course PHY 121 7/8/09 10:15 pm If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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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: ‘dt a v0 ‘dv vf vAve ‘ds Confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 3 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: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: You can use v0 and vf to find vAve and ‘dv. You can use vAve and ‘dt to find ds. And you can use ‘dt and ‘dv to find a. 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: The third fundamental equation of motion is `ds = v0 `dt + .5 a `dt^2 Since you know v0, a, and ‘dt use can plug in those variables to find ‘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): I believed I had answered the question, however, after seeing the solution I am not sure. The question was asking about the third equation, however, the answer looks as though it incorporates all the equartions. Self-critique Rating:2 ********************************************* Question: Why do we think in terms of seven fundamental quantities while we model uniformly accelerated motion in terms of five? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The four equations incorporate the five quantities. Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 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: If the incline is constant then the change in velocity is the same. But if the distance travled changes then the velocity will change. 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): ok 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 altitude of the trapezoid is the vAve. The area that the trapezoid covers is the displacement during the interval. 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: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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: "