly accelerating object. The two are then added to get the total `ds.
.................................................
......!!!!!!!!...................................
23:49:29 ** In this situation we know v0, a and `ds. From v0 and a we cannot draw any conclusions, and the same is true for v0 and `ds and also for a and `ds. No pair of variables allows us to draw any additional information. **
......!!!!!!!!...................................
RESPONSE --> None of these variables let us draw any conclusions
.................................................
......!!!!!!!!...................................
23:54:43 ** We can use the fourth equation vf^2 = v0^2 + 2 a `ds to obtain vf, then knowing the values of v0, vf, a and `ds we easily find `dt either by direct reasoning or by using one of the fundamental equations. **
......!!!!!!!!...................................
RESPONSE --> We can use the fourth equation vf^2 = v0^2 + 2 a `ds to get vf, then after we get the values of v0, vf, a and `ds we can then find by using one of the other fundamental equations.
.................................................
......!!!!!!!!...................................
23:58:32 ** You can use equations without understanding much of anything. To use the equations you don't even need to understand things like average velocity or change in velocity. You just have to be able to identify the right numbers and plug them in, which is an important task in itself but which doesn't involve understanding of the physical concepts behind the equations. **
......!!!!!!!!...................................
RESPONSE --> You can use equations without understanding much of anything about the reasoning behind with whatyou are doing. You just have to know the right numbers to plug in and where.
.................................................
......!!!!!!!!...................................
00:01:03 ** We can sketch a straight line as close as possible to our data points. Then we use the average slope of that graph; this average slope is the acceleration of gravity. **
......!!!!!!!!...................................
RESPONSE --> I can sketch a straight line as close as I can to my points, then use the average slope of that graph which is the acceleration of gravity
.................................................
......!!!!!!!!...................................
00:02:22 ** STUDENT ANSWER: This error causes the slope to increase at an increasing rate rather than form a linear line. INSTRUCTOR COMMENT: Good answer. A systematic error would do that. Even random, non-systematic errors affect the placement of points on the graph, and this tends to affect the slope of the straight line approximating the graph, and also to reduce the accuracy of this slope. **
......!!!!!!!!...................................
RESPONSE --> The error causes the line to increase at an increasing rate rather than being straight like it should be
.................................................
......!!!!!!!!...................................
00:05:10 ** GOOD STUDENT ANSWER: The only effect this systematic error has on the graph is to change the m coordinate of each point by an amount equal to the slope of the table, which is always the same.Since it is the graph slope that comprises our final result, a small table slope would have no effect on our conclusions. **
......!!!!!!!!...................................
RESPONSE --> The only effect this systematic error has on the graph is to change the m coordinate of each point by an amount equal to the slope of the table, which is always the same.Since it is the graph slope that comprises our final result, a small table slope would have no effect on our conclusions
.................................................
......!!!!!!!!...................................
00:07:09 ** GOOD STUDENT ANSWER: The timer is started with a slight delay due to the reaction time of the person doing the timing. This would be OK if the individual's reaction time caused the individual to stop the timer with the same delay. However, the person doing the timing often anticipates the instant when the cart reaches the end of the ramp, so that the delay is not added onto the end time as it was to the starting time. The anticipating individual often triggers the timer slightly before the cart reaches the end, compounding the error even further and also causing the graph to curve rather than remain linear. **
......!!!!!!!!...................................
RESPONSE --> because then person doing the timing's delay in starting the timer and hastiness to stop it in time so it won't go overs욭wʲo㏅Ըay assignment #007 ~zWJ˘FچŸ Physics I Class Notes 11-03-2005
......!!!!!!!!...................................
07:57:59 ** When acceleration is uniform average velocity is the average of initial and final velocities, (vf + v0) / 2. Average velocity is `ds / `dt (whether accel is uniform or not). Setting `ds / `dt = (vf + v0) / 2 we obtain `ds = (vf + v0) / 2 * `dt, which is the first equation of uniformly accelerated motion. So the definition of average velocity is equivalent to the definition of average velocity. Average acceleration is aAve = `dv / `dt = (vf - v0) / `dt. Since for uniform acceleration the acceleration is constant, we can just say that in this case a = (vf - v0) / `dt. This equation is easily rearranged to give the second equation of motion, vf = v0 + a `dt. **
......!!!!!!!!...................................
RESPONSE --> because the first one has to do do with the formula for the avereage velocity while the second one has to do with the formula of acceleration
.................................................
......!!!!!!!!...................................
08:11:37 ** If acceleration is uniform then the v vs. t graph is linear, so that the average velocity over any time interval must be equal to the velocity at the midpoint of that interval. It follows that the average velocity must be midway between the initial and final velocities. (vf + v0) / 2 is the average of the initial and final velocities, and is therefore halfway between the v0 and vf. **
......!!!!!!!!...................................
RESPONSE --> the graph of time vs. velocity is linear if acceleration is uniform. The average velocity has to be equal to the velocity at the midpoint of the interval. So the average velocity must be midway between the first and last velocity
.................................................
......!!!!!!!!...................................
08:24:12 ** If the uniform acceleration is the same in both cases, then assuming that both initial velocity and acceleration are positive, a greater initial velocity will result in a shorter time interval to cover the given distance. A shorter time interval leaves less time for velocity to change, resulting in a smaller change in velocity. **
......!!!!!!!!...................................
RESPONSE --> A greater beginning velocity will cause a shorter time interval, and smaller time interval makes a shorter time for velocity to change, making it a smaller change in velocity.
.................................................
wȬlj}| assignment #008 ~zWJ˘FچŸ Physics I Class Notes 11-03-2005
.................................................