course Phy 121 I am submitting everything I have for assignment one including a partially incomplete query. I have already emailed you about it.
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09:45:52 `questionNumber 10000 What do we mean by velocity?
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RESPONSE --> Velocity can be determined by measuring the time required to travel through a known displacement. The velocity of the ball continues to change as the ball rolls down the incline.
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09:47:10 01-18-2006 09:47:10 ** 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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NOTES -------> Velocity is the rate at which position is changing. change in position/change in clock time.
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09:49:23 `questionNumber 10000 How can we determine the velocity of a ball rolling down an incline?
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RESPONSE --> Measure the distance that the ball rolls and divide that number by how long it took the ball to roll.
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09:50:08 01-18-2006 09:50:08 ** 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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NOTES -------> displacement/time=average velocity
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09:55:48 `questionNumber 10000 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 --> Roll the ball down a variety of different inclines all the same length. Measure the time it takes the ball to roll down the incline. If the time required to go down an incline is always less on a steeper slope then it can be concluded that the ball travels down a steeper slope faster.
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09:56:07 `questionNumber 10000 ** 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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10:01:17 `questionNumber 10000 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 --> Measure specific points on the ramp and time the ball each time it passes one of these points. Divide the distance by the time.
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10:04:17 01-18-2006 10:04:17 ** 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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NOTES -------> To obtain an accurate reading you would have to figure the velocity function and take its derivative.
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10:10:14 `questionNumber 10000 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 --> Divide the change in velocity by the change in time. Create a graph. The rate at which velocity changes is the slope of the velocity versus time.
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10:10:50 `questionNumber 10000 ** 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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ۤqa|`˓}ܑZ assignment #002 [뺗܃ܝ Physics I Class Notes 01-18-2006
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10:18:53 `questionNumber 10000 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 of two points can be divided to give you the slope between those to points. The area of a trapezoid defined by a v vs. t graph is the position change.
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10:20:49 `questionNumber 10000 ** 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 --> Instead of slope rise over run here represents the average velocity over the corresponding time interval.
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10:20:50 `questionNumber 10000 ** 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 -->
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10:29:44 `questionNumber 10000 What does the graph of position vs. clock time look like for constant-acceleration motion?
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RESPONSE --> The graph is concave up.
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10:30:32 `questionNumber 10000 ** 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
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10:34:28 `questionNumber 10000 How can we obtain a graph of velocity vs. clock time from a position vs. clock time graph?
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RESPONSE --> The slope between two points gives you the average velocity between clock times. The slope of a position vs. clock time graph represents average velocity. Tje graph of area under the velocity graph vs. clock time has slope which gives us back average velocity.
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10:34:58 `questionNumber 10000 ** 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
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10:36:08 `questionNumber 10000 How can we obtain a graph of position vs. clock time from a velocity vs. clock time graph
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RESPONSE --> This is the same as the previous question I just answered.
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10:37:13 `questionNumber 10000 ** 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 --> I still thought the question was the same as the previous so why does it have a different answer?
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\TnTАWwC{ assignment #001 [뺗܃ܝ Physics I Class Notes 01-18-2006
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16:22:40 `questionNumber 10000 How can we obtain a graph of velocity vs. clock time from an acceleration vs. clock time graph
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RESPONSE --> Didn't mean to leave a blank answer on this question. Acceleration is the rate of change in velocity which is represented by the slope on the graph. This would be graphed similarly to the way we plotted the slope of the position vs. clock time.
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16:24:05 `questionNumber 10000 ** 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 --> ok
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ȞsҔ]ݦz assignment #001 [뺗܃ܝ Physics I Class Notes 01-18-2006
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16:38:32 `questionNumber 10000 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 the change in velocity over clock time we will use that slope and graph the picture similarly to the velocity vs. clock time graph.
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16:43:40 `questionNumber 10000 ** 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 --> the approximate change in area under the acceleration vs. clock time graph for the interval. Add change in velocity to the previous velocity and make the graph interval by interval.
