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15:35:42 Intro Prob 6 Intro Prob 6 How do you find final velocity and displacement given initial velocity, acceleration and time interval?

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RESPONSE --> To find the displacement, multiply change in time by the acceleration. To find the final velocity, add the displacement to the initial velocity.

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15:35:48 ** 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**

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RESPONSE --> ok

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15:40:38 Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.

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RESPONSE --> Given v0 anf vf, you can find `dv (you can also get average velocity from v0 and vf). With average velocity and `dt, you can find displacement. From `dv and `dt, yo can find acceleration. (With the associated lines running from each value.)

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15:40:50 ** The flow diagram shows us the flow of information, what we get from what, usually by combining two quantites 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. **

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RESPONSE --> ok

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15:56:55 Principles 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.

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RESPONSE --> The internet claims there is about 3000 miles from New York to California (coast to coast). A km is equal to around .62 miles. 3000 miles * 1 km / (.62 miles) = 4839 km At 10 km / hr, the time required would be 4839 km / (10 km / hr) = 489 km / (km/hr) = 489 hr.

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15:58:05 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.

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RESPONSE --> I did not round. And, I couuld never run that far!

You never know your limits until you try, and it's been done (not all on the same day), but very few people could hold up from coast to coast.

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16:03:35 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?

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RESPONSE --> My average heart rate is 65 beats/min. If I lived a lifetime ~80 + years (I hope), then 65 beats/min * 60 min/hour =3900 beats/hour * 24 hr/day =93600 beats/hr * 365 days/yr =34164000 * ~80 + years =2733120000 heartbeats/lifetime

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16:03:44 ** 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, approximately. **

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RESPONSE --> ok

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16:03:52 University Physics Students Only: Problem 1.52 (i.e., Chapter 1, Problem 52): Angle between -2i+6j and 2i - 3j. What angle did you obtain between the two vectors?

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RESPONSE --> n/a

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16:03:59 ** 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.. **

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RESPONSE --> n/a

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16:05:03 Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE --> Flow diagrams confuse me, but I think I've got them figured out.

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16:05:15 ** 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. **

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RESPONSE --> ok

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Your work on this assignment is very good. Let me know if you have questions.