sol03 - ( 29 ( 29 75 mph 2 h 150 miles d s t d st = = = =...

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PHYSICS 103 Chapter 2 -- HW #3 Questions 20. The information that we are given allows us to calculate the marathoner’s average speed. However, the average speed does not tell us anything about the details of the run. Therefore, we cannot determine how fast he was running when he passed the 10-mile marker. 24. We are being given a velocity because both the speed (65 mph) and the direction (east) are given. Exercise 10. Because the definition of average speed involves the distance traveled and the elapsed time, we