course Phy 121 The assignments page said that the lab for Asst 5 would be posted this week. I did not see an update, so I did not do a lab for this assignment.
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15:58:51 `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?
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RESPONSE --> Change in velocity = `dv = 25m/s - 5m/s = 20m/s Ave acceleration= aAve = `dv/`dt = (20m/s)/(4s) = 5m/s^2 When acceleration is uniform, Ave velocity, vAve = (vf + vo)/2 = (30m/s)/2 = 15m/s
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15:59:23 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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16:00:16 `q002. How far does the object of the preceding problem travel in the 4 seconds?
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RESPONSE --> Going at an average of 15m/s it would travel 60 meters in 4 seconds. (4s)(15m/s) = 60 m
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16:00:40 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 `ds = vAve x `dt
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16:03:35 `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 --> Acceleration is the rate of change in velocity. So we would find the change in velocity between the initial and final and divide them by the time duration, `dt. The distance traveled, `ds, is the change in position from the beginning of the interval to the end. It is found by multiplying the average velocity (vf + vo / 2) by the time interval, `dt.
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16:03:44 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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16:05:57 `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 -vo) / `dt vAve = (vf +vo)/2 `ds = [(vf +vo)/2] (`dt)
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16:06:11 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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16:06:24 `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 --> vAve = (vf +vo)/2
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16:06:35 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 +vo)/2] (`dt)
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16:10:50 `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?
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RESPONSE --> The graph of velocity vs. clock time would have a linear slope since the acceleration is uniform. The y-axis represents the change in velocity, and the x-axis represents the change in time. The first point would be at the the x = 0 point because the initial velocity occurs at t = 0.The initial velocity was 5m/s. Therefore the first coordinates would be (0,5). The final coordinates would be (4, 25) since the time interval is 4 seconds and the final velocity is 25m/s.
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16:12:29 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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16:12:50 `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 --> The graph would be increasing at a constant rate.
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16:12:58 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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16:15:01 `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 = acceleration = (y2 - y1) / (x2 -x1) = `dv/ `dt = (25 - 5) / (4 - 0) = 20 / 4 = 5m/s^2 The slope is the rate of change of velocity over the given time interval. This is also known as acceleration.
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16:17:03 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 --> Ok. NOTE: Rise = `dv = 25 m/s - 5m/s = 20 m/s Run = `dt = 4s -0s = 4 s
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16:20:46 `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 --> The average altitude represents the average velocity of the object. Altitude = vAve = (vf +vo)/2 = (25m/s + 5m/s) / 2 = 15m/s The area of the trapezoid represents the change in position of the object over that time interval at a given average rate of velocity. Area = `ds = vAve x `dt = (15m/s)(4s) = 60m
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16:20:53 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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ʹvָյݡ assignment #006 ۶mSۥ} Physics I Class Notes 06-11-2006
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16:58:09 How do flow diagrams help us see the structure of our reasoning processes?
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RESPONSE --> A flow diagram helps us analyze information by visibly noting in what order they are worked out. It also helps us see how each component is connected.
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16:58:22 ** 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. **
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RESPONSE --> Ok
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16:58:58 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 --> Velocity is the rate that position changes over a certain time interval. It is defined by the equation: vAve = (vf + vo)/2. Acceleration is the rate at which that velocity changes. Uniform Acceleration is defined by the equation: aAve = `dv/`dt = (vf - vo) /`dt
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17:03:52 ** 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. **
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RESPONSE --> OK. NOTES: (a)( `dt) is the change in velocity, which when added to the initial velocity gives us the final velocity. Change in position = `ds = (vAve)(`dt) = (vf + vo)/2 x `dt.
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17:04:46 How can we interpret the third fundamental equation of uniformly accelerated motion?
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RESPONSE --> The third equation states that displacement comes from initial velocity and acceleration. The formula defines displacement as the initial velocity multiplied by the time interval added to half of the acceleration times the time interval squared.`ds = (vo)(`dt) + 1/2 (a)(`dt^2) (vo)(`dt) = object's position change when velocity is uniform at vo. 1/2 (a)(`dt^2) = distance traveled at a uniform acceleration.
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17:09:04 ** 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. **
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RESPONSE --> OK.
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17:09:14 Why can we not directly reason out the basic 'impossible situation'?
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RESPONSE --> No, we cannot evaluate the situation from these quantities without knowing something about the variables that are not known.
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17:10:22 ** 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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17:10:38 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 --> The fourth equation helps us evaluate the impossible situation because all we need to know is vo, vf, a and `ds. It reads, vf^2 = vo^2 + 2(a)(`ds).
