course phy 121 I found the Phy CD # 2 but I am not sure I have the complete set of stuff in the lab kit. oڒ]źassignment #001
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14:50:12 What do we mean by velocity?
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RESPONSE --> By velocity we mean the rate by which an object changes positions.
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14:50:50 ** 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 --> ok
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14:57:26 How can we determine the velocity of a ball rolling down an incline?
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RESPONSE --> We can determine the average velocity of an object by determining the time required to travel through a known displacement.
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14:57:37 ** 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 --> ok
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14:58:52 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 will roll a ball down a variety of different inclines all of the same length. We time the ball on each incline, from a fixed starting point to another fixed point. If the time required to go down an incline is always less on any greater incline, then we conclude that the ball picked up more speed on the steeper incline.
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14:59:16 ** 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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15:03:00 ** 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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15:07:36 How do we determine the rate at which the velocity changes? How can we understand the concept of the rate at which velocity changes?
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RESPONSE --> The rate at which velocity changes is the slope of the velocity vs. time graph. The dv / dt is the rise / run of the slope
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15:08:19 ** 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 --> Change in velocity and divide by change in clock time... ok
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15:16:48 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 --> We multiply average height (which represents average velocity) by average width to get the area of a trapezoid.
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15:17:47 ** 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 --> rise (change in velocity) and run (change in clock time) dv / dt = Ave velocity.
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15:22:30 What does the graph of position vs. clock time look like for constant-acceleration motion?
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RESPONSE --> I am not sure, but I am guessing the graph would look like a slope that has the same rise/ run ratio between each of its points given it is a constant acceleration.
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15:24:37 ** 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 --> ds is change in position which represents the slope and dt is change in clocktime ds / dt rise / run = the velocity
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15:44:41 How can we obtain a graph of velocity vs. clock time from a position vs. clock time graph?
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RESPONSE --> We have started with a velocity vs. time graph, used its areas to obtain a position vs. time graph and used the slopes of this position vs. time graph to get back to the velocities we started with.
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15:46:33 ** 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 --> velocity is CHANGE in position / CHANGE in clock time NOT POSITION / TIME okay.
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15:50:12 How can we obtain a graph of position vs. clock time from a velocity vs. clock time graph
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RESPONSE --> With the velocity = change in position / change in time the position vs clock time doesn't show the average it shows the actual graph.
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15:51:03 ** 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
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15:56:15 How can we obtain a graph of acceleration vs. clock time from a velocity vs. clock time graph?
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RESPONSE -->
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15:57:40 ** 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 -->
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15:57:47 How can we obtain a graph of velocity vs. clock time from an acceleration vs. clock time graph
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RESPONSE -->
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15:58:34 ** 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 -->
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Nez assignment #001 zrԯTߛ m Physics I Vid Clips 09-17-2006
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16:12:12 Physics video clip 01: A ball rolls down a straight inclined ramp. It is the velocity the ball constant? Is the velocity increasing? Is the velocity decreasing?
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RESPONSE --> The velocity would be constant since the ramp is considered level on an incline.
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16:13:13 ** It appears obvious, from common experience and from direct observation, that the velocity of the ball increases. A graph of position vs. clock time would be increasing, indicating that the ball is moving forward. Since the velocity increases the position increases at an increasing rate, so the graph increases at an increasing rate. **
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RESPONSE --> I meant that the velocity is increasing, but the graph position would appear to be a straight line.
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16:13:58 If the ball had a speedometer we could tell. What could we measure to determine whether the velocity of the ball is increase or decreasing?
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RESPONSE --> the change in distance (or displacement) / change in time
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16:14:23 ** STUDENT RESPONSE: By measuring distance and time we could calculate velocity. INSTRUCTOR COMMENTS: The ball could be speeding up or slowing down--all you could get from the calculation you suggest is the average velocity. You could measure the time to travel the first half and the time to travel the second half of the ramp; if the latter is less then we would tend to confirm increasing velocity (though those are still average velocities and we wouldn't get certain proof that the velocity was always increasing). You would need at least two velocities to tell whether velocity is increasing or decreasing. So you would need two sets of distance and time measurements. **
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RESPONSE --> ok
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16:15:10 What is the shape of the velocity vs. clock time graph for the motion of the ball?