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snխ{ assignment #001 [뺗܃ܝ Physics I Class Notes 01-18-2006 H礔QL|yޣ assignment #002 [뺗܃ܝ Physics I Class Notes 01-18-2006
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16:53:05 `questionNumber 20000 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 --> The shape of the corresponding position vs. clock time graph shows that the time required to travel the first half of the incline is greater than that required to travel the second half because the first half of the graph is not as steep as the second half.
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16:54:14 01-18-2006 16:54:14 ** 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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NOTES -------> Over time the curve is increasing at an increasing rate.
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16:56:05 `questionNumber 20000 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 --> Not all of the question is visible and it is unclear as to what the question is therefore, I cannot answer this question.
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16:58:29 `questionNumber 20000 object?
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RESPONSE --> ok so the question is on two slides. measure the changes.
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16:59:32 `questionNumber 20000 ** 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 don't understand this answer.
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17:09:13 `questionNumber 20000 position vs. clock time?
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RESPONSE --> Note the average velocity. Calculate the position change as a change in area. The height corresponds to the velocity and the width corresponds to time by which we multiply average velocity.
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17:10:26 01-18-2006 17:10:26 ** 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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NOTES -------> You can use trapezoids to get a series of areas.
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17:15:40 `questionNumber 20000 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 --> Height corresponds to velocity width corresponds with time and altitude corresponds with average velocity.
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17:16:11 `questionNumber 20000 ** 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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17:24:38 `questionNumber 20000 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 --> The graph of area und the velocity vs. clock time has slopes which gives us average velocities. You can use a series of these velocities to help create a position vs. clock time graph.
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17:26:08 01-18-2006 17:26:08 ** 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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NOTES -------> The position change for a time interval corresponds to the area of the graphed trapezoid for that time interval. The greater the area the greater the change in position.
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Ŧؤ assignment #003 [뺗܃ܝ Physics I Class Notes 01-18-2006
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17:37:52 `questionNumber 30000 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 --> To find the average velocity average the initial velocity and the final velocity. Find the displacement by multiplying the average velocity by the time interval. The displacement will be the change in position. To find the acceleration divide the velocity by the time duration.
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17:38:08 `questionNumber 30000 ** 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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17:52:25 `questionNumber 30000 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 slope represents the time rate of change of the velocity known as acceleration. Multiply the average altitude by the width of the trapezoid, which is the same as multiplying the average velocity by the time duration to obtain the displacement.
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17:52:52 `questionNumber 30000 ** 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 --> Okay.
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דȊlj\ٵڒБ Student Name: assignment #001 001. Areas
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17:57:30 `q001. There are 11 questions and 7 summary questions in this assignment. What is the area of a rectangle whose dimensions are 4 m by 3 meters.
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RESPONSE --> Area=lenght*width 4*3=12 meters
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17:58:22 01-18-2006 17:58:22 A 4 m by 3 m rectangle can be divided into 3 rows of 4 squares, each 1 meter on a side. This makes 3 * 4 = 12 such squares. Each 1 meter square has an area of 1 square meter, or 1 m^2. The total area of the rectangle is therefore 12 square meters, or 12 m^2. The formula for the area of a rectangle is A = L * W, where L is the length and W the width of the rectangle. Applying this formula to the present problem we obtain area A = L * W = 4 m * 3 m = (4 * 3) ( m * m ) = 12 m^2. Note the use of the unit m, standing for meters, in the entire calculation. Note that m * m = m^2.
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NOTES -------> I forgot to square the meters
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18:04:51 `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?
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RESPONSE --> Area of a triangle is 1/2bh. A=1/2*4*3 A= 6m^2
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18:05:03 A right triangle can be joined along its hypotenuse with another identical right triangle to form a rectangle. In this case the rectangle would have dimensions 4.0 meters by 3.0 meters, and would be divided by any diagonal into two identical right triangles with legs of 4.0 meters and 3.0 meters. The rectangle will have area A = L * W = 4.0 m * 3.0 m = 12 m^2, as explained in the preceding problem. Each of the two right triangles, since they are identical, will therefore have half this area, or 1/2 * 12 m^2 = 6.0 m^2. The formula for the area of a right triangle with base b and altitude h is A = 1/2 * b * h.
