course phy 201 2/27/2010 1:08 p.m. I am behind, my back went out and they put me on some heavy duty stuff to releive it so I have not posted anything in a couple of days...I am going to get the major quiz done this week and try to get caught up. Sorry for the delay in work.
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Given Solution: We start with v0, vf and `dt on the first line of the diagram. We use v0 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. ** STUDENT COMMENT i dont understand how you answer matches up with the question INSTRUCTOR RESPONSE All quantities are found from basic definitions where possible; where this is possible each new quantity will be the result of two other quantities whose value was either given or has already been determined. Using 'dt and a, find 'dv (since a = `dv / `dt, we have `dv = a `dt). Using 'dv and v0, find vf, indicated by lines from `dv and v0 to vf (vf = v0 + `dv). Using vf and v0, find vAve, indicated by lines from vf and v0 to vAve ( (vf + v0) / 2 = vAve, for uniform acceleration). Using 'dt and vAve, find 'ds, indicated by lines from `dt and vAve to `ds (vAve = `ds / `dt so `ds = vAve * `dt). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok, it would have been easier to get the vave first to get to the 'ds, and then obtain the a by the subtraction of the v0'- vf and and dividing that by the 'dt ------------------------------------------------ Self-critique rating #$&*:2 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: Describe the flow diagram you would use for the uniform acceleration situation in which you are given `dt, a, v0 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The first line would have 'dt a and v0. On the second line we can find the 'dv from the a * dt, and the Vf can be found by the Vo + the dv. On the third line we can find the vave from the average of the V0 and the Vf and and we can find the 'ds by the average of the v0 and vf * 'dt. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Student Solution: Using 'dt and a, find 'dv. Using 'dv and v0, find vf, indicated by lines from `dv and v0 to vf. Using vf and v0, find vAve, indicated by lines from vf and v0 to vAve Using 'dt and vAve, find 'ds, indicated by lines from `dt and vAve to `ds. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok, I think I had pretty much the same I had started by first line with 'dt , a and V0 the second line I had the 'dv with lines connected to a and 'dt. Also on the second line I had vf with lines connected to 'dv and V0. Also on the third line to the left of the flow chart I had the 'ds with line drawn from the 'dt to the vave. so It looks like a square almost. ------------------------------------------------ Self-critique rating #$&*:3 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: Check out the link flow_diagrams and give a synopsis of what you see there. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Ok, I tried to follow this link, but I did not find anything. I have,however, gone to the link given in a solution from a previous assignment http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/flow_diagrams.htm...I found this to be very helpful and it explained step by step how the flow chart is created. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: You should have seen a detailed explanation of a flow diagram, and your 'solution' should have described the page. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I liked the link it help me alot ------------------------------------------------ Self-critique rating #$&*:3 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: Explain in detail how the flow diagram for the situation in which v0, vf and `dt are known gives us the two most fundamental equations of motion. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The flow diagram would start on the first line with V0, Vf, 'dt, from the V0 and the Vf you can determine the 'dv (v0 - Vf = 'dv). Also from the v0 and the vf you can fine the vave (V0 +Vf/2 = vave) , from the vave and 'dt you can find the 'ds (vave * 'dt = 'ds). From the 'dv and the 'dt you can find the a (a = 'dv / 'dt). confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. ** @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: From the V0, a and ' dt on the first line, you can determine the 'dv from a * 'dt, from the vo + dv you can get the vf, from the average of the vf and the V0 you have tge vave, and from the vave times the 'dt you have the 'ds. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: a and `dt give you `dv. `dv and v0 give you vf. v0 and vf give you vAve. vAve and `dt give you `ds. In symbols, `dv = a `dt. Then vf = v0 + `dv = v0 + a `dt. Then vAve = (vf + v0)/2 = (v0 + (v0 + a `dt)) / 2) = v0 + 1/2 a `dt. Then `ds = vAve * `dt = [ v0 `dt + 1/2 a `dt ] * `dt = v0 `dt + 1/2 a `dt^2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok that is what I had. ------------------------------------------------ Self-critique rating #$&*:3 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: Why do we think in terms of seven fundamental quantities while we model uniformly accelerated motion in terms of five? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because ther must be two or more quantities available to find the other five. