course Phy 241
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20:12:54 What do we mean by velocity?
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RESPONSE --> A known displacement over a change in time.
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20:13:42 ** 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 --> Yes direction is also included.
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20:14:26 How can we determine the velocity of a ball rolling down an incline?
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RESPONSE --> Time the ball and record the distance it travels down the ramp. The velocity is the distance per unit time.
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20:29:27 ** 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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20:30:55 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 --> Take velocity measurements a variety of times between the traveled distance. Graph the results on a velocity vs time graph. This will show acceleration properties or velocity changes per second.
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20:35:25 ** 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 --> Is this referring to the same ramp? I am a little confused on the question now. If there are two ramps, and two balls why would the one with the greatest average velocity necessarily imply anything about acceleration?
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20:36:07 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 --> Well if we are not using derivatives, we can estimate it by using the triangles and get a slope that is approaching that of the tangent line on the velocity vs time graph.
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20:36:44 ** 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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20:46:01 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 --> Know the velocity at two points in time. Take the difference or dv and divide by the dt. The rate at which velocity changes is the same as acceleration.
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20:46:30 ** 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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20:50:41 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 can be used to determine the average acceleration. If you draw a tangent line to a specific point on a v vs t graph you can find a specific accelertion value with the slope approaching the tangent line.
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20:54:59 ** 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 --> How will a rise and run make a trapezoid?
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20:55:17 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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20:59:35 ** 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 --> Oh i was thinking the velocity vs time graph would show a straight line if the acceleration is constant.
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22:07:07 How can we obtain a graph of velocity vs. clock time from a position vs. clock time graph?
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RESPONSE --> If the velocity is constant, the velocity is a horizontal line and can be found by the slope of the position vs clock time graph. If the velocity is not constant, different points must be chosen fromt he graph and calculate the change in position per change in time to obtain points for the velocity vs time graph.
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22:11:21 ** 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 --> Ok I understand.
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22:14:03 How can we obtain a graph of position vs. clock time from a velocity vs. clock time graph
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RESPONSE --> We can calculate change in position by finding the average velocities and multiplying by the time passed.
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22:16:33 ** 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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22:17:02 How can we obtain a graph of acceleration vs. clock time from a velocity vs. clock time graph?
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RESPONSE --> The acceleration is found as the tangent slope to points on the v vs t graph
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22:18:53 ** 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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22:19:59 How can we obtain a graph of velocity vs. clock time from an acceleration vs. clock time graph
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RESPONSE --> Divide the graph into sections and find the area of each, much like obtaining position v time from the velocity v time graph.
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22:22:05 ** 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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ܝȧy҆RϮ assignment #002 q} Physics I Class Notes 06-04-2006
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22:29:12 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 --> If the second half of the graph is more so curved the increase over a certain time interval is greater, therefore it takes less time to travel the same distance.
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22:31:41 ** 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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22:37:45 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 --> If acceleration is known and the initial velocity is known, we can break the times into sections and add the change in velocity to the initial velocity. Then the positions can be found by finding the area under the graph made by these velocity points. This is done by multiplying each velocity section by the duration of time.
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22:37:50 object?
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22:47:44 ** 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 am not sure if my reasoning is completely wrong or not. This makes sense though to multiply acceleration or dv/dt x the time to get the change in velocity. Average velocity if found by with the initial and final found from that. Then you can use the ds/dt to multiply by time to get a change in position.
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22:47:51 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
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22:52:02 position vs. clock time?
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RESPONSE --> If you have velocity over a certain period of time just multiply the velocity by the time and you obtain a change in position. Then use the intervals of time corresponding to each position to create the position vs time graph.
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22:55:42 ** 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 --> The velocity of the certain period of times I spoke about are the dimensions of the trapezoid in the answer. Also adding how to construct the final graph- the individual position changes within the trapezoid are added to the previous total of the change.
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22:57:17 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 --> The time is the width and the altitude represents the average velocity. By multiplying the width (dt) by the velocity (ds/dt) you can obtain the area of ds (the change of position).
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22:57:53 ** 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 --> My response was correct.
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22:59:53 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 --> Each trapezoid of the v vs t graph represents a change in position. If we start with the initial postition and add each change to the previous we can construct the position vs clock graph. The quantity of the position change dictates the steepness of the graph. If the change is quick and large the graph will show a sharp increasing trend.
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23:00:48 ** 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 --> My answer is also correct.
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Ę̄|䐃JThfq assignment #003 q} Physics I Class Notes 06-04-2006
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23:04:35 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 --> We can calculate the average change of velocity by adding the initial and final velocities and dividing by two. To find the average acceleration we can divide by the time duration. The change in the position of the object can be found by finding the area under the velocity function or taking the integral.
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23:07:31 ** 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 --> yes it is much easier to multiply the ds/dt by dt to get ds of the change in position.
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23:12:10 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 --> The altitude of the trapezoid in the v vs time graph represents the average velocity. The width represents the change in time. The area that is created here is the change in position. It can be evaulated by multiplying the width and the avg altitude. The slope of the trapezoid signifies the changing velocity. If the velocity were constant the slope would be 0 and the area could still be found by multiplying the sides of the rectangle. The x axis still being time and the y being velocity. If this is the case, the acceleration is zero, but if the slope is not horizontal the acceleration at a point can be found by the slope of the tangent line at that point or the triangle slope method in which the triangles becoming smaller and enclosing on that certain point.
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23:13:33 ** 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 --> yes, i understand the slope of the graph represents the acceleration or the rate of change of the velocity.
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