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RESPONSE --> ok
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18:09:11 `q003. What is the area of a parallelogram whose base is 5.0 meters and whose altitude is 2.0 meters?
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RESPONSE --> A=b*h A=5*2 A=10 m^2
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18:09:22 A parallelogram is easily rearranged into a rectangle by 'cutting off' the protruding end, turning that portion upside down and joining it to the other end. Hopefully you are familiar with this construction. In any case the resulting rectangle has sides equal to the base and the altitude so its area is A = b * h. The present rectangle has area A = 5.0 m * 2.0 m = 10 m^2.
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RESPONSE --> ok
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18:12:13 `q004. What is the area of a triangle whose base is 5.0 cm and whose altitude is 2.0 cm?
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RESPONSE --> A=1/2b*h A=1/2*5*2 A=5 cm^2
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18:12:18 It is possible to join any triangle with an identical copy of itself to construct a parallelogram whose base and altitude are equal to the base and altitude of the triangle. The area of the parallelogram is A = b * h, so the area of each of the two identical triangles formed by 'cutting' the parallelogram about the approriate diagonal is A = 1/2 * b * h. The area of the present triangle is therefore A = 1/2 * 5.0 cm * 2.0 cm = 1/2 * 10 cm^2 = 5.0 cm^2.
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RESPONSE --> ok
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18:15:19 `q005. What is the area of a trapezoid with a width of 4.0 km and average altitude of 5.0 km?
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RESPONSE --> Area a trapezoid=1/2(b1+b2)h A=1/2(4+4)5 A=20km^2
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18:15:24 Any trapezoid can be reconstructed to form a rectangle whose width is equal to that of the trapezoid and whose altitude is equal to the average of the two altitudes of the trapezoid. The area of the rectangle, and therefore the trapezoid, is therefore A = base * average altitude. In the present case this area is A = 4.0 km * 5.0 km = 20 km^2.
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RESPONSE --> ok
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18:18:59 `q006. What is the area of a trapezoid whose width is 4 cm in whose altitudes are 3.0 cm and 8.0 cm?
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RESPONSE --> 1/2(b1+b2)h A=1/2(3+8)4 A=22cm^2
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18:19:19 The area is equal to the product of the width and the average altitude. Average altitude is (3 cm + 8 cm) / 2 = 5.5 cm so the area of the trapezoid is A = 4 cm * 5.5 cm = 22 cm^2.
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RESPONSE --> ok
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18:22:25 `q007. What is the area of a circle whose radius is 3.00 cm?
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RESPONSE --> A=pi*r^2 A=pi*3^2 A=28.27cm^2
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18:22:39 01-18-2006 18:22:39 The area of a circle is A = pi * r^2, where r is the radius. Thus A = pi * (3 cm)^2 = 9 pi cm^2. Note that the units are cm^2, since the cm unit is part r, which is squared. The expression 9 pi cm^2 is exact. Any decimal equivalent is an approximation. Using the 3-significant-figure approximation pi = 3.14 we find that the approximate area is A = 9 pi cm^2 = 9 * 3.14 cm^2 = 28.26 cm^2, which we round to 28.3 cm^2 to match the number of significant figures in the given radius. Be careful not to confuse the formula A = pi r^2, which gives area in square units, with the formula C = 2 pi r for the circumference. The latter gives a result which is in units of radius, rather than square units. Area is measured in square units; if you get an answer which is not in square units this tips you off to the fact that you've made an error somewhere.
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NOTES -------> I went ahead and multiplied pi out.
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18:24:29 `q008. What is the circumference of a circle whose radius is exactly 3 cm?
