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11:47:58 `q001. Note that there are 9 questions in this assignment.
If the acceleration of an object is uniform, then the following statements apply: 1. A graph of velocity vs. clock time forms a straight line, either level or increasing at a constant rate or decreasing at a constant rate. 2. The average velocity of the object over any time interval is equal to the average of its velocity at the beginning of the time interval (called its initial velocity) and its velocity at the end of the time interval (called its final velocity). 3. The velocity of the object changes at a constant rate (this third statement being obvious since the rate at which the velocity changes is the acceleration, which is assumed here to be constant). 4. The acceleration of the object at every instant is equal to the average acceleration of the object. Suppose that an object increases its velocity at a uniform rate, from an initial velocity of 5 m/s to a final velocity of 25 m/s during a time interval of 4 seconds. By how much does the velocity of the object change? What is the average acceleration of the object? What is the average velocity of the object?......!!!!!!!!...................................
RESPONSE --> 'dv = v(f) - f(0) 'dv = (25 m/s) - (5 m/s) 'dv = 20 m/s
aAve = 'dv / 'dt aAve = (20 m/s) / (4 s) aAve = 5 m/s/s vAve = [(v(f) + f(0)) / 2 vAve = [(25 m/s)+ (5 m/s)] / 2 vAve = (30 m/s) / 2 vAve = 15 m/s.................................................
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11:48:40 The velocity of the object changes from 5 meters/second to 25 meters/second so the change in velocity is 20 meters/second. The average acceleration is therefore (20 meters/second) / (4 seconds) = 5 m / s^2. The average velocity of the object is the average of its initial and final velocities, as asserted above, and is therefore equal to (5 meters/second + 25 meters/second) / 2 = 15 meters/second (note that two numbers are averaged by adding them and dividing by 2).
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RESPONSE --> OK.
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11:50:18 `q002. How far does the object of the preceding problem travel in the 4 seconds?
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RESPONSE --> 'ds = vAve * 'dt 'ds = (15 m/s) * (4 s) 'ds = 60 m
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11:50:29 The displacement `ds of the object is the product vAve `dt of its average velocity and the time interval, so this object travels 15 m/s * 4 s = 60 meters during the 4-second time interval.
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RESPONSE --> OK.
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11:52:44 `q003. Explain in commonsense terms how we determine the acceleration and distance traveled if we know the initial velocity v0, and final velocity vf and the time interval `dt.
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RESPONSE --> We find the acceleration by dividing how much the velocity changed by the amount of time it took to change. We find the distance traveled by multiplying the average velocity by the time interval.
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11:52:59 In commonsense terms, we find the change in velocity since we know the initial and final velocities, and we know the time interval, so we can easily calculate the acceleration. Again since we know initial and final velocities we can easily calculate the average velocity, and since we know the time interval we can now determine the distance traveled.
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RESPONSE --> OK.
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11:54:52 `q004. Symbolize the situation by first giving the expression for the acceleration in terms of v0, vf and `dt, then by giving the expression for vAve in terms of v0 and vf, and finally by giving the expression for the displacement in terms of v0, vf and `dt.
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RESPONSE --> aAve = (vf - v0) / 'dt
vAve = (vf + v0) / 2 'ds = (vf - v0) / 'dt.................................................
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11:55:11 The acceleration is equal to the change in velocity divided by the time interval; since the change in velocity is vf - v0 we see that the acceleration is a = ( vf - v0 ) / `dt.
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RESPONSE --> OK.
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11:55:27 `q005. The average velocity is the average of the initial and final velocities, which is expressed as (vf + v0) / 2.
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RESPONSE --> OK.
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11:57:43 When this average velocity is multiplied by `dt we get the displacement, which is `ds = (v0 + vf) / 2 * `dt.
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RESPONSE --> 'ds = [(vf + v0) / 2] / 'dt
I understand this formula and simply used the wrong variable for the numerator. I know it is average velocity and I used change in velocity..................................................