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ONE WAY OF PUTTING IT: The four equations are expressed in terms of five fundamental quantities, v0, vf, a, `dt and `ds. However to think in terms of meanings we have to be able to think not only in terms of these quantities but also in terms of average velocity vAve and change in velocity `dv, which aren't among these five quantities. Without the ideas of average velocity and change in velocity we might be able to use the equations and get some correct answers but we'll never understand motion. ANOTHER WAY: The four equations of unif accelerated motion are expressed in terms of five fundamental quantities, v0, vf, a, `dt and `ds. The idea here is that to intuitively understand uniformly accelerated motion, we must often think in terms of average velocity vAve and change in velocity `dv as well as the five quantities involved in the four fundamental equations. one important point is that we can use the five quantities without any real conceptual understanding; to reason things out rather than plugging just numbers into equations we need the concepts of average velocity and change in velocity, which also help us make sense of the equations. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok, so that we may understand the equations we need to think in terms of change in velocity and average change of velocity, we must know the concepts, not just how to work the equations. ------------------------------------------------ Self-critique rating #$&*:3 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: 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 accelerate down a constant incline for the same time, but not when we accelerate down the same incline for a constant distance? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because if the object is traveling at a constant rate, down a constant incline the velocities will not change until they have reached a certain distance at which they either will speed up or slow down. When the object starts out down the ramp it may be at a certain speed until the object get to the end and begins to slow down until it reaches equilibrium with the force of gravity and stops. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If we accelerate down a constant incline our rate of change of velocity is the same whatever our initial velocity. So the change in velocity is determined only by how long we spend coasting on the incline. Greater `dt, greater `dv. If you travel the same distance but start with a greater speed there is less time for the acceleration to have its effect and therefore the change in velocity will be less. You might also think back to that introductory problem set about the car on the incline and the lamppost. Greater initial velocity results in greater average velocity and hence less time on the incline, which gives less time for the car to accelerate. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): So if acceleration is constant a, the rate of change of velocity 'dv, is the same as the initial velocity v0, If you have increase in time 'dt you have an imcrease in velocity. Change of velocity is determined by the time on the incline. ------------------------------------------------ Self-critique rating #$&*:3 @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ ********************************************* Question: Explain how the v vs. t trapezoid for given quantities v0, vf and `dt leads us to the first two equations of linearly accelerated motion. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: the average of vo and vf (vave) times the 'dt gives us the 'ds ( change in distance). And the change in V0 and Vf ('dv) divided by the ' dt gives us the acceleration. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If acceleration is uniform then the v vs. t graph is linear. So the average velocity on the interval is vAve = (vf + v0) / 2. From the definition of average velocity we conclude that `ds = vAve * `dt. Thus `ds = (vf + v0) / 2 * `dt. This is the first equation of uniformly accelerated motion. Note that the trapezoid can be rearranged to form a rectangle with 'graph altitude' vAve and 'graph width' equal to `dt. The area of a rectangle is the product of its altitude and its width. Thus the product vAve * `dt represents the area of the trapezoid. More generally the area beneath a v vs. t graph, for an interval, represents the displacement during that interval. For University Physics, this generalizes into the notion that the displacement during a time interval is equal to the definite integral of the velocity function on that interval. The definition of average acceleration, and the fact that acceleration is assumed constant, leads us to a = `dv / `dt. `dv = vf - v0, i.e., the change in the velocity is found by subtracting the initial velocity from the final Thus a = (vf - v0) / `dt. `dv = vf - v0 represents the 'rise' of the trapezoid, while `dt represents the 'run', so that a = `dv / `dt represents the slope of the line segment which forms the top of the trapezoid. For University Physics, this generalizes into the notion that the acceleration of an object at an instant is the derivative of its velocity function, evaluated at that instant. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok, so the if accerleration is uniform then the graph is linear so we find vave, and from there we find the 'ds. Vave is the same as the graph altitutde and 'dt is the graph width, which leads us to the rise/ run or slope of the graph. The area is the L * W or vave * 'dt of the graph which is the same as displacement. Ok, so the accleration is the same as the slope(rise / run) or (vave / 'dt) and the area is the same as (Vave * 'dt) or (L * W) which is also the displacement. ------------------------------------------------ Self-critique rating #$&*:3