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RESPONSE --> circumference=pi*diameter C=pi*6 you take the radius and times it by 2 to get the diameter
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18:24:54 The circumference of this circle is C = 2 pi r = 2 pi * 3 cm = 6 pi cm. This is the exact area. An approximation to 3 significant figures is 6 * 3.14 cm = 18.84 cm. Note that circumference is measured in the same units as radius, in this case cm, and not in cm^2. If your calculation gives you cm^2 then you know you've done something wrong.
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RESPONSE --> ok
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18:26:07 `q009. What is the area of a circle whose diameter is exactly 12 meters?
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RESPONSE --> A=pi*r^2 A=pi*6^2 A=pi*36 Get the radius by dividing the diameter by 2.
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18:26:57 The area of a circle is A = pi r^2, where r is the radius. The radius of this circle is half the 12 m diameter, or 6 m. So the area is A = pi ( 6 m )^2 = 36 pi m^2. This result can be approximated to any desired accuracy by using a sufficient number of significant figures in our approximation of pi. For example using the 5-significant-figure approximation pi = 3.1416 we obtain A = 36 m^2 * 3.1416 = 113.09 m^2.
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RESPONSE --> I left off my units they should be m^2
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18:30:41 `q010. What is the area of a circle whose circumference is 14 `pi meters?
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RESPONSE --> Use the diameter from the Circumference to find the radius then plug it in the area formula. A=pi*r^2 A=pi*7pi^2
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18:32:23 01-18-2006 18:32:23 We know that A = pi r^2. We can find the area if we know the radius r. We therefore attempt to use the given information to find r. We know that circumference and radius are related by C = 2 pi r. Solving for r we obtain r = C / (2 pi). In this case we find that r = 14 pi m / (2 pi) = (14/2) * (pi/pi) m = 7 * 1 m = 7 m. We use this to find the area A = pi * (7 m)^2 = 49 pi m^2.
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NOTES -------> Further the simplication A=pi*7^2 A= 49 pi m^2
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18:38:34 `q011. What is the radius of circle whose area is 78 square meters?
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RESPONSE --> A=pi*r^2 78=pi*r^2 78/pi=r^2 24.8=r^2 (sqrt)24.8=(sqrt)r^2 4.99 m^2 = r
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18:38:51 01-18-2006 18:38:51 Knowing that A = pi r^2 we solve for r. We first divide both sides by pi to obtain A / pi = r^2. We then reverse the sides and take the square root of both sides, obtaining r = sqrt( A / pi ). Note that strictly speaking the solution to r^2 = A / pi is r = +-sqrt( A / pi ), meaning + sqrt( A / pi) or - sqrt(A / pi). However knowing that r and A are both positive quantities, we can reject the negative solution. Now we substitute A = 78 m^2 to obtain r = sqrt( 78 m^2 / pi) = sqrt(78 / pi) m.{} Approximating this quantity to 2 significant figures we obtain r = 5.0 m.
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NOTES -------> I did not round the number up.
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18:40:56 `q012. Summary Question 1: How do we visualize the area of a rectangle?
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RESPONSE --> Draw a rectangle. The area is all of the space within the confines of the rectangle.
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18:42:17 01-18-2006 18:42:17 We visualize the rectangle being covered by rows of 1-unit squares. We multiply the number of squares in a row by the number of rows. So the area is A = L * W.
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NOTES -------> Visualize the rectangle filled with blocks then multiply the number of squares in a row by the number of rows. Hence the equation A=l*w
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18:44:14 `q013. Summary Question 2: How do we visualize the area of a right triangle?
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RESPONSE --> Draw a right triangle. Then imagine another triangle with their hypotenuse touching. This makes a square. So the area is 1/2b*h
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18:44:26 We visualize two identical right triangles being joined along their common hypotenuse to form a rectangle whose length is equal to the base of the triangle and whose width is equal to the altitude of the triangle. The area of the rectangle is b * h, so the area of each triangle is 1/2 * b * h.
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RESPONSE --> ok
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18:45:46 `q014. Summary Question 3: How do we calculate the area of a parallelogram?
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RESPONSE --> It is best to first draw a picture of a parallelogram. Next label the bases and the height. Proceed by plugging into the formula. A=1/2(b1+b2)h
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18:46:02 The area of a parallelogram is equal to the product of its base and its altitude. The altitude is measured perpendicular to the base.
