course PHY 121
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01:33:29 Describe the flow diagram you would use for the uniform acceleration situation in which you are given v0, vf, and `dt.
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RESPONSE --> On the first level you would have `dt v0 vf Then on the next level you would get the change in velocity from vo and vf. so you would have `dv = vf - v0. Then from `dt and `dv you can calculate the acceleration. `dv / `dt.
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02:12:34 ** We start with v0, vf and `dt on the first line of the diagram. We use vO and vf to find Vave, indicated by lines from v0 and vf to vAve. Use Vave and 'dt to find 'ds, indicated by lines from vAve and `dt to `ds. Then use `dv and 'dt to find acceleration, indicated by lines from vAve and `dt to a. **
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RESPONSE --> I did not take the diagram down far enough I stopped at the acceleration. But from the vo and vf you could also calculate the vAve, which would allow you to calculate the `ds.
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02:22:28 Describe the flow diagram you would use for the uniform acceleration situation in which you are given `dt, a, v0
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RESPONSE --> On the first level we would have `dt, a, v0. From this we can get `dv by a * `dt. Then the next level we obtain vf from the initial velocity plus the change in velocity. on the 4th level we can get the avg. velocitythen we can calculate the displacement.
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02:22:34 ** Student Solution: Using 'dt and a, find 'dv. Using 'dv and v0, find vf. Using vf and vO, find vave. Using 'dt and Vave, find 'ds. **
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RESPONSE --> ok
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02:31:42 Explain in detail how the flow diagram for the situation in which v0, vf and `dt are known gives us the two most fundamental equations of motion.
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RESPONSE --> we can determine the change in velocity from the final and initial velocity and from the initial and final velocities we can also determine the avg. velocity. This allows us to then determine the acceleration from the change in velocity and time interval. `dv/`dt. From the acceleration, we can get a = (vf - v0)/ `dt. This formula allows us to get vf = v0 + a `dt. Which is the 1st equation of motion. Then second equation `ds = (vf + v0) / `dt can be formulated from the time interval and average velocity.
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02:31:49 **Student Solution: v0 and vf give you `dv = vf - v0 and vAve = (vf + v0) / 2. `dv is divided by `dt to give accel. So we have a = (vf - v0) / `dt. Rearranging this we have a `dt = vf - v0, which rearranges again to give vf = v0 + a `dt. This is the second equation of motion. vAve is multiplied by `dt to give `ds. So we have `ds = (vf + v0) / 2 * `dt. This is the first equation of motion Acceleration is found by dividing the change in velocity by the change in time. v0 is the starting velocity, if it is from rest it is 0. Change in time is the ending beginning time subtracted by the ending time. **
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RESPONSE --> ok
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02:42:36 Explain in detail how the flow diagram for the situation in which v0, a and `dt are known gives us the third fundamental equations of motion.
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RESPONSE --> v0, a and `dt allows us to obtain vf and v0. 'dv = a and vf = v0 + `dv in the first two levels. This allows us to obtain the avg. velocity which is vAve = vf - vo / 2 the vf becomes vo = `dv. to get vAve = v0 + `dv + v0 / 2. then dv = a and v0+ v0 = 2v0. Which gives us VAve = 2 v0 + a `dt / 2. Then through multiplication we obtain the 3rd equation: `ds = vAve * `dt = v0 * `dt + 1/2 * a * `dt ^2.
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02:42:41 ** a and `dt give you `dv. `dv and v0 give you vf. v0 and vf give you vAve. vAve and `dt give you `ds. In symbols, `dv = a `dt. Then vf = v0 + `dv = v0 + a `dt. Then vAve = (vf + v0)/2 = (v0 + (v0 + a `dt)) / 2) = v0 + 1/2 a `dt. Then `ds = vAve * `dt = [ v0 `dt + 1/2 a `dt ] * `dt = v0 `dt + 1/2 a `dt^2. **
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RESPONSE -->
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02:44:09 Why do we think in terms of seven fundamental quantities while we model uniformly accelerated motion in terms of five?
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RESPONSE --> Beucae we can always define the two of the seven.
