course Phy121
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22:59:39 What do we mean by velocity?
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RESPONSE --> Velocity signifies how fast an object is moving and also where it is moving to. How fast and how far (so to speak).
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23:01:00 ** STUDENT RESPONSE:Velocity is the speed and direction an object is moving. INSTRUCTOR COMMENT: Good. More succinctly and precisely velocity is the rate at which position is changing. obtained by dividing change in position by change in clock time **
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RESPONSE --> Change in position, so its displacement, not only distance. Distance is always positive, whereas displacement can include positive and negative.
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23:01:25 How can we determine the velocity of a ball rolling down an incline?
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RESPONSE --> We can determined the velocity of a ball rolling down an incline by measuring the distance of this incline, and timing how long it takes the ball to travel this incline.
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23:02:43 ** 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. **
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RESPONSE --> Displacement divided by time = average velocity. I phrased it as distance, but perhaps, more accurate to say displacement, or position change.
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23:03:25 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?
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RESPONSE --> We can determine if a ball traveling down different slopes will experience a greater change in velocity by comparing the amount of time it takes to travel the same distance at the various slopes. We can also determine this change on an individual slope by comparing the velocities between the initial part, and a portion downslope. The velocity should increase as the ball travels from a greater incline, and will increase as it reaches the bottom of this incline compared to its velocity at the starting point.
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23:03:43 ** 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. **
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RESPONSE --> Ok.
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23:04:29 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.
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RESPONSE --> y utilizing slope triangles to find this information. We do the intial slope triangle from the point we are interested in (t0), to a point later in time. We find its slope (rise/run), and this shows the average acceleration from t0 to a point in the future. If we create this same triangle,and others after this, but at points closer in time to t0, the slopes of these triangles, approach the slope of the tangent line, or the limiting value of the acceleration, and thus, the value of t0, which is the acceleration at this instant.
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23:05:21 ** 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. **
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RESPONSE --> Ok.
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23:05:52 How do we determine the rate at which the velocity changes? How can we understand the concept of the rate at which velocity changes?
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RESPONSE --> We determine the rate at which velocity changes by looking at the change in position over time. We can understand this by conducting experiements, by solving problems, by graphing our solutions to illustrate this change. By determining the rise over run value, we can determine the rate of change for an object.
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23:06:02 ** 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. **
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RESPONSE --> Ok.
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23:09:54 It is essential to understand what a trapezoid on a v vs. t graph represents. Give the meaning of the rise and run between two points, and the meaning of the area of a trapezoid defined by a v vs. t graph.
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RESPONSE --> The rise and run between two points is the slope, which represents the average acceleration. The area of a trapezoid defined by a v vs t graph represents the the dispacement of an object during a time interval.
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23:11:46 ** 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. **
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RESPONSE --> Ok. The explanation of this at the end of #3 notes is very helpful, thanks for including it.
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23:12:59 What does the graph of position vs. clock time look like for constant-acceleration motion?
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RESPONSE --> A straight line.
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23:17:38 ** 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. **
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RESPONSE --> Ok, I realize now you were looking for more information. Yes, the graph for constant positive acceleration would be increasing at an increasing rate. There are also other shapes depending upon intial and ending velocity. If a car is slowly decelerating the farther it travels, it would be increasing at a decreasing rate.
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23:20:47 How can we obtain a graph of velocity vs. clock time from a position vs. clock time graph?
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RESPONSE --> A position vs clock time graph is basically average velocity vs. time. From these data we can determine the velocity data and graph accordingly.
