#$&* course Phy 121 9/19 9:50 pm 007. `query 7
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Given Solution: We start with v0, vf and `dt on the first line of the diagram. We use v0 and vf to find Vave, indicated by lines from v0 and vf to vAve. Use Vave and 'dt to find 'ds, indicated by lines from vAve and `dt to `ds. Then use `dv and 'dt to find acceleration, indicated by lines from vAve and `dt to a. ** STUDENT COMMENT i dont understand how you answer matches up with the question INSTRUCTOR RESPONSE All quantities are found from basic definitions where possible; where this is possible each new quantity will be the result of two other quantities whose value was either given or has already been determined. Using 'dt and a, find 'dv (since a = `dv / `dt, we have `dv = a `dt). Using 'dv and v0, find vf, indicated by lines from `dv and v0 to vf (vf = v0 + `dv). Using vf and v0, find vAve, indicated by lines from vf and v0 to vAve ( (vf + v0) / 2 = vAve, for uniform acceleration). Using 'dt and vAve, find 'ds, indicated by lines from `dt and vAve to `ds (vAve = `ds / `dt so `ds = vAve * `dt). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: Describe the flow diagram you would use for the uniform acceleration situation in which you are given `dt, a, v0 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First line: `dt a v0 Second line: dv with lines back to `dt and a Third line: `ds with lines back to `dt and `dv. vf with lines back to `dv and v0 Fourth line: vAve with lines back to vf and v0 confidence rating #$&*: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. STUDENT QUESTION Can you only have two lines that connect to one variable because i utilized the formula vf=v0 +a `dt and connected all three to find vf? I do see how it could be done using two in the above solution. INSTRUCTOR RESPONSE The idea is to use the definitions of velocity and acceleration whenever possible. This is possible in this case: If you know `dt and a you can use the definition of acceleration to find `dv (which is equal to a `dt). Then you can use v0 and `dv to get vf (which is equal to v0 + `dv; from this you could conclude that vf = v0 + a `dv). You end up with the same result you would have gotten from the formula, but you are using insight into the nature of velocity and acceleration by using the definitions, as opposed to a memorized formula that can be applied whether or not you understand its meaning. The only exceptional cases are when you know v0 or vf (but not both), acceleration a and displacement `ds. In that case you need to start with the third or fourth equation, where I recommend that you start with the fourth. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I erroneously had `ds connected back to `dt and `dv. This should have been ""lines back to vAve and `dt"" ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: Check out the link flow_diagrams and give a synopsis of what you see there. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The link begins with a student response asking for additional help with flow diagrams. Then the rest of the link shows a thorough and methodical work through of a flow diagram to include a few graphical representations of a flow diagram in the steps of being created. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: You should have seen a detailed explanation of a flow diagram, and your 'solution' should have described the page. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: Explain in detail how the flow diagram for the situation in which v0, vf and `dt are known gives us the two most fundamental equations of motion. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First line: vo vf `dt Second line: vAve with lines back to vO and vf. `dv with lines back to v0 anc vf Third line `ds with lines back to vAve and `dt. a with lines back to `dv and `dt. This diagram gives us the two most fundatmental equations of motion: vAVe = `ds/`dt (our third line shows us this relationship) a = `dv/`dt (our third line also shows us this relationship) confidence rating #$&*: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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First line: v0 a and `dt Second line: `dv with lines going back to a and `dt Third line: vf with lines back to v0 and `dv Fourth line: vAve with lines back to v0 and vf Fifth line: `ds with lines back to vAve and `dt If `ds = vAVe*`dt adns vAve is derived from v0 and vf we can write `ds =( v0+vf)/2 *dt but to get that vf we needed vO and `dv `ds = [v0 +(`dv+v0)]/2 *`dt but to get that `dv we needed a and `dt `ds = [v0 + (a*`dt +v0)]/2 *`dt `ds =( 2v0 + a*`dt)/2 *`dt ds =( v0 + .5a`dt) *dt (distribute `dt) `ds = v0`dt + .5a dt^2 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. ** STUDENT COMMENT: I do not understand how to get the equation out of the flow diagram or calculations. INSTRUCTOR RESPONSE: Presumably the flow diagram was the basis for your responses 'You can get the `dv from the `dt and a by: a * `dt = `dv Then, you can get the vf by: `dv + v0 = vf Next, you can get the vAve by: (v0 + vf) / 2 = vAve Then, you can get the `ds by: vAve * `dt = `ds The change in position is what is being solved for in the equation: `ds = v0 * `dt + .5 a `dt^2.' Using your responses as a basis: You can get the `dv from the `dt and a by: a * `dt = `dv Then, you can get the vf by: `dv + v0 = vf. Since `dv = a * `dt, we have a * `dt + v0 = vf Next, you can get the vAve by: (v0 + vf) / 2 = vAve Then, you can get the `ds by: vAve * `dt = `ds v0 is considered to be one of the given quantities, and vf = v0 + a `dt from the line before the preceding line. So vAve * `dt = (v0 + vf) / 2 * `dt = (v0 + (v0 + a `dt) ) / 2 * `dt = (2 v0 + a `dt) / 2 * `dt = (v0 + 1/2 a `dt) * `dt = v0 `dt + 1/2 a `dt^2. It is the change in position for which we are solving the equation: `ds = v0 * `dt + .5 a `dt^2. the preceding showed that `ds = v0 `dt + 1/2 a `dt^2 STUDENT COMMENT used direct reasoning for my answer, which was Assuming to say here that v0 and vf will give us dv. Which with dt gives aAve. And vAve comes from dt and ds. INSTRUCTOR RESPONSE Your response was 'Assuming to say here that v0 and vf will give us dv. Which with dt gives aAve. And vAve comes from dt and ds. ' All these statements are correct, but it will turn out that they don't work for the given information. First let's look at the details we can get from your reasoning: You are correct that v0 and vf will give us dv. Specifically, `dv = vf - v0. So if v0 and vf were given quantities, we would now have an expression for `dv in terms of given quantities (however notice that vf isn't a given quantity). You follow with 'Which with dt gives aAve'. Specifically, aAve = `dv / `dt so aAve = (vf - v0) / `dt. So had the given quantities been v0, vf and `dt, we would at this point have an expression for aAve in terms of given quantities (however note once more that vf isn't given). Then you say 'And vAve comes from dt and ds.'. The specifics: vAve = `ds / `dt. Since `ds and `dt are given quantities, we do have an expression for vAve. However if we assume v0, a and `dt as given quantities, we would not yet have sufficient information to get vAve. In your solution you didn't specify what the initial information is. In this case you are asked to reason from v0, a and `dt. As shown above, the reasoning you give doesn't work for this given information. One sequence of reasoning that does work for this information is given in the first line of the 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. The given solution then fills in the details, using the following relationships: `dv = a `dt. vf = v0 + `dv vAve = (vf + v0)/2 `ds = vAve * `dt Be sure you understand the remaining details of the given solution. I'll welcome more questions if you have them. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK It was a bit of a slog, but using the flow diagram and working backward really helped. There wasn't a self-critique for the previous problem. I feel like I did the flow diagrams correctly for the previous problem but I misinterpretered the two most fundatmental equations of motion to be vAve = `ds/`dt and a=`dv/`dt instead of the first two equations of uniformly acclerated motion. I get this now and understand who to derive these from reasoning out the flow diagram. ------------------------------------------------ Self-critique Rating:OK
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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. ** STUDENT QUESTION I understand how to make flow diagrams and use all of the concepts to figure out the missing variable from the equation. I even understand `dv and vAve are intuitive but don't these still show up in the flow diagrams? Aren't they still in a sense being modeled? Good question. They show up in the diagrams but not in the four equations of uniformly accelerated motion. The point is that in the process of reasoning out a situation, we must always use `dv and vAve, both of which are part of our definitions of velocity and acceleration. However we can write a set of equations that do not include vAve and `dv as variables. These equations involve only v0, vf, a, `ds and `dt. Given any three of these five we can use the equations to find the other two, and we never have to think about `dv and vAve to do so. We reduce the physics to a mechanical process involving only simple algebra, unconnected to the basic definitions. The five-variable formulation is very nice and easy to use. We can use it to solve problems in fewer steps than the direct-reasoning-from-definitions approach, and this is something we very much want to be able to do. The trick in a first-semester physics course is to achieve a very basic understanding of uniformly accelerated motion, eventually learning to use the equations without using them as a crutch to bypass understanding. So we learn to reason using the seven quantities, then we learn to use the