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02:49:26 ** ONE WAY OF PUTTING IT: The four equations are expressed in terms of five fundamental quantities, v0, vf, a, `dt and `ds. However to think in terms of meanings we have to be able to think not only in terms of these quantities but also in terms of average velocity vAve and change in velocity `dv, which aren't among these five quantities. Without the ideas of average velocity and change in velocity we might be able to use the equations and get some correct answers but we'll never understand motion. ANOTHER WAY: The four equations of unif accelerated motion are expressed in terms of five fundamental quantities, v0, vf, a, `dt and `ds. The idea here is that to intuitively understand uniformly accelerated motion, we must often think in terms of average velocity vAve and change in velocity `dv as well as the five quantities involved in the four fundamental equations. one important point is that we can use the five quantities without any real conceptual understanding; to reason things out rather than plugging just numbers into equations we need the concepts of average velocity and change in velocity, which also help us make sense of the equations. **
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RESPONSE --> ok
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02:51:54 Accelerating down an incline through a given distance vs. accelerating for a given time Why does a given change in initial velocity result in the same change in final velocity when we accelerated down a constant incline for the same time, but not when we accelerated down the same incline for a constant distance?
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RESPONSE --> For the change in velocity it would vary in distances, and not in the time.
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02:52:27 ** If we accelerate down a constant incline our rate of change of velocity is the same whatever our initial velocity. So the change in velocity is determined only by how long we spend coasting on the incline. Greater `dt, greater `dv. If you travel the same distance but start with a greater speed there is less time for the acceleration to have its effect and therefore the change in velocity will be less. You might also think back to that introductory problem set about the car on the incline and the lamppost. Greater initial velocity results in greater average velocity and hence less time on the incline, which gives less time for the car to accelerate. **
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RESPONSE --> ok
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02:52:50 Acceleration of gravity experiment: STUDENT COMMENT: Just followed directions. I was bothered by the crude way of measuring from the start. I would have preferred to be able to see the millimeters, but I couldn't. I used this fact to calculate the error. That was the only bit of error, except human error that is impossible to get rid of since we need you to get us through the rest of the course.
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RESPONSE --> ??
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02:52:54 ** INSTRUCTOR RESPONSE: The millimeters don't turn out to be that important at these velocities when we are using a pendulum for the timer. The error in synchronizing the pendulum is much more significant than a half-centimeter error in the distance measurements. It's an instance of the old saying that you don't worry about draining the swamp when you're up to your neck (or whatever) in alligators. We're up to our necks (or at least our knees) in pendulum error here. **
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RESPONSE --> ??
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03:01:57 Why do you think the slope of the graph obtained in this experiment should be equal, or very nearly equal to the acceleration of gravity?
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RESPONSE --> The acceleration should be proportional to the slope or at least very close
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03:03:24 ** What this one comes down to is that since for small slopes the force accelerating the object down the incline appears to be in the same proportion to the weight of the cart as the rise to the run of the graph, if we extrapolate the small-slope graph of acceleration vs. ramp slope to a slope of 1 we should get the acceleration that would result from the full force of gravity on the cart. COMMON MISCONCEPTION: Using a small slope, the effects of friction and other variables are minimal, so the object behaves much as if it were falling, both of which circumstance have the same force acting upon them, gravity. RESPONSE: Friction isn't minimal, but it doesn't vary much from slope to slope. The result is that it shifts the graph of accel vs. slope downward (less accel for a given ramp slope) by about the same amount for each slope, but that as a result it maintains the same graph slope. STUDENT COMMENT: I honestly do not understand at this point. RESPONSE: ** Understandable at this point. Our experiments show that F vs. slope is linear with graph slope equal to total mass, a vs. slope graph is linear, a vs. F is linear. Put them together and we draw the conclusion that a vs. slope is linear. Reasons: Gravitational force component along the incline is m g sin(`theta), where `theta is angle in radians. For small angles sin(`theta) is very close to `theta (observe that the derivative of sin(`theta) is cos(`theta), which is very nearly 1 for small angles, so slope of sin(`theta) vs. `theta graph is 1; hence the approximation). Note also that the slope is tan(`theta), which is close to sin(`theta) for small `theta. So net F along incline = m g sin(`theta) -frict = m g * `theta - frict, for small slopes. frict, force of friction, doesn't vary much for small slopes (normal force remains close to entire weight, etc.) Thus accel = Fnet / m = g sin(`theta) - frict / m = g * slope - frict / m, to a good approximation. On a graph of accel vs. slope this graph has slope g. **
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RESPONSE --> I uderstand the first part of the answer, however I get a bit confused with the net force stuff and the end.
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