#$&* course PHY 201 6/1/2012 10:30 Samantha RogersPHY 201
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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 andQf = 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: We know vAve by (vf+v0)/2 and since we know `dt then we can solve for `ds by multiplying `dt and vAve. 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): ok ------------------------------------------------ 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: we obtain vAve from (vf+v0)/2 when we multiply `dt by vAve then we get `ds. We can also find `dv from vf-v0=`dv. Then we are able to find acceleration which is `dv/`dt. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The first level in the diagram would contain `dt, v0 and vf. From v0 and vf we can easily reason out `dv, so v0 and vf would connect to `dv in the second level. The second level would also contain vAve, also obtained from v0 and vf and therefore connected from vf in the first level to v0 in the first level. The third level would contain an a, which is reasoned out from `dv and `dt and so is connected to `dv in the second level and `dt in the first level. The third level would also contain `ds, which follows from vAve and `dt and is therefore connected to vAve in the second level and `dt in the first level. ** STUDENT QUESTION: I'm not sure what is meant by a flow diagram. I know that we can determine 'ds from the equation 'ds=(v0+vf)/2* 'dt. Then I can use 'ds to find other possible information by plugging this and other information into other equations. INSTRUCTOR RESPONSE The instructor's response developed into an entire document, a bit too long to include in this query without interrupting the flow. The document has been posted at http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/flow_diagrams.htm and should be very useful to anyone who is having trouble with the idea of flow diagrams. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I completely understand the flow chart but I did not describe them very clearly. I also didn't use levels. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qDescribe the flow diagram we obtain for the situation in which we know v0, vf and `dt. Your solution: this is the same questions as before… confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: The trapezoid is the initial velocity to the final velocity, then down to the horizontal axis, and straight back to the beginning. vf-v0. this is the difference in the final velocity and the initial velocity. This is represented by the rise in the slope The aspect of the graph that represents the change in clock time is the width of the trapezoid, or the run of the slope. The slope for the graph represents the acceleration for the interval because it is change in velocity / clock time. The area of the graph represents the displacement and can be found by using the average velocity multiplied by the time interval. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `qPrinciples of Physics and General College Physics Students: Prob. 1.26: Estimate how long it would take a runner at 10 km / hr to run from New York to California. Explain your solution thoroughly. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: distance between NY and Cali is about 300 miles. 1 km = 0.6214 mi 300mi * km/0.6214mi = 4827.8 km `dt = `ds / vAve 4827.8 km / 10km/h = 482.7 hr or 4.8 *10^2 confidence rating #$&*: 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): I didn't estimate enough. My numbers were more exact. ------------------------------------------------ Self-critique Rating: ok