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23:24:22 ** We can find the slope of the position vs. clock time graph at a series of clock times, which will give us the velocities at those clock times. We can put this information into a velocity vs. clock time table then plot the velocities vs. clock time as a 'guidepost points', and fill in the connecting curve in such a way as to be consistent with the trend of the slopes of the position vs. clock time graph. COMMON MISCONCEPTION: To get velocity vs. clock time find average velocity, which is position (m) divided by time (s). Plot these points of vAvg on the velocity vs. time graph. INSTRUCTOR RESPONSE: Ave velocity is change in position divided by change in clock time. It is not position divided by time. Position can be measured from any reference point, which would affect a position/time result, but which would not affect change in position/time. Graphically velocity is the slope of the position vs. clock time graph. If it was just position divided by time, it would be the slope of a line from the origin to the graph point. **
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RESPONSE --> I am guilty of the common misconception. Rather than attempting to convert back to velocities, we can find the velocities directly from the position vs. clock time graph by calculating slopes at various times. Could you provide an example of this? I will try and do one on my own as well, but I just want to make sure I'm following this correctly.
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23:27:50 How can we obtain a graph of position vs. clock time from a velocity vs. clock time graph
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RESPONSE --> To find the change of position, we need to multiply the average velocity by the time interval. We find the average velocity by dividing the change in velocity over the change in time.
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23:29:36 ** 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. **
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RESPONSE --> Ok. This utilizes the trapezoid method to discover the displacement. I was using the first method we were shown in the lecture.
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23:33:06 How can we obtain a graph of acceleration vs. clock time from a velocity vs. clock time graph?
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RESPONSE --> Acceleration consists of the change in velocity over the change in time. It is the slope of this graph. We can obtain the data by finding the height at the midpoint between the time intervals.
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23:34:02 ** 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. **
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RESPONSE --> OK.
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23:37:20 How can we obtain a graph of velocity vs. clock time from an acceleration vs. clock time graph
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RESPONSE --> Since acceleration is determined by the change in velocity over the change in time,
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23:39:19 ** STUDENT RESPONSE: Take your acceleration and multiply by time to find the change in velocity. Start with initial velocity and graph your velocity by increasing initial velocity by the slope, or change in velocity. INSTRUCTOR COMMENT: Good. More precisely we can approximate change in velocity during a given time interval by finding the approximate area under the acceleration vs. clock time graph for the interval. We can then add each change in velocity to the existing velocity, constructing the velocity vs. clock time graph interval by interval. A velocity vs. clock time graph has slopes which are equal at every point to the vertical coordinate of the acceleration vs. clock time graph. University Physics students note: These two statements are equivalent, and the reason they are is at the heart of the Fundamental Theorem of Calculus. **
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RESPONSE --> Oops, became trigger happy. As I was saying. Acceleration is the change in velocity over the change in time. We can figure out velocity from this equation. The graph for acceleration will be the slope of the velocity versus time graph.
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Ƒ̂|圫e֎R assignment #002 ̐輻Iyʴ Physics I Class Notes 01-29-2006
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23:42:34 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?
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RESPONSE --> As more time passes, going from 0 to 1 to 2 to 3 to 4, etc. the graph is increasing, not decreasing. It is also increasing at an increasing rate. The longer the time, the farther it travels, thus, it must be traveling faster as time passes.
