course Phy 201 ≿f̏Nmassignment #007
......!!!!!!!!...................................
23:38:29 Describe the flow diagram you would use for the uniform acceleration situation in which you are given v0, vf, and `dt.
......!!!!!!!!...................................
RESPONSE --> v0, vf, and `dt are on the first level v0 and vf are used to find `dv and vAve on the second level. vAve and `dt are used to fin `ds on the third level. `dv and `dt are used to find aAve on the third level. confidence assessment: 2
.................................................
......!!!!!!!!...................................
23:39:02 ** We start with v0, vf and `dt on the first line of the diagram. We use vO 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. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
23:42:54 Describe the flow diagram you would use for the uniform acceleration situation in which you are given `dt, a, v0
......!!!!!!!!...................................
RESPONSE --> `dt, a, and v0 are on the 1st level. We can use these three in a uniform accleration problem to find `ds and vf, which both go on the 2nd level. confidence assessment: 2
.................................................
......!!!!!!!!...................................
23:44:02 ** Student Solution: Using 'dt and a, find 'dv. Using 'dv and v0, find vf. Using vf and vO, find vave. Using 'dt and Vave, find 'ds. **
......!!!!!!!!...................................
RESPONSE --> Does my way work? I just assumed we had to use the uniform acceleration equations for this one, even though its much easier just to think about it this way for me. self critique assessment: 2
.................................................
......!!!!!!!!...................................
23:46:46 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.
......!!!!!!!!...................................
RESPONSE --> We can find out average velocity by (vf-vo) / 2. We can find out average acceleration by dividing vAve/`dt. confidence assessment: 1
.................................................
......!!!!!!!!...................................
23:47:14 **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. **
......!!!!!!!!...................................
RESPONSE --> ok, this is obvious self critique assessment: 3
.................................................
......!!!!!!!!...................................
23:56:23 qaExplain 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.
......!!!!!!!!...................................
RESPONSE --> We plug in v0, a and `dt to find `ds in the equation `ds = (v0)(`dt)+.5(a)(`dt^2) self critique assessment: 1
.................................................
......!!!!!!!!...................................
23:56:34 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
23:57:31 Why do we think in terms of seven fundamental quantities while we model uniformly accelerated motion in terms of five?
......!!!!!!!!...................................
RESPONSE --> The others (vAve and `dv) are simple to figure out based on the 5 of v0, vf, `dt, `ds and a. confidence assessment: 2
.................................................
......!!!!!!!!...................................
23:57:44 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
00:21:57 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 accelerated down a constant incline for the same time, but not when we accelerated down the same incline for a constant distance?
......!!!!!!!!...................................
RESPONSE --> The change in v will be constant when we accelerate for the same time because a `dt = vf - v0. The change in v will not be constant when we accerlerate for a constant distance becase a `ds = (vf^2 - v0^2) / 2. The wording of this question is very confusing for someone like me. I had to look at the notes and guess what question you were asking. confidence assessment: 0
.................................................
......!!!!!!!!...................................
00:22:42 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok this part makes sense: 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. self critique assessment: 2
................................................."