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12:08:07 `q006. This situation is identical to the previous, and the conditions implied by uniformly accelerated motion are repeated here for your review: If the acceleration of an object is uniform, then the following statements apply:
1. A graph of velocity vs. clock time forms a straight line, either level or increasing at a constant rate or decreasing at a constant rate. 2. The average velocity of the object over any time interval is equal to the average of its velocity at the beginning of the time interval (called its initial velocity) and its velocity at the end of the time interval (called its final velocity). 3. The velocity of the object changes at a constant rate (this third statement being obvious since the rate at which the velocity changes is the acceleration, which is assumed here to be constant). 4. The acceleration of the object at every instant is equal to the average acceleration of the object. Describe a graph of velocity vs. clock time, assuming that the initial velocity occurs at clock time t = 0. At what clock time is the final velocity then attained? What are the coordinates of the point on the graph corresponding to the initial velocity (hint: the t coordinate is 0, as specified here; what is the v coordinate at this clock time? i.e., what is the velocity when t = 0?). What are the coordinates of the point corresponding to the final velocity?......!!!!!!!!...................................
RESPONSE --> The graph would form a straight line, and would be increasing at a constant rate.
vf is attained at 4 seconds. (0, 5) corresponds to the initial velocity. (4, 25) corresponds to the final velocity..................................................
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12:08:37 The initial velocity of 5 m/s occurs at t = 0 s so the corresponding graph point is (0 s, 5 m/s). The final velocity of 25 meters/second occurs after a time interval of `dt = 4 seconds; since the time interval began at t = 0 sec it ends at at t = 4 seconds and the corresponding graph point is ( 4 s, 25 m/s).
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RESPONSE --> OK.
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12:08:56 `q007. Is the graph increasing, decreasing or level between the two points, and if increasing or decreasing is the increase or decrease at a constant, increasing or decreasing rate?
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RESPONSE --> Increasing at a constant rate.
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12:09:05 Since the acceleration is uniform, the graph is a straight line. The graph therefore increases at a constant rate from the point (0, 5 m/s) to the point (4 s, 25 m/s).
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RESPONSE --> OK.
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12:10:58 `q008. What is the slope of the graph between the two given points, and what is the meaning of this slope?
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RESPONSE --> slope = rise / run slope = 20 / 4 slope = 5
The slope represents the average acceleration..................................................
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12:11:53 The rise of the graph is from 5 m/s to 25 m/s and is therefore 20 meters/second, which represents the change in the velocity of the object. The run of the graph is from 0 seconds to 4 seconds, and is therefore 4 seconds, which represents the time interval during which the velocity changes. The slope of the graph is rise / run = ( 20 m/s ) / (4 s) = 5 m/s^2, which represents the change `dv in the velocity divided by the change `dt in the clock time and therefore represents the acceleration of the object.
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RESPONSE --> Is the slope supposed be a unitless value? Or should I have included m/s/s.
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12:16:50 `q009. The graph forms a trapezoid, starting from the point (0,0), rising to the point (0,5 m/s), then sloping upward to (4 s, 25 m/s), then descending to the point (4 s, 0) and returning to the origin (0,0). This trapezoid has two altitudes, 5 m/s on the left and 25 m/s on the right, and a base which represents a width of 4 seconds. What is the average altitude of the trapezoid and what does it represent, and what is the area of the trapezoid and what does it represent?
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RESPONSE --> (25 m/s + 5 m/s) / 2 = 15 m/s This represents the average velocity.
a = 15 m/s * 4 s a = 60 m This represents the displacement..................................................
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12:17:15 The two altitudes are 5 meters/second and 25 meters/second, and their average is 15 meters/second. This represents the average velocity of the object on the time interval. The area of the trapezoid is equal to the product of the average altitude and the base, which is 15 m/s * 4 s = 60 meters. This represents the product of the average velocity and the time interval, which is the displacement during the time interval.
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RESPONSE --> OK.