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RESPONSE --> I would predict that the shape is a straight line.
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16:15:53 ** If the ramp has an increasing slope, the velocity would increase at an increasing rate and the graph would curve upward, increasing at an increasing rate. If the ramp has a decreasing slope, like a hill that gradually levels off, the graph would be increasing but at a decreasing rate. On a straight incline it turns out that the graph would be linear, increasing at a constant rate, though you aren't expected to know this at this point. All of these answers assume an absence of significant frictional forces such as air resistance. **
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RESPONSE --> oh okay. It would be curved because it is increasing at an increasing rate (since it is on a slope)
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16:16:29 A ball rolls down ramp which curves upward at the starting end and otherwise rests on a level table. What is the shape of the velocity vs. clock time graph for the motion of the ball?
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RESPONSE --> I think it would be increasing at a constant rate.
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16:16:55 ** While on the curved end the ball will be speeding up, and the graph will therefore rise. By the time the ball gets to the level part the velocity will no longer be increasing and the graph will level off; because of friction the graph will actually decrease a bit, along a straight line. As long as the ball is on the ramp the graph will continue on this line until it reaches zero, indicating that the ball eventually stops. In the ideal frictionless situation on an infinite ramp the line just remains level forever. **
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RESPONSE --> ok
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16:18:08 For the ball on the straight incline, we would certainly agree that the ball's velocity is increasing. Is the velocity increasing at a constant, an increasing, or a decreasing rate? What does the graph of velocity vs. clock time look like?
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RESPONSE --> I am sure that it is not increasing at an increasing rate, but does it remain constant or is it decreasing?
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16:18:37 ** It turns out that on a straight incline the velocity increases at a constant rate, so the graph is a straight line which increases from left to right. Note for future reference that a ball on a constant incline will tend to have a straight-line v vs. t graph; if the ball was on a curved ramp its velocity vs. clock time graph would not be straight, but would deviate from straightness depending on the nature of the curvature (e.g., slope decreasing at increasing rate implies v vs. t graph increasing at increasing rate).**
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RESPONSE --> ok so it is constant
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]\K assignment #001 zrԯTߛ m Physics I 09-17-2006
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16:23:51 Briefly state what you think velocity is and how you think it is an example of a rate.
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RESPONSE --> Velocity is the rate at which an object changes positions. a Rate is something per something else such as miles per hour which in this case can also represent velocity since it is changing positions. But velocity cannot be the same as a rate as you cannot get the velocity of dollars per hour.
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16:24:03 ** A rate is a change in something divided by a change in something else. This question concerns velocity, which is the rate of change of position: change in position divided by change in clock time. **
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RESPONSE --> ok
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16:27:07 Given average speed and time interval how do you find distance moved?
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RESPONSE --> the average speed is distance traveled/ time of travel but I am not sure how to work this backwards using the average time to get the total distance.
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16:28:06 ** You multiply average speed * time interval to find distance moved. For example, 50 miles / hour * 3 hours = 150 miles. **
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RESPONSE --> I thought it was an average time I must have misread it. But I do understand how to get to this answer.
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16:29:20 Given average speed and distance moved how do you find the corresponding time interval?
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RESPONSE --> Given the average speed / the distance = time Ex. 50mph / 50 miles = 1 hour
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16:30:12 ** time interval = distance / average speed. For example if we travel 100 miles at 50 mph it takes 2 hours--we divide the distance by the speed. In symbols, if `ds = vAve * `dt then `dt = `ds/vAve. Also note that (cm/s ) / s = cm/s^2, not sec, whereas cm / (cm/s) = cm * s / cm = s, as appropriate in a calculation of `dt. **
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RESPONSE --> lol okay I think I am a little backwards today. I meant distance/ ave speed.
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16:30:59 Given time interval and distance moved how do you get average speed?
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RESPONSE --> the distance traveled / time traveled = Ave speed
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16:31:07 ** Average speed = distance / change in clock time. This is the definition of average speed. For example if we travel 300 miles in 5 hours we have been traveling at an average speed of 300 miles / 5 hours = 60 miles / hour. **
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RESPONSE --> ok
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"
course phy 121 I found the Phy CD # 2 but I am not sure I have the complete set of stuff in the lab kit. oڒ]źassignment #001
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14:50:12 What do we mean by velocity?
