course PHY 121 7/7/09 11:45 pm If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qWhat is the displacement `ds associated with uniform acceleration from velocity v0 to velocity vf in clock time `dt? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If acceleration is uniform then vAve = (v0 + vf) /2. Once you know vAve then you can find displacement by using ‘ds = vAve * ‘dt. Confidence assessment 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ‘dt v0 vf ‘dv vAve a ‘ds Confidence assessment 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 fourth level and `dt in the first level. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qIntro Prob 6 Intro Prob 6 How do you find final velocity and displacement given initial velocity, acceleration and time interval? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ‘dv = a * ‘dt vf = ‘dv + v0 Confidence assessment rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 3 `a** To find final velocity from the given quantities initial velocity, acceleration and `dt: • Multiply `dt by accel to get `dv. • Then add change in velocity `dv to init vel , and you have the final velocity** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `qDescribe the flow diagram we obtain for the situation in which we know v0, vf and `dt. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ‘dt v0 vf ‘dv vAve a ‘ds Confidence assessment rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: 3 ********************************************* 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: There is around 3000 miles between New York and California. 1km = .62 miles. 3000 miles / .62 = 4839 km. Take 4839 km / 10 km/hr, since the runner is running 10 km/hr. 4839 km / 10 km/hr = 484 hrs is how long it would take the runner to make it to California. Confidence assessment 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating:3 ********************************************* 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: If you measure a person’s heart rate and if it is 80 heart beats per mintues and a person live to age 80 you then have to think about time. 60 minutes = 1 hour 24 hours = 1,440 minutes 365 days = 525,600 minutes 80 years = 42,048,000 minutes 42,048,000 mintues * 80 heart beats = 3,363,840,000 beat / minute
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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): ok Self-critique Rating:3 ********************************************* 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 assessment 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: Add comments on any surprises or insights you experienced as a result of this assignment. ** I had to get a little help from a friend on vectors, but now I think I understand them. They are not as difficult to deal with as I thought. ** "