course Phy 201
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12:09:19 NOTE PRELIMINARY TO QUERY:
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RESPONSE --> Ok
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12:09:34 ** Questions about velocity, average velocity, acceleration, etc. are very confusing because so many of the concepts have similar definitions. People have trouble distinguishing things like average velocity, which for uniform acceleration can be obtained in a process that adds two velocities, from average acceleration, which involves subtracting two velocities; one of these processes involves dividing by 2 and the other dividing by the time interval `dt.
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RESPONSE --> Ok
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12:09:44 It is essential to keep the definitions and the meanings of the terms very clear and to work everything from definitions. It is equally important to have a good common-sense understanding of every definition so you can develop the intuition to make sense of everything you do.
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RESPONSE --> Ok
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12:09:53 That inevitably takes people a little time. But in the process you develop the habits you will need to succeed in the course. **
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RESPONSE --> Ok
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몚ʐȄ¼ assignment #003 呌騶~Q Physics I Class Notes 06-07-2006
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13:24:47 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 --> Given the initial velocity, final velocity and time duration of a uniformly accelerating object, we reason out the corresponding acceleration and change in the position of an object by calculating the difference between the initial and final velocities and divide it by the time duration to find the acceleration.
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13:25:18 ** 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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13:29:02 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 velocity vs. clock time trapezoid represents a change in position and acceleration by rise/run *2 where height = a and width is clock time and base = position of acceleration
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13:29:46 ** 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 --> I was incorrect where width(area) and where slope = rise / run represents change in velocity / time interval = acceleration.
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Cԁ᫅ assignment #002 呌騶~Q Physics I Vid Clips 06-07-2006
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13:40:07 Given a graph of velocity vs. time how do we determine the rate which velocity increases?
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RESPONSE --> We determine the rate which velocity increases by looking at the slope
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13:40:47 ** To find the average rate of change of velocity we use the slope of the graph between two points, which indicates rise / run = change in velocity/change in clock time. This is equivalent to using the slope of the graph, which indicates rise / run = change in velocity/change in clock time. COMMON ERROR: rate = Velocity/time INSTRUCTOR COMMENT: If you divide v by t you do not, except in certain special cases, get the average rate at which velocity increases. You have to divide change in velocity by change in clock time. For example if you're traveling at a constant 60 mph for 1 hour and divide 60 mph by 1 hour you get a nonzero acceleration of 60 mph / hr while in fact you are traveling at a constant rate and have acceleration 0. **
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RESPONSE --> Correct- however I could have gone into more detail of rise/run = change in velocity/change in clock time.
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13:47:19 If we know the velocities at two given clock times how do we calculate the average rate at which the velocity is changing during that time interval?
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RESPONSE --> we calculate the average rate at which the velocity by finding the difference between the two times and calculating vAve = 'ds/'dt.
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13:47:24 ** We divide the difference `dv in velocities by the duration `dt of the time interval. **
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RESPONSE --> Correct
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13:48:27 Does the rate at which a speedometer needle moves tell us how fast the vehicle is moving?
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RESPONSE --> The speedometer needle tells us the change in velocity regarding the amoutn of distance travelled overa given time.
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13:48:46 ** The rate at which the speedometer needle moves does not tell us how fast the vehicle is moving. The fact that the speedometer needle moves tells us that the car is either speeding up or slowing down, not at what rate the car is moving. The reading on the speedometer tells us how fast we are moving. A speedometer that simply sits there on 60 mph doesn't move at all, but presumably the vehicle is moving pretty fast. If the speedometer moves, that indicates that velocity is changing. If the speedometer moves quickly, the velocity is changing quickly. If the speedometer moves slowly, velocity is changing slowly. A quickly moving speedometer needle doesn't imply a quickly moving vehicle. The speedometer might go past the 5 mph mark very fast. But at the instant it passes the mark the car isn't moving very quickly. The needle can move quickly from 1 mph to 2 mph, or from 100 mph to 101 mph. The speed of the needle has little to do with the speed of the vehicle. Of course if the needle is moving quickly for an extended period of time, this implies a large change in the velocity of the vehicle. However this information still does not tell you what that velocity actually is. **
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RESPONSE --> Correct
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13:49:21 Does a stationary speedometer needle implies stationary vehicle?
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RESPONSE --> No the stationary speedometer needle may not necessarily imply a stationary vehicle- the vehicle could be moving fast but at a constant speed
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13:49:29 ** A stationary speedometer needle does not imply a stationary vehicle. The speedometer could be constant at the 60 mph mark while the car is moving 60 mph. **
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RESPONSE --> Correct
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13:51:22 Does a quickly moving speedometer needle imply a quickly moving vehicle?
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RESPONSE --> a quickly moving speedometer needle implies a moving vehicle is changing velocities at a quick rate. A car could go from 5 mph to 10 mph and back down to 5 mph in a few seconds and be moving quickly.
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13:51:30 ** A quickly moving speedometer needle does not imply a quickly moving vehicle. It merely implies that the velocity of car is changing quickly. **
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RESPONSE --> Correct
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13:51:56 What does it feel like inside a car when the speedometer needle is moving fast?
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RESPONSE --> There is a feel of acceleration or deacceleration as a speedometer needle moves fast.
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13:52:58 ** When the speedometer is moving fast, as for example when you are first starting out in a hurry, in a powerful vehicle, you feel yourself pushed back in your seat. The speedometer can also move quickly when you press hard on the brakes. In this case you tend to feel pressed forward toward the steering wheel. **
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RESPONSE --> Correct
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13:56:18 What does the speed of the speedometer needle tell us?
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RESPONSE --> The speed of the speedometer tells us the velocity in which the car is changing. The car could have the speedometer at 60 and still at 20 mph, indicating that pretty soon the car will be going much fast as the velocity increases.
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13:56:23 ** The position on the speedometer tells us how fast we are moving--e.g., 35 mph, 50 mph. However the speed of the needle tells us at what rate we are speeding up or slowing down--specifically the speed of the speedometer needle gives us the rate at which velocity changes. **
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RESPONSE --> correct
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