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17:11:50 ** 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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17:12:24 What is the difference between understanding uniformly accelerated motion and analyzing it with the use of equations?
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RESPONSE --> The difference in understanding and analyzing accelerated motion with equations is that you can simply plug and chug, so to speak, when you use the formulas. But the analyzation allows you to see every angle and understand every part of the process.
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17:13:44 ** 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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17:13:50 How do we extrapolate our acceleration vs. ramp slope data to obtain an estimate of the acceleration of gravity?
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RESPONSE --> Make a line of best fit through the data points. This will be a straight line that goes through and is close to as many points as possible. The slope of that line is the gravitational acceleraion.
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17:15:06 ** 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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17:15:35 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 --> Timing errors that result from the TIMER program are known as a random errors because they result in the timing being too short or too long. When this happens the slope can not be linear, but it will be curved. This type of error may have nothing to do with the individual or the process used to find the slope.
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17:19:16 ** 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. **
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RESPONSE --> Ok.
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17:21:23 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 --> The acceleration points that we graph would be too high or too low depending on the slope of the table. But since the estimate of gravitational acceleration is simply an average, the exact points on the graph would not matter and the average would account for the error.
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17:21:56 ** 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. It is a systematic error.
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17:26:04 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 --> The time taken at the instant of release will be a bit behind the actual time. If we anticipate the cart reaching the end of the ramp, we should have an accurate time recorded. This means that the delay is unaccounted for. If we stopped the timer with the same delay, this would not be a problem. This is an unavoidable systematic error. In this case, the graph would be curved and therefore, the acceleration of gravity would not be expressed as a straight line.
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17:26:48 ** 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. **
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RESPONSE --> Ok
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yN۟ assignment #003 ۶mSۥ} Physics I Vid Clips 06-11-2006
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18:31:50 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?
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RESPONSE --> The rate at which velocity changes is represented by the slope of the graph of velocity vs. time: Slope = rise/run = `dv/`dt = Rate of change in velocity
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18:32:48 ** 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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18:33:30 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 --> By finding the slopes for different time intervals. This would show up on the graph as many different sized triangles. The smallest being close to t = 0 and the slope of the tangent line.
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18:34:54 ** 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. **
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RESPONSE --> NOTE: This limiting value is the instantaneous rate as well as the slope of the tangent line.
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18:36:22 How do we depict the instantaneous rate of velocity change on a graph of velocity vs. time?
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RESPONSE --> The instantaneous rate of velocity change (as it gets closer to 0) on the graph is depicted by the limiting value or the slope of the tangent line.
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18:36:30 ** See the preceding answer. **
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RESPONSE --> Ok.
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18:38:08 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?
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RESPONSE --> Displacement is the change in position over a time interval. Distance is the length covered over a time interval. The main difference is that displacement can be positive or negative, but distance can only be positive.
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18:38:15 ** 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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18:39:28 How do we calculate displacement from velocity and time interval? What is the common sense of this calculation?
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RESPONSE --> Displacement, `ds, can be calculated by finding the average velocity and multiplying it by the time interval. vAve = (vf + vo)/2 `ds = vAve x `dt = (vf + vo)/2 x `dt By multiplying the average velocity by the time interval we find the object's position change over that given time duration at that particular velocity.
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18:40:39 ** 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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18:41:28 How do we use a graph to help visualize the calculation of displacement when velocity is constant?
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RESPONSE --> The graph helps us visualize the calculation of displacement because we can see that the area under the graph for that particular time interval is equal to the change in velocity (because velocity is constant) times the time interval which is displacement. Area = h x w Height = change in velocity, width = change in time.
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18:42:42 ** 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. **
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RESPONSE --> Ok.
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18:42:57 What aspect of a v vs. t graph tells us the displacement?
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RESPONSE --> The area of the rectangle/square that is in that time interval. Area = h x w Height = change in velocity, width = change in time. `ds = `dv/`dt
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18:43:25 ** 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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ѕµ{TϞJϝLw assignment #005 ۶mSۥ} Physics I 06-11-2006
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20:38:54 Intro Prob 6 Intro Prob 6 How do you find final velocity and displacement given initial velocity, acceleration and time interval?