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{??[·???????????? assignment #006 }_?M??_??·????|G???? Physics I Class Notes 09-22-2005
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15:01:11 How do flow diagrams help us see the structure of our reasoning processes?
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RESPONSE --> They help us see the connections between the variables we are given.
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15:01:20 ** They help us to visualize how all the variables are related.
Flow diagrams can also help us to obtain formulas relating the basic kinematic quantities in terms of which we have been analyzing uniformly accelerated motion. **......!!!!!!!!...................................
RESPONSE --> OK.
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15:08:21 How do the two most fundamental equations of uniformly accelerated motion embody the definitions of average velocity and of acceleration?
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RESPONSE --> fundamental equations: vf = v0 + a 'dt 'ds = [(vf + v0) / 2] * 'dt
vAve = 'ds / 'dt aAve = 'dv / 'dt The fundamental equations manipulate the formulas for average velocity and acceleration to find vf or 'ds..................................................
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15:08:43 ** Velocity tells us the rate at which the position changes whereas the acceleration tells us the rate at which the velocity is changing.
If acceleration is uniform ave velocity is the average of initial and final velocities. The change in position is found by taking the average velocity vAve = (vf+ v0) / 2 and multiplying by the'dt to get the first fundamental equation `ds = (v0 + vf)/2 * `dt. The acceleration is accel = rate of change of velocity = change in velocity / `dt = (vf - v0) / `dt. In symbols this equation is a = (vf + v0) / `dt. Algebraic rearrangement gives us this equation in the form vf = v0 + a `dt. This form also has an obvious interpretation: a `dt is the change in velocity, which when added to the initial velocity gives us the final velocity. **......!!!!!!!!...................................
RESPONSE --> OK.
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15:12:52 How can we interpret the third fundamental equation of uniformly accelerated motion?
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RESPONSE --> First 2 fundamental equations: vf = v0 + a 'dt 'ds = [(vf + v0) / 2] * 'dt
Third: 'ds = v0 'dt + 1/2 a 'dt^2 Third fundamental equation is derived from the first two..................................................
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15:13:18 ** The third equation says that `ds = v0 `dt + .5 a `dt^2.
This means that the displacement `ds arises independently from initial velocity v0 and acceleration a: v0 `dt is the displacement of an object with uniform velocity moving at velocity v0, and 1/2 a `dt^2 the distance moved from rest by a uniformly accelerating object. The two contributions are added to get the total `ds. **......!!!!!!!!...................................
RESPONSE --> OK.
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15:14:06 Why can we not directly reason out the basic 'impossible situation'?
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RESPONSE --> Because we cannot draw conclusions from any pair of the given quantities.
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15:14:12 ** In this situation we know v0, a and `ds. From v0 and a we cannot draw any conclusions, and the same is true for v0 and `ds and also for a and `ds. No pair of variables allows us to draw any additional information. **
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RESPONSE --> OK.
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15:14:43 What strategy will we use to reconcile the basic 'impossible situation'? WE cann write down the 2 most fundamental equations and see what we do know.
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RESPONSE --> Use the system of equations.
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15:14:50 ** We can use the fourth equation vf^2 = v0^2 + 2 a `ds to obtain vf, then knowing the values of v0, vf, a and `ds we easily find `dt either by direct reasoning or by using one of the fundamental equations. **
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RESPONSE --> OK.
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15:16:07 What is the difference between understanding uniformly accelerated motion and analyzing it with the use of equations?
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RESPONSE --> We use 7 quantities to understand uniformly accelerated motion, but we only use 5 to analyze it.
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15:16:18 ** You can use equations without understanding much of anything. To use the equations you don't even need to understand things like average velocity or change in velocity. You just have to be able to identify the right numbers and plug them in, which is an important task in itself but which doesn't involve understanding of the physical concepts behind the equations. **
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RESPONSE --> OK.
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15:18:36 How do we extrapolate our acceleration vs. ramp slope data to obtain an estimate of the acceleration of gravity?