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RESPONSE --> ok
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18:49:09 `q015. Summary Question 4: How do we calculate the area of a trapezoid?
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RESPONSE --> On the previous problem I mistakenly entered the area of a trapezoid instead of the area of a parallelogram which is A=B*H The trapezoid is 1/2(b1+b2)h
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18:49:16 We think of the trapezoid being oriented so that its two parallel sides are vertical, and we multiply the average altitude by the width.
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RESPONSE --> ok
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18:49:56 `q016. Summary Question 5: How do we calculate the area of a circle?
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RESPONSE --> Calculate the area of a circle by squaring the radius and multiplying by pi
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18:50:03 We use the formula A = pi r^2, where r is the radius of the circle.
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RESPONSE --> ok
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18:51:57 `q017. Summary Question 6: How do we calculate the circumference of a circle? How can we easily avoid confusing this formula with that for the area of the circle?
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RESPONSE --> Calculate the circumference by multiplying the diameter by pi. You can avoid confusion by remembering that the area of a circle uses the radius squared and the circumference only uses diameter and it is not squared.
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18:52:06 We use the formula C = 2 pi r. The formula for the area involves r^2, which will give us squared units of the radius. Circumference is not measured in squared units.
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RESPONSE --> ok
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18:54:19 `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> All of the formulas have come back to memory and are refreshed in my brain as the areas for various shapes. This exercise was a very nice refresher on area. In the process of doing the exercise I wrote down all of the formulas for future use.
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18:54:33 This ends the first assignment.
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RESPONSE --> Yeeeesssssss.
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ܫsapɶ assignment #001 [뺗܃ܝ Physics I Vid Clips 01-18-2006
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19:40:16 `questionNumber 10000 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 is increasing.
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19:40:23 `questionNumber 10000 ** 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 --> ok
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19:41:30 `questionNumber 10000 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 --> When could measure various points on the ramp and calculate the velocity by measuring the distance and time and then comparing the information to determine if the velocity is increasing, decreasing, or constant.
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19:41:58 `questionNumber 10000 ** 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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19:42:35 `questionNumber 10000 What is the shape of the velocity vs. clock time graph for the motion of the ball?
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RESPONSE --> The graph curves upward to the right because the velocity is increasing at an increasing rate.
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19:42:41 `questionNumber 10000 ** 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 --> ok
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19:44:31 `questionNumber 10000 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 --> The curved end will go faster but once it evens out on the level ramp it will have an even speed. The graph is curved to the right and then it levels out.
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19:44:41 `questionNumber 10000 ** 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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19:46:18 `questionNumber 10000 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 --> The velocity is increasing at a constant rate. The graph is a straight line upward going from left to right.
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19:46:29 `questionNumber 10000 ** 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
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`zEdSz{^؏ assignment #001 [뺗܃ܝ Physics I 01-18-2006
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19:52:02 `questionNumber 10000 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 distance divided by time. It is an example of rate because it is measuring an quantity with respect to another measured quantity.
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19:52:12 `questionNumber 10000 ** 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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19:55:51 `questionNumber 10000 Given average speed and time interval how do you find distance moved?
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RESPONSE --> Divide the average speed by the time interval to find the distance moved.
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19:56:41 01-18-2006 19:56:41 ** You multiply average speed * time interval to find distance moved. For example, 50 miles / hour * 3 hours = 150 miles. **
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NOTES -------> You actually have to multiply the average speed by the time interval to find distance moved.
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19:57:57 `questionNumber 10000 Given average speed and distance moved how do you find the corresponding time interval?
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RESPONSE --> Divide distance by the average speed to find the time interval.
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19:58:05 `questionNumber 10000 ** 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 --> ok
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20:06:21 `questionNumber 10000 Given time interval and distance moved how do you get average speed?
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RESPONSE --> To get the average speed divide distance by time.
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20:07:01 `questionNumber 10000 ** 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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