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RESPONSE --> By velocity we mean the rate by which an object changes positions.
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14:50:50 ** 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 --> ok
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14:57:26 How can we determine the velocity of a ball rolling down an incline?
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RESPONSE --> We can determine the average velocity of an object by determining the time required to travel through a known displacement.
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14:57:37 ** 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 --> ok
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14:58:52 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 will roll a ball down a variety of different inclines all of the same length. We time the ball on each incline, from a fixed starting point to another fixed point. If the time required to go down an incline is always less on any greater incline, then we conclude that the ball picked up more speed on the steeper incline.
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14:59:16 ** 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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15:03:00 ** 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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15:07:36 How do we determine the rate at which the velocity changes? How can we understand the concept of the rate at which velocity changes?
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RESPONSE --> The rate at which velocity changes is the slope of the velocity vs. time graph. The dv / dt is the rise / run of the slope
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15:08:19 ** 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 --> Change in velocity and divide by change in clock time... ok
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15:16:48 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 --> We multiply average height (which represents average velocity) by average width to get the area of a trapezoid.
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15:17:47 ** 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 --> rise (change in velocity) and run (change in clock time) dv / dt = Ave velocity.
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15:22:30 What does the graph of position vs. clock time look like for constant-acceleration motion?
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RESPONSE --> I am not sure, but I am guessing the graph would look like a slope that has the same rise/ run ratio between each of its points given it is a constant acceleration.
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15:24:37 ** 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 --> ds is change in position which represents the slope and dt is change in clocktime ds / dt rise / run = the velocity
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15:44:41 How can we obtain a graph of velocity vs. clock time from a position vs. clock time graph?
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RESPONSE --> We have started with a velocity vs. time graph, used its areas to obtain a position vs. time graph and used the slopes of this position vs. time graph to get back to the velocities we started with.
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15:46:33 ** 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 --> velocity is CHANGE in position / CHANGE in clock time NOT POSITION / TIME okay.
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15:50:12 How can we obtain a graph of position vs. clock time from a velocity vs. clock time graph
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RESPONSE --> With the velocity = change in position / change in time the position vs clock time doesn't show the average it shows the actual graph.
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15:51:03 ** 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
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15:56:15 How can we obtain a graph of acceleration vs. clock time from a velocity vs. clock time graph?
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RESPONSE -->
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15:57:40 ** 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 -->
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15:57:47 How can we obtain a graph of velocity vs. clock time from an acceleration vs. clock time graph
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RESPONSE -->
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15:58:34 ** 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 -->
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Nez assignment #001 zrԯTߛ m Physics I Vid Clips 09-17-2006
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16:12:12 Physics video clip 01: A ball rolls down a straight inclined ramp. It is the velocity the ball constant? Is the velocity increasing? Is the velocity decreasing?
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RESPONSE --> The velocity would be constant since the ramp is considered level on an incline.
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16:13:13 ** It appears obvious, from common experience and from direct observation, that the velocity of the ball increases. A graph of position vs. clock time would be increasing, indicating that the ball is moving forward. Since the velocity increases the position increases at an increasing rate, so the graph increases at an increasing rate. **
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RESPONSE --> I meant that the velocity is increasing, but the graph position would appear to be a straight line.
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16:13:58 If the ball had a speedometer we could tell. What could we measure to determine whether the velocity of the ball is increase or decreasing?
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RESPONSE --> the change in distance (or displacement) / change in time
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16:14:23 ** STUDENT RESPONSE: By measuring distance and time we could calculate velocity. INSTRUCTOR COMMENTS: The ball could be speeding up or slowing down--all you could get from the calculation you suggest is the average velocity. You could measure the time to travel the first half and the time to travel the second half of the ramp; if the latter is less then we would tend to confirm increasing velocity (though those are still average velocities and we wouldn't get certain proof that the velocity was always increasing). You would need at least two velocities to tell whether velocity is increasing or decreasing. So you would need two sets of distance and time measurements. **
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RESPONSE --> ok
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16:15:10 What is the shape of the velocity vs. clock time graph for the motion of the ball?
