#$&* course Phy 201 2/26/13 151pm Query 5If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `a**You would use accel. and `dt to find `dv: a * `dt = `dv. Adding `dv to initial vel. v0 you get final vel. Then average initial vel. and final vel. to get ave. vel.: (v0 + vf) / 2 = ave. vel. You would then multiply ave. vel. and `dt together to get the displacement For example if a = 3 m/s^2, `dt = 5 s and v0 = 3 m/s: 3 m/s^2 * 5 s = 15 m/s = `dv 15 m/s + 3 m/s = 18 m/s = fin. vel. (18 m/s + 3 m/s) / 2 = 10.5 m/s = vAve 10.5 m/s * 5 s = 52.5 m = displacement In more abbreviated form: a * `dt = `dv v0 + `dv = vf (vf + v0) /2 = vAve vAve * `dt = `ds so `ds = (vf + v0) / 2 * `dt. ** STUDENT QUESTION If we have the formula vf= v0 + a * dt, then we would substract the v0 from both sides to isolate the a * dt algebraically, so our formula would be vf-v0= a* `dt, how is this in comparison to the initial velocity v0 + the change in velocity(dv) = to the final velocity(vf). If we multiply the acceleration(a) times time(dt) we find the change in velocity(dv).......we then add the initial to the change to find the final....... Why do we add the initial to the change in velocity to find the final? INSTRUCTOR RESPONSE The initial velocity is v0, the final velocity is vf, so the change in velocity is `dv = vf - v0. Thus your early result vf-v0= a* `dt shows that a * `dt is equal to `dv. In general the change in any quantity is equal to its final value minus its initial value. It follows immediately from this that if you add the change in the quantity to its original value, you get its final value. The following two statements say the same thing: statement 1: If the temperature starts at 20 degrees and ends up at 35 degrees then it changed by +15 degrees. statement 2: If the temperature starts at 20 degrees and changes by +15 degrees then it ends up at 35 degrees. We generalize this to the two symbolic statements If a quantity Q changes from Q0 to Qf then the change is `dQ = Qf - Q0. If a quantity Q starts out at Q0 and changes by `dQ, then it ends up at Qf. These statements can be expressed as two equations `dQ = Qf - Q0 and Qf = Q0 + `dQ These two equations are algebraically equivalent: you can get the second by adding Q0 to both sides of the first, or you can get the first by subtracting Q0 from both sides and reversing sides. A third equation also follows: Q0 = Qf - `dQ, which can be interpreted in terms of the preceding examples into obvious statements. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qWhat is the displacement `ds associated with uniform acceleration from velocity v0 to velocity vf in time interval `dt? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ds= v0*dt + .5a *dt^2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since accel is uniform vAve = (v0 + vf) / 2. Thus displacement is `ds = vAve * `dt = (v0 + vf) / 2 * `dt, which is the first equation of uniformly accelerated motion. ** STUDENT QUESTION I failed to make reference to uniformly accelerated motion. What exactly is the difference between uniformly accelerated motion and average acceleration??? Will we be asked to differentiate between the two for problems, or is this something we should be able to determine on our own easily??? INSTRUCTOR RESPONSE Uniformly accelerated motion is motion in which the acceleration is uniform, unchanging. If motion is uniformly accelerated, then the acceleration is constant, so the acceleration at any instant is equal to the average acceleration. If motion is uniformly accelerated, then since the slope of the velocity vs. clock time graph represents acceleration, the slope is constant; i.e., the graph is a straight line. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): For my answers I have been using the formulas given under idea 4. I do understand how to reason them out as well even though it is a little easier for me to do so when given numbers and units. ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I dont understand what you mean by flow diagram confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The first level in the diagram would contain `dt, v0 and vf. From v0 and vf we can easily reason out `dv, so v0 and vf would connect to `dv in the second level. The second level would also contain vAve, also obtained from v0 and vf and therefore connected from vf in the first level to v0 in the first level. The third level would contain an a, which is reasoned out from `dv and `dt and so is connected to `dv in the second level and `dt in the first level. The third level would also contain `ds, which follows from vAve and `dt and is therefore connected to vAve in the second level and `dt in the first level. ** STUDENT QUESTION: I'm not sure what is meant by a flow diagram. I know that we can determine 'ds from the equation 'ds=(v0+vf)/2* 'dt. Then I can use 'ds to find other possible information by plugging this and other information into other equations. INSTRUCTOR RESPONSE The instructor's response developed into an entire document, a bit too long to include in this query without interrupting the flow. The document has been posted at http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/flow_diagrams.htm and should be very useful to anyone who is having trouble with the idea of flow diagrams. STUDENT COMMENT Flow diagrams are useful in that they give us something to logically grind out. It's not enough to know that there are formulas to find variables. True learning is when a person can take whats given, twist it and manipulate it, and find other answers. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): The link did not work, but isnt this basically the same as manipulating variables in the equations for a desired outcome?
