course Phy 231 6/15 8 007. `query 7 Question: `qDescribe the flow diagram you would use for the uniform acceleration situation in which you are given v0, vf, and `dt.
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I did not have my quantities in the same levels, but I believe I got all of the relationships correct. Self-critique rating #$&* 3 Question: Describe the flow diagram you would use for the uniform acceleration situation in which you are given `dt, a, v0 Your solution: Dt, a and v0 are on the first level. Dt and a connect with dv to form a triangle to the second level. Then the dv and the v0 connect to make another triangle with vf. On the third level, vf and v0 connect to form a triangle with vAve. Finally, the vAve and dt combine to get ds. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 Self-critique rating #$&* OK ********************************************* Question: Check out the link flow_diagrams and give a synopsis of what you see there. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The document gives step-by-step instructions on how to make flow diagrams. It says that the known quantities are in the first level, and then you can use relationships to put in second-level quantities: ones that are formed from only the first level. Finally, you find the third level quantities by combining numbers from the first and second levels. The page also has diagrams to explain the set up and reasons for using the diagrams and the equations. confidence rating #$&* 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): OK Self-critique rating #$&* OK Question: Explain in detail how the flow diagram for the situation in which v0, vf and `dt are known gives us the two most fundamental equations of motion. Your solution: From v0 and vf, you find the average and this gives you vAve. Then you multiply vAve by dt and you get the displacement: #1 Then, you use vf-v0 = dv and dv = a*dt to make the next equation. Set them equal to each other and solve for vf and you get vf = v0+a(dt): #2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. Your solution: For the third equation of motion, you have to use both the first and second equations. You know that the final velocity is equal to v0+a dt. You substitute this expression in for vf in the second equation. Ds = (v0+a dt +v0)/2 dt Ds = (2v0+adt)/2 dt Ds = v0+.5a dt dt = v0+.5a dt^2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: a and `dt give you `dv. `dv and v0 give you vf. v0 and vf give you vAve. vAve and `dt give you `ds. In symbols, `dv = a `dt. Then vf = v0 + `dv = v0 + a `dt. Then vAve = (vf + v0)/2 = (v0 + (v0 + a `dt)) / 2) = v0 + 1/2 a `dt. Then `ds = vAve * `dt = [ v0 `dt + 1/2 a `dt ] * `dt = v0 `dt + 1/2 a `dt^2. ** Self-critique (if necessary): Ok Self-critique rating #$&* Ok Question: Why do we think in terms of seven fundamental quantities while we model uniformly accelerated motion in terms of five? Your solution: There are 7 quantities, but vAve and dv are really easy to find with the other five quantities in the equations. Also, you dont want an equation with too many variables because then you would not be able to solve it. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The four equations are expressed in terms of five fundamental quantities, v0, vf, a, `dt and `ds. However to think in terms of meanings we have to be able to think not only in terms of these quantities but also in terms of average velocity vAve and change in velocity `dv, which aren't among these five quantities. Without the ideas of average velocity and change in velocity we might be able to use the equations and get some correct answers but we'll never understand motion. The four equations of unif accelerated motion are expressed in terms of five fundamental quantities, v0, vf, a, `dt and `ds. The idea here is that to intuitively understand uniformly accelerated motion, we must often think in terms of average velocity vAve and change in velocity `dv as well as the five quantities involved in the four fundamental equations. one important point is that we can use the five quantities without any real conceptual understanding; to reason things out rather than plugging just numbers into equations we need the concepts of average velocity and change in velocity, which also help us make sense of the equations. ** Self-critique (if necessary): Not having the important variables in the equations makes you think about the concepts and the quantities, so that you are not merely plugging numbers and getting results. You have to understand the equations to use them. Self-critique rating #$&* 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? Your solution: When you do the test for the same time interval, the object has the same amount of time to accelerate, so the acceleration on the object is the same and the change in velocity is the same. If you do it for a constant distance and the initial velocity is faster, then the object will cover the distance faster and will not have as much time to accelerate, making the change in velocity less. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If we accelerate down a constant incline our rate of change of velocity is the same whatever our initial velocity. So the change in velocity is determined only by how long we spend coasting on the incline. Greater `dt, greater `dv. If you travel the same distance but start with a greater speed there is less time for the acceleration to have its effect and therefore the change in velocity will be less. You might also think back to that introductory problem set about the car on the incline and the lamppost. Greater initial velocity results in greater average velocity and hence less time on the incline, which gives less time for the car to accelerate. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* 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 second equation: you take the average of v0 and vf, which will give you the average velocity and is the short side times the ling side divided by two. Then you multiply that by dt, which is the base of the trapezoid. For the first equation, you do the acceleration times dt, which will give you the change in velocity. You add this number to the initial velocity (it extends the v0 side to the level of vf) and you find the value of vf. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If acceleration is uniform then the v vs. t graph is linear. So the average velocity on the interval is vAve = (vf + v0) / 2. From the definition of average velocity we conclude that `ds = vAve * `dt. Thus `ds = (vf + v0) / 2 * `dt. This is the first equation of uniformly accelerated motion. Note that the trapezoid can be rearranged to form a rectangle with 'graph altitude' vAve and 'graph width' equal to `dt. The area of a rectangle is the product of its altitude and its width. Thus the product vAve * `dt represents the area of the trapezoid. More generally the area beneath a v vs. t graph, for an interval, represents the displacement during that interval. For University Physics, this generalizes into the notion that the displacement during a time interval is equal to the definite integral of the velocity function on that interval. The definition of average acceleration, and the fact that acceleration is assumed constant, leads us to a = `dv / `dt. `dv = vf - v0, i.e., the change in the velocity is found by subtracting the initial velocity from the final Thus a = (vf - v0) / `dt. `dv = vf - v0 represents the 'rise' of the trapezoid, while `dt represents the 'run', so that a = `dv / `dt represents the slope of the line segment which forms the top of the trapezoid. For University Physics, this generalizes into the notion that the acceleration of an object at an instant is the derivative of its velocity function, evaluated at that instant. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating #$&* OK Question: (required only of University Physics students): If s(t) = .3 m/s^3 * t^3 - 2 m/s^2 * t^2 + 5 m/s * t + 12 m then what are the velocity and acceleration functions? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: V(t) = .9m/s^3 t^2-4m/s^2 t+5m/s A(t) = 1.8m/s^3 t-4m/s^2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If s(t) = .3 m/s^3 * t^3 - 2 m/s^2 * t^2 + 5 m/s * t + 12 m then: The derivative of .3 m/s^3 * t^3 is (.3 m/s^3 * t^3 ) ' = (.3 m/s^3) * (t^3) ' = (.3 m/s^3) * (3 t^2) = .9 m/s^3 * t^2. Note that .3 m/s^2 is a constant, and also that if t is in seconds the units of the result are m/s^3 * (s)^2 = m/s, which is the unit of velocity. Similarly the derivatives for the other terms are (-2 m/s^2 * t^2 ) ' = -4 m/s^2 * t (5 m/s * t) ' = 5 m/s and (12 m) ' = 0 Thus the derivative of s(t) is v(t) = s ' (t) = .9 m/s^3 * t^2 - 4 m/s^2 * t + 5 m/s The acceleration function is the derivative of v(t): a(t) = v ' (t) = 1.8 m/s^3 * t - 4 m/s^2 You should check to be sure you understand that the units of each of these terms are m/s^2, which agrees with the unit for acceleration. Self-critique (if necessary): OK Self-critique rating #$&* OK "