#$&* course PHY 121 6/19 1 007. `query 7
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Given Solution: We start with v0, vf and `dt on the first line of the diagram. We use v0 and vf to find Vave, indicated by lines from v0 and vf to vAve. Use Vave and 'dt to find 'ds, indicated by lines from vAve and `dt to `ds. Then use `dv and 'dt to find acceleration, indicated by lines from vAve and `dt to a. ** STUDENT COMMENT i dont understand how you answer matches up with the question INSTRUCTOR RESPONSE All quantities are found from basic definitions where possible; where this is possible each new quantity will be the result of two other quantities whose value was either given or has already been determined. Using 'dt and a, find 'dv (since a = `dv / `dt, we have `dv = a `dt). Using 'dv and v0, find vf, indicated by lines from `dv and v0 to vf (vf = v0 + `dv). Using vf and v0, find vAve, indicated by lines from vf and v0 to vAve ( (vf + v0) / 2 = vAve, for uniform acceleration). Using 'dt and vAve, find 'ds, indicated by lines from `dt and vAve to `ds (vAve = `ds / `dt so `ds = vAve * `dt). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: Describe the flow diagram you would use for the uniform acceleration situation in which you are given `dt, a, v0 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The first line would have `dt, a and v0. Since I know the elapsed time and acceleration, I would use that to find the change in velocity on the next line. `dt and a would attach to `dv. Then I would use the change in velocity and the initial velocity to calculate the final velocity. `dv and v0 would attach to fv. Next I would use v0 and vf to find the vAve. Finally, I would use the vAve and `dt to find `ds. confidence rating #$&*: 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. 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): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: Check out the link flow_diagrams and give a synopsis of what you see there. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I was able to open them this time. I have printed them off so that I can use them as a reference. I have seven pages explaining how to thing through what you know and how to find what you don't know. It details where to start and how to build the levels based on what you can find out from what you know. It shows how to build each of the three levels of the diagram. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: You should have seen a detailed explanation of a flow diagram, and your 'solution' should have described the page. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: Explain in detail how the flow diagram for the situation in which v0, vf and `dt are known gives us the two most fundamental equations of motion. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using v0, vf, and `dt,you can find `ds (displacement of the object) You do this by adding v0 and vf and then dividing the sum by 2 to find the average of the velocity. Then you multiply that by the elapsed time (`dt) to find how far the object moved. You can also use it to find acceleration. To do this, you subtract vf-v0 and then divide the difference by the elapsed time. That gives you a = (vf-v0)/`dt. To work it into the equation, you would have to do the following: (vf-v0)/`dt = a vf - v0 = a`dt vf = v0 + a`dt The two equations are: `ds=(v0+vf)/2 * `dt vf = v0 + a * `dt confidence rating #$&*: 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: `dt and a can find `dv v0 and `dv can find vf v0 and vf can find vAve vAve and dt can find `ds The equation is: `ds = v0`dt + .5a`dt^2 You would find it as follows: `ds = vAve * `dt vAve can be found by averaging the intial velocity and the final velocity. You can find the final velocity by finding the change in velocity and adding it to the initial velocity. You find the change in velocity by multiplying the acceleration by the time elapsed. This means that vAve actually is v0 + v0 +a*`dt all divided by 2. If you take all of this and plug it into the formula, you will get: (v0 + v0 + a*`dt)/2 * `dt. This will give you (2v0 + a`dt)/2 * `dt. You can change this into (2v0/2 + a`dt/2)*`dt. The 2v0/2 becomes v0, and the a`dt/2 can become .05a`dt. This gives you (v0 + 0.5a`dt)*`dt. Using the distributive property, you will get `ds = v0`dt + 0.5a`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 took me a while but I figured it out!! ------------------------------------------------ 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: You and find the average velocity and the change in velocity using both intial velocity (v0) and final velocity (vf). Therefore, instead of putting vAve into a formula, you could substitute (v0+vf)/2. Instead of putting `dv into a formula, you could replace it with (vf-v0). That kind of makes vAve and `dv shortcuts, but the other qualities don't have those kinds of simple substitutions that will always work. 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. ** 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: 3 ********************************************* 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: Because the incline stays the same, it will change the velocity of the object in the same amount for the first example, when measuring time. What is changing is the distance traveled. When you are measuring for distance, the object can travel farther at a higher velocity, so the change in velocity is different. confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I did better than I thought. The incline is what affects the rate of change of velocity. So, if you time 30 seconds at one initial velocity and 30 seconds at another initial velocity, the change in velocity will stay the same. However, measuring the distance is going to change the `dv because at a higher velocity, the object will be able to gain more speed and will have more space to do it. ------------------------------------------------ Self-critique Rating: 2 ********************************************* 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 first equation is `ds = (v0+vf)/2 * `dt v0 is the initial velocity, which is the y coordinate of the first point plotted. vf is the final velocity, which is the y coordinate of the final point plotted. 'dt is the amount of time between x2 and x1 of those same points. The `ds is the area of the trapezoid. The second equation is vf = v0 + a*`dt In this, v0 is the y coordinate of the first point. `dt is the amount of time that is elapsed, as determined by x2-x1. Acceleration is the change in velocity divided by the change in time. This is shown by the difference y2-y1. Acceleration is `dv/`dt, so the `dt will cancel out and you will have initial velocity plus the change in velocity, which will be a triangle that you can create by making at horizontal line at x1. 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 ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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!