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Given Solution: `a** The flow diagram shows us the flow of information, what we get from what, usually by combining two quantities at a time. How we get each quantity may also be included. From vf and v0 we get `dv, shown by lines from vf and v0 at the top level to `dv. From vf and v0 we also get and vAve, shown by similar lines running from v0 and vf to vAve. Then from vAve and `dt we get `ds, with the accompanying lines indicating from vAve and `dt to `ds, while from `dv and `dt we get acceleration, indicated similarly. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: Suppose we have two points on a straight-line graph of velocity vs. clock time. How do we construct a trapezoid to represent the motion on the intervening interval? What aspect of the graph represents the change in velocity for the interval, and why? What aspect of the graph represents the change in clock time for the interval, and why? What aspect of the graph represents the acceleration for the interval, and why? What aspect of the graph represents the displacement for the given interval, and why? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Draw a line from each point to the x axis to construct a trapezoid The rise represents change in velocity The run of the graph represents change in clock time The slope of the graph represents acceleration The area of the trapezoid represents displacement confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `qPrinciples of Physics and General College Physics Students: Prob. 1.26: Estimate how long it would take a runner at 10 km / hr to run from New York to California. Explain your solution thoroughly. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Im going to guess it is about 3000 miles from New York to California. There are roughly 1.609 km in a mile. So 3000miles = 4827km. At 10km/hr it would take a runner about 482.7hours to run from NY to CA. 4827miles/ 10km/hr = 482.7hr confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: It is about 3000 miles from coast to coast. A km is about .62 mile, so 3000 miles * 1 km / (.62 miles) = 5000 km, approximately. At 10 km / hr, the time required would be 5000 km / (10 km / hr) = 500 km / (km/hr) = 500 km * (hr / km) = 500 (km / km) * hr = 500 hr. Be sure you understand the units of this calculation. Units should be used at every step of every calculation. The corresponding symbolic solution: vAve = `ds / `dt; we want to find `dt so we solve to get `dt = `ds / vAve. Substituting `ds = 5000 km and vAve = 10 km/hr we have `dt = 5000 km / (10 km/hr) = 500 hr. STUDENT SOLUTION (with some inconsistencies in units) The student's estimate of the distance was 4000 km, which is perfectly OK: To find out how much time it takes to travel this far, I took 4000 km and divided it by 10 km/h. This was set up as follows: 4000 km / 10 km This becomes 400 km * 1 hr Our kilometers cancel out and we are left with 400 hours to run from New York to California. INSTRUCTOR RESPONSE I would certainly accept your solution, with little or no penalty at the level of Phy 121. However your use of units does have some contradictions, and you will understand units better if you understand them: In the first place, 4000 km / (10 km) = 400, not 400 km. The km divide out. 400 represented the number of 10 km intervals in a 4000 km trip. Since average speed is 10 km/hr, meaning that a 10 km interval is covered each hour, it therefore takes about 400 hours to complete the trip. Note also that the calculation given in your solution as 400 km * 1 hr would be 400 km * hr, not the 400 hr you intend. Finally, to use the fact that v_Ave = `ds / `dt: The time to cover distance `ds at average speed v_Ave is `dt = `ds / v_Ave, and that the units of v_Ave are km / hr. So to be entirely correct, the correct calculation could read `dt = `ds / v_Ave = 4000 km / (10 km/hr) = 400 hr. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: All Students: Estimate the number heartbeats in a lifetime. What assumptions did you make to estimate the number of heartbeats in a human lifetime, and how did you obtain your final result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Heart beats per minute depends on fitness level, but I am going to guess that an average heart beats at 70 beats per minute. 70bpm *60minutes in an hour *24hours in a day *365 days in a year = 36792000 beats per year. If an average person lives to be 80 years old, then we would multiply 80 times 36792000 to get 2,943,360,000 beats in a lifetime. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Typical assumptions: At 70 heartbeats per minute, with a lifetime of 80 years, we have 70 beats / minute * 60 minutes/hour * 24 hours / day * 365 days / year * 80 years = 3 billion beats, approximately. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Interesting that we had the same estimations. I promise that I didnt look at the solution before doing the problem. ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: University Physics Students Only: Problem 1.55 (11th edition 1.52) (i.e., Chapter 1, Problem 52): Angle between -2i+6j and 2i - 3j. What angle did you obtain between the two vectors? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: For the given vectors we have dot product =-2 * 2 + 6 * (-3) = -22 magnitude of first vector = sqrt( (-2)^2 + 6^2) = sqrt(40) magnitude of second vector = sqrt( 2^2 + (-3)^2 ) = sqrt(13) Since dot product = magnitude of 1 st vector * magnitude of 2d vector * cos(theta) we have cos(theta) = dot product / (magnitude of 1 st vector * magnitude of 2d vector) so that theta = arccos [ dot product / (magnitude of 1 st vector * magnitude of 2d vector) ] = arccos[ -22 / ( sqrt(40) * sqrt(13) ) ] = arccos ( -.965) = 164 degrees, approx.. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: University Physics Students Only: Problem 1.55 (11th edition 1.52) (i.e., Chapter 1, Problem 52): Angle between -2i+6j and 2i - 3j. What angle did you obtain between the two vectors? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: For the given vectors we have dot product =-2 * 2 + 6 * (-3) = -22 magnitude of first vector = sqrt( (-2)^2 + 6^2) = sqrt(40) magnitude of second vector = sqrt( 2^2 + (-3)^2 ) = sqrt(13) Since dot product = magnitude of 1 st vector * magnitude of 2d vector * cos(theta) we have cos(theta) = dot product / (magnitude of 1 st vector * magnitude of 2d vector) so that theta = arccos [ dot product / (magnitude of 1 st vector * magnitude of 2d vector) ] = arccos[ -22 / ( sqrt(40) * sqrt(13) ) ] = arccos ( -.965) = 164 degrees, approx.. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: #*&!