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RESPONSE --> I would predict that the shape is a straight line.
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16:15:53 ** If the ramp has an increasing slope, the velocity would increase at an increasing rate and the graph would curve upward, increasing at an increasing rate. If the ramp has a decreasing slope, like a hill that gradually levels off, the graph would be increasing but at a decreasing rate. On a straight incline it turns out that the graph would be linear, increasing at a constant rate, though you aren't expected to know this at this point. All of these answers assume an absence of significant frictional forces such as air resistance. **
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RESPONSE --> oh okay. It would be curved because it is increasing at an increasing rate (since it is on a slope)
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16:16:29 A ball rolls down ramp which curves upward at the starting end and otherwise rests on a level table. What is the shape of the velocity vs. clock time graph for the motion of the ball?
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RESPONSE --> I think it would be increasing at a constant rate.
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16:16:55 ** While on the curved end the ball will be speeding up, and the graph will therefore rise. By the time the ball gets to the level part the velocity will no longer be increasing and the graph will level off; because of friction the graph will actually decrease a bit, along a straight line. As long as the ball is on the ramp the graph will continue on this line until it reaches zero, indicating that the ball eventually stops. In the ideal frictionless situation on an infinite ramp the line just remains level forever. **
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RESPONSE --> ok
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16:18:08 For the ball on the straight incline, we would certainly agree that the ball's velocity is increasing. Is the velocity increasing at a constant, an increasing, or a decreasing rate? What does the graph of velocity vs. clock time look like?
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RESPONSE --> I am sure that it is not increasing at an increasing rate, but does it remain constant or is it decreasing?
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16:18:37 ** It turns out that on a straight incline the velocity increases at a constant rate, so the graph is a straight line which increases from left to right. Note for future reference that a ball on a constant incline will tend to have a straight-line v vs. t graph; if the ball was on a curved ramp its velocity vs. clock time graph would not be straight, but would deviate from straightness depending on the nature of the curvature (e.g., slope decreasing at increasing rate implies v vs. t graph increasing at increasing rate).**
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RESPONSE --> ok so it is constant
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]\K assignment #001 zrԯTߛ m Physics I 09-17-2006
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16:23:51 Briefly state what you think velocity is and how you think it is an example of a rate.
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RESPONSE --> Velocity is the rate at which an object changes positions. a Rate is something per something else such as miles per hour which in this case can also represent velocity since it is changing positions. But velocity cannot be the same as a rate as you cannot get the velocity of dollars per hour.
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16:24:03 ** A rate is a change in something divided by a change in something else. This question concerns velocity, which is the rate of change of position: change in position divided by change in clock time. **
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RESPONSE --> ok
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16:27:07 Given average speed and time interval how do you find distance moved?
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RESPONSE --> the average speed is distance traveled/ time of travel but I am not sure how to work this backwards using the average time to get the total distance.
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16:28:06 ** You multiply average speed * time interval to find distance moved. For example, 50 miles / hour * 3 hours = 150 miles. **
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RESPONSE --> I thought it was an average time I must have misread it. But I do understand how to get to this answer.
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16:29:20 Given average speed and distance moved how do you find the corresponding time interval?
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RESPONSE --> Given the average speed / the distance = time Ex. 50mph / 50 miles = 1 hour
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16:30:12 ** time interval = distance / average speed. For example if we travel 100 miles at 50 mph it takes 2 hours--we divide the distance by the speed. In symbols, if `ds = vAve * `dt then `dt = `ds/vAve. Also note that (cm/s ) / s = cm/s^2, not sec, whereas cm / (cm/s) = cm * s / cm = s, as appropriate in a calculation of `dt. **
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RESPONSE --> lol okay I think I am a little backwards today. I meant distance/ ave speed.
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16:30:59 Given time interval and distance moved how do you get average speed?
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RESPONSE --> the distance traveled / time traveled = Ave speed
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16:31:07 ** Average speed = distance / change in clock time. This is the definition of average speed. For example if we travel 300 miles in 5 hours we have been traveling at an average speed of 300 miles / 5 hours = 60 miles / hour. **
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RESPONSE --> ok
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