four-equation model. There is an additional approach for University Physics students, which involves calculus and is not relevant (and not accessible) to anyone who doesn't know calculus. We first understand how the derivative is an instantaneous rate-of-change function, so that the velocity function is the derivative of the position function, and the acceleration function the derivative of the velocity function. Then, understanding how the integral is the change-in-quantity function, we integrate the acceleration function with respect to clock time to get the velocity function, and finally integrate the velocity function to get the position function. STUDENT QUESTION Had an issue explaining it clearly, therefore posted the given solution. Five quantities are used to explain four fundamental equations relative to constant acceleration? INSTRUCTOR RESPONSE (summary of the use of definitions and equations) The equations involve the five quantities v0, vf, a, `ds, `dt. Each equation contains four of the five. If you know the values of three of the five quantities, there is always at least one equation that contains those three and can hence be solved to get the value of a fourth. Knowing four of the five you can just reason from the definitions to find the fifth; alternatively you also have your choice of at least two equations which could be solved for the fifth. Most situations can be reasoned out from the definitions without using the equations. In reasoning you generally need to think in terms of the quantities vAve and `dv, in addition to the five quantities represented in the equations. Graphs can be helpful in the reasoning process. To fully understand uniformly accelerated motion you need to be able to reason in terms of the definitions, and you need to be able to use the equations. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: Accelerating down an incline through a given distance vs. accelerating for a given time If we accelerate down a constant incline for `dt seconds, starting at some initial velocity, then repeat the process, accelerating for `dt second but with another initial velocity, the change `dv in velocity will be the same for both trials. If we accelerate through displacement `ds on a constant incline, starting at some initial velocity, then repeat the process, accelerating through displacement `ds but with another initial velocity, the change `dv in velocity will be different for the two trials. 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: In the fist situation we are dealing with three variables a, `dt, and `dv. These three are all related in the equation a=`dv/`dt . If as we are told in the problem that a and `dt are kept constant than so shall `dv be kept constant. In this case the final velocity will increase or decrease the same way the intitial velocity increases or decreases thus keep `dv constant. In the second situation `ds is not related to `dv and a in the same way `dt was. Thus we can have a `dv that varies. A change in intitial velocity will affect how much time the object will be travelling and how long it has to be accelerating. A high intitial velocity affords less time on the incline to increase velocity and therefore will result in a lower change in velocity. A lower initial velocity will allow more time on the incline and therefore will result in more time for acceleration and a greater change in velocity. 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: With trapezoid we have a linear slope of `dv/`dt so we know it is uniformly accelerating. The area of the graph underneath the trapezoid represnts `ds and this is found from vAVe * `dt or the average altitude times width. `ds = vAve *`dt But we need to use the terms in the equations of uniformly accelerated motion so we substitute (v0+vf)/2 for vAVE `ds = (v0+vf)/2 * `dt For this same trapezoid a = `dv/`dt which is rise/run or slope. We need to express `dv in the terms used in the equations of uniformly accelerated motion so we substitute vf-v0 for `dv a = (vf-v0)/`dt since the second equation of uniformly accelerated motion isolates vf, we will do the same. a*`dt = vf-v0 a*`dt +v0 = vf Reverse this and we get vf = v0 + a*`dt 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: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. STUDENT QUESTION Using the given equation s(t) = .3 m/s^3 * t^3 - 2 m/s^2 * t^2 + 5 m/s * t + 12 m the velocity and acceleration functions are found my taking the derivative of the above expression giving me (.3 m/s^3) * (3 t^2) = .9 m/s^3 * t^2. This is all I know to do?? Im stuck here!
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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. STUDENT QUESTION Using the given equation s(t) = .3 m/s^3 * t^3 - 2 m/s^2 * t^2 + 5 m/s * t + 12 m the velocity and acceleration functions are found my taking the derivative of the above expression giving me (.3 m/s^3) * (3 t^2) = .9 m/s^3 * t^2. This is all I know to do?? Im stuck here!