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RESPONSE --> We create a graph and then draw the closes straight line and find the slope.
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15:18:49 ** We can sketch a straight line as close as possible to our data points. Then we use the average slope of that graph; this average slope is the acceleration of gravity. **
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RESPONSE --> OK.
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15:19:56 How do the unavoidable timing errors due to the uncertainty in the computer timer affect our estimate of the acceleration of gravity?
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RESPONSE --> It affects the slope of the graph.
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15:20:10 ** STUDENT ANSWER: This error causes the slope to increase at an increasing rate rather than form a linear line.
INSTRUCTOR COMMENT: Good answer. A systematic error would do that. Even random, non-systematic errors affect the placement of points on the graph, and this tends to affect the slope of the straight line approximating the graph, and also to reduce the accuracy of this slope. **......!!!!!!!!...................................
RESPONSE --> OK.
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15:20:39 How could the slight slope of the table on which the ramp rests, if not accounted for, affect our graph of acceleration vs. ramp slope but not our estimate of the acceleration of gravity?
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RESPONSE --> It would move our line to the right but the slope would stay the same.
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15:20:49 ** GOOD STUDENT ANSWER: The only effect this systematic error has on the graph is to change the m coordinate of each point by an amount equal to the slope of the table, which is always the same.Since it is the graph slope that comprises our final result, a small table slope would have no effect on our conclusions. **
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RESPONSE --> OK.
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15:21:30 How could anticipation of the instant at which a cart reaches the end of the ramp, but not of the instant at which it is released, affect our graph as well as our estimate of the acceleration of gravity?
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RESPONSE --> It would cause errors in the slope and thus affect the estimate of the acceleration of gravity.
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15:21:40 ** GOOD STUDENT ANSWER: The timer is started with a slight delay due to the reaction time of the person doing the timing.
This would be OK if the individual's reaction time caused the individual to stop the timer with the same delay. However, the person doing the timing often anticipates the instant when the cart reaches the end of the ramp, so that the delay is not added onto the end time as it was to the starting time. The anticipating individual often triggers the timer slightly before the cart reaches the end, compounding the error even further and also causing the graph to curve rather than remain linear. **......!!!!!!!!...................................
RESPONSE --> OK.
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T?{ y??????D???? assignment #007 }_?M??_??·????|G???? Physics I Class Notes 09-22-2005
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16:29:47 Why do we say that the first equation of uniformly accelerated motion expresses the definition of average velocity, while the second expresses the definition of acceleration?
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RESPONSE --> The equation vf = v0 + a 'dt is derived from the formula for acceleration. The equation 'ds = (vf + v0)/2 * 'dt is derived from the velocity formula.
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16:30:16 ** When acceleration is uniform average velocity is the average of initial and final velocities, (vf + v0) / 2. Average velocity is `ds / `dt (whether accel is uniform or not). Setting `ds / `dt = (vf + v0) / 2 we obtain `ds = (vf + v0) / 2 * `dt, which is the first equation of uniformly accelerated motion. So the definition of average velocity is equivalent to the definition of average velocity.
Average acceleration is aAve = `dv / `dt = (vf - v0) / `dt. Since for uniform acceleration the acceleration is constant, we can just say that in this case a = (vf - v0) / `dt. This equation is easily rearranged to give the second equation of motion, vf = v0 + a `dt. **......!!!!!!!!...................................
RESPONSE --> OK.
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16:32:40 Why, for uniform acceleration, is vAve = (vf + v0) / 2, while this is not usually true for nonuniform acceleration?
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RESPONSE --> Because the acceleration is constant, the average of vf and v0 represents each velocity point. For nonuniform, the velocity points vary so the average would not be represented by vf and v0.
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16:32:55 ** If acceleration is uniform then the v vs. t graph is linear, so that the average velocity over any time interval must be equal to the velocity at the midpoint of that interval. It follows that the average velocity must be midway between the initial and final velocities. (vf + v0) / 2 is the average of the initial and final velocities, and is therefore halfway between the v0 and vf. **
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RESPONSE --> OK.
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16:34:27 In commonsense terms, why does change of velocity over a given distance, with a given uniform acceleration, differ with initial velocity?
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RESPONSE --> Because the velocity is increasing the final velocity must be more.
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16:34:41 ** If the uniform acceleration is the same in both cases, then assuming that both initial velocity and acceleration are positive, a greater initial velocity will result in a shorter time interval to cover the given distance. A shorter time interval leaves less time for velocity to change, resulting in a smaller change in velocity. **
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RESPONSE --> OK.
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?????]????????·assignment #003 }_?M??_??·????|G???? Physics I Vid Clips 09-22-2005
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16:46:30 Video Clip 7: Slope triangle, instantaneous acceleration, tangent line
How do we represent the calculation of the rate which velocity changes on graph of velocity vs. time?......!!!!!!!!...................................
RESPONSE --> a = 'dv / 'dt
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16:46:44 ** Correct response from student: We can represent the rate of velocity change by calculating the slope. Slope is rise/run which is change in velocity/change in time **
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RESPONSE --> OK.
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16:47:40 If we know the velocity at every clock time, that how we find the precise rate at which velocity changes adding given clock time?
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RESPONSE --> Find the slope at the given clock time and velocity.
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16:47:55 ** STUDENT RESPONSE: I am not sure I understand, however I think you would make a triangle between the two times and velocities--i.e., between the two points on the v vs. t graph. Then find slope.
INSTRUCTOR CRIQITUE: You would do this, but you would calculate the rate or slope over smaller and smaller intervals containing the clock time at the specified instant. You would try to determine the limiting value as the interval approaches 0. This limiting value is the instantaneous rate as well as the slope of the tangent line. **......!!!!!!!!...................................
RESPONSE --> OK.
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16:48:34 How do we depict the instantaneous rate of velocity change on a graph of velocity vs. time?
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RESPONSE --> Find the tangent line by finding the slope of smaller and smaller triangles.
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16:48:41 ** See the preceding answer. **
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RESPONSE --> OK.
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16:52:31 Physics video clips 08: Finding displacement: define displacement, then calculate by common sense, by formula, by area
What is the difference between displacement and distance?......!!!!!!!!...................................
RESPONSE --> Displacement is the change in position.
It is the average velocity multiplied by the time. 'ds = vAve * 'dt a = (vf + v0)/2 * 'dt.................................................
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16:52:35 ** Displacement tells the change in position, which is the same as distance but displacement can have a negative or positive value whereas distance is always positive. **
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RESPONSE --> OK.
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16:54:12 How do we calculate displacement from velocity and time interval?? What is the common sense of this calculation?
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RESPONSE --> We calculate displacement by multiplying the average velocity by the time interval.
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16:54:20 ** Correct response from student: The vAve multiplied by the time interval. If the vAve tells us how far it travels one second, then if we multiply it by the time interval it will tell us how far it will travel in that amt. of time. **
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RESPONSE --> OK.
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16:54:41 How do we use a graph to help visualize the calculation of displacement when velocity is constant?
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RESPONSE --> It is the area under the slope.
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16:54:53 ** Since the vel. is constant, a graph of velocity vs. clock time over a given time interval will form a rectangle whose 'height' represents the velocity and whose width represents the time interval. Multiplying velocity * time interval gives displacement.
A graph of position vs. clock time will be a straight-line graph, and the displacement corresponding to two clock times will be the change in the y coordinate between those clock times. **......!!!!!!!!...................................
RESPONSE --> OK.
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16:55:02 What aspect of a v vs. t graph tells us the displacement?
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RESPONSE --> The area.
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16:55:06 ** The area of the rectangle formed between the two given clock times will indicate the displacement (see preceding comment). **
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RESPONSE --> OK.
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Randomized Problems: Week 2 Quiz 2