course PHY 201 Question: `qIntro Prob 6 given init vel, accel, `dt find final vel, dist
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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):
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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): ------------------------------------------------ Self-critique Rating: ********************************************* 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: Given v0, a, and `dt, you can easily find the final velocity. First you must find `dv, which can be done by using the formula a = `dv/`dt So a * `dt = `dv This means that v0+ dv = vf confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `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: ok ********************************************* Question: `qDescribe the flow diagram we obtain for the situation in which we know v0, vf and `dt. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 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): ------------------------------------------------ Self-critique 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: vAve = `ds / `dt vAve * dt = ds dt = ds / vAve dt = 3884km / 10km per hour dt = 388.4 hours I googled the distance from California to New York, which was problem measured from the inward most borders of the states. This is why I believe my answer was different from yours, since you said you used the distance from coast to coast. 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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: 65 beats in 1 minute * 60 minutes * 24 hours * 365 days * 85 years = 2.9 million beats in a lifetime. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: 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: 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. ** "