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23:44:09 ** STUDENT RESPONSE: The shape of the graph, with its upward curve, shows us that it is increasing at an increasing rate over the duration of time. The average velocity between initial time and t1 is `ds/`dt = 1/1 or 1 m/s. From t1 to t2, `ds/`dt = (4-1)/1 = 3 m/s. The next 1 second increment produces a velocity of 5 m/s, and finally from t3 to t4, velocity is shown to be 7 m/s. It is simple to visually see that from the first half of the time duration, the slope is less, than for the second half of the time duration. Also, finding the average s, (16-0)/2 = 8, we find that the average or halfway distance is reached at 3 seconds, or three-quarters of the way through the total time duration, and in the next second, the distance doubles, therefore it does take longer to travel the first half of the incline, and much less time to travel the second half. It is easier to see than to explain. INSTRUCTOR COMMENT: Everything you say is correct. The halfway position is indicated by the position on the vertical or y axis halfway between initial and final position. The instant at which the halfway position is attained is the corresponding coordinate on the horizontal or t axis. The time required for the change in position from the initial to the halfway position is represented by the interval between the two corresponding clock times. On a graph which is concave down, this position will occur more than halfway between initial and final clock times. ** ANOTHER GOOD STUDENT RESPONSE: We could mark on the graph the initial and final points, then move over to the y axis and mark the position at the intial point and at the final point. We could then mark the y coordinate halfway between these and move over to the graph to obtain the graph point which corresponds to the halfway point. We can then construct a line segment from the graph point corresponding to the initial point, to the graph point corresponding to the halfway point, and finally to the graph point corresponding to the final point. Since the graph increases at an increasing rate the slope of the second segment is greater than that of the first. Since the change in the y coordinate is the same for both, it follows that the run of the second segment is shorter than the first. Since the run corresponds to the time interval, we see that the time interval corresponding to the second half is shorter than that corresponding to the first half. **
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RESPONSE --> Wow, these are very thorough and detailed answers. I see what is needed.
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23:48:21 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
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RESPONSE --> From the velocity changes, and change in time, we can calculate average velocity or acceleration of the object. From this we can then determine the displacement at each time interval.
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23:50:26 ** 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. **
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RESPONSE --> I forgot to mention that you multiply acceleration by the time interval.
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23:56:41 position vs. clock time?
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RESPONSE --> We can determine position changes over specified time intervals from a graph of velocity vs. clock time, by figuring out the area of a trapezoid between two points. We multiple the average altitude by the width of the trapezoid, we're multiplying the average velocity by time, which results in the displacement over that time interval. We can then plot these points along the graph.
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23:58:17 ** 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. **
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RESPONSE --> I only did this for one time interval, and did not carry it out for what would be needed to construct a graph.
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23:59:51 In terms of the meanings of altitudes, area and width, how does a velocity vs. clock time trapezoid represent change in position?
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RESPONSE --> Altitude represents the height of the trapezoid, as well as the velocity. The width represents the time. Area is found by the average altitude multipled by the time interval. The average altitude is also the average velocity. Multiplying the average velocity by the time equals the change in position.
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00:00:12 ** 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.
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RESPONSE --> Ok.
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00:03:13 How can a series of velocity vs. clock time trapezoids help us to calculate and visualize position vs. clock time information?
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RESPONSE --> Several trapezoids will provide the change in position from the start of the first trapezoid to the end of the last. This will provide us with a group or collection of position changes, since the area of each trapezoid is a position change.
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00:23:07 ** 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 **
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RESPONSE --> I did not think how visually the trapezoids would demonstrate increasing heights as time increased. I also keep forgetting to include a statement about adding the successive displacements.
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x|QKNϚxK assignment #003 ̐輻Iyʴ Physics I Class Notes 01-30-2006
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00:27:09 Given the initial velocity, final velocity and time duration of a uniformly accelerating object, how do we reason out the corresponding acceleration and change in the position of an object?
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RESPONSE --> From the initial velocity, final velocity and time duration ,we are able to find the average velocity/acceleration of the object. We divide the change in velocity (vf-v0) by the change in time (tf-t0). Then at each time interval, we multiply the average velocity by the time to produce the displacement for that interval. We can then add the displacement figures for each time interval for the accumulated change in position.
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00:27:53 ** 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. **
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RESPONSE --> Ok.
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00:30:22 In terms of the meanings of altitudes, area, slope and width, how does a velocity vs. clock time trapezoid represent change in position and acceleration?
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RESPONSE --> a velocity vs. clock time trapezoid represents change in position and acceleration, in that the area of this trapezoid equals the change in position. The height of the trapezoid = average altitude or average velocity (acceleration), and the width of the trapezoid = time interval. By multiplying the height times the width, we produce the area of the displacement.
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00:32:07 ** 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. **
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RESPONSE --> I forgot the piece about the slope and triangle.
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