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RESPONSE --> First we find the change in velocity, `dv = aAve x `dt `dv = (7m/s/s)(6s) = 42m/s Then we can use `dv and vo to find vf: vf = vo + `dv = 8m/s + 42m/s = 50m/s Then we use vf and vo to find average velocity, vAve: vAve = (vf + vo)/2 = (50m/s + 8m/s)/2 = 29m/s Then we use the vAve and `dt to find displacement: `ds = (vAve)(`dt) = (29m/s)(6s) = 174m
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20:39:01 ** To find final velocity from the given quantities initial velocity, acceleration and `dt: Multiply `dt by accel to get `dv. Then add change in velocity `dv to init vel , and you have the final velocity**
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RESPONSE --> Ok
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20:42:43 Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.
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RESPONSE --> First level: vo, vf, and `dt Second level: Use vf and vo to find `dv --> `dv = vf - vo Also use vf and vo to find vAve --> (vo + vf)/2 Third level: Use `dv and `dt to find aAve --> aAve = `dv / `dt Fourth level: Use vAve and `dt to find `ds --> `ds = vAve x `dt
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20:42:54 ** The flow diagram shows us the flow of information, what we get from what, usually by combining two quantites at a time. How we get each quantity may also be included. From vf and v0 we get `dv, shown by lines from vf and v0 at the top level to `dv. From vf and v0 we also get and vAve, shown by similar lines running from v0 and vf to vAve. Then from vAve and `dt we get `ds, with the accompanying lines indicating from vAve and `dt to `ds, while from `dv and `dt we get acceleration, indicated similarly. **
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RESPONSE --> Ok
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20:48:27 Principles of Physics and General College Physics Students: Prob. 1.26: Estimate how long it would take a runner at 10 km / hr to run from New York to California. Explain your solution thoroughly.
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RESPONSE --> I looked up the distance from New York, NY to Los Angeles, CA. The distance was about 2800 miles. I then converted miles to kilometers using the conversion factor, 1 mile = 1.609 km. (2800mi)(1.609km/mi) = 4505.2 km Then I divided the distance in kilometers by the rate at which the runner is traveling to find the time duration. (4505.2km) / (10km/hr) = 450.52 hrs So the trip from New York City to Los Angeles would have taken the runner about 450 hours.
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20:49:15 It is about 3000 miles from coast to coast. A km is about .62 mile, so 3000 miles * 1 km / (.62 miles) = 5000 km, approximately. At 10 km / hr, the time required would be 5000 km / (10 km / hr) = 500 km / (km/hr) = 500 km * (hr / km) = 500 (km / km) * hr = 500 hr.
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RESPONSE --> I did this problem correctly, my distance was a little different just because I chose different locations in New York and California.
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20:54:34 All Students: Estimate the number heartbeats in a lifetime. What assumptions did you make to estimate the number of heartbeats in a human lifetime, and how did you obtain your final result?
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RESPONSE --> I looked up the following information in a Medical Encyclopedia on the internet: The average human lives 80 years. The average human heart beats about 70 times in a minute. (70bt/min)(60min/1hr)(24hr/1day)(365d/1yr) = 36792000 beats per year (36792000bt/yr)(80yrs) = 2,943,360,000 = about 3,000,000,000 beats per lifetime
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20:54:55 ** Typical assumptions: At 70 heartbeats per minute, with a lifetime of 80 years, we have 70 beats / minute * 60 minutes/hour * 24 hours / day * 365 days / year * 80 years = 3 billion, approximately. **
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RESPONSE --> I understand this problem.
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20:55:10 University Physics Students Only: Problem 1.52 (i.e., Chapter 1, Problem 52): Angle between -2i+6j and 2i - 3j. What angle did you obtain between the two vectors?
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RESPONSE --> I am a Principles of Physics student.
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20:55:15 ** For the given vectors we have dot product =-2 * 2 + 6 * (-3) = -22 magnitude of first vector = sqrt( (-2)^2 + 6^2) = sqrt(40) magnitude of second vector = sqrt( 2^2 + (-3)^2 ) = sqrt(13) Since dot product = magnitude of 1 st vector * magnitude of 2d vector * cos(theta) we have cos(theta) = dot product / (magnitude of 1 st vector * magnitude of 2d vector) so that theta = arccos [ dot product / (magnitude of 1 st vector * magnitude of 2d vector) ] = arccos[ -22 / ( sqrt(40) * sqrt(13) ) ] = arccos ( -.965) = 164 degrees, approx.. **
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RESPONSE -->
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20:57:06 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I found some of the conversions difficult. When I had to think 3-dimensionally, things were a bit confusing. Could you please walk me through #25 of Chapter 1 in the text? I have tried to find the correct answer (which is 7 x 10^5 books) but I just can't get it.
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20:57:19 ** I had to get a little help from a friend on vectors, but now I think I understand them. They are not as difficult to deal with as I thought. **
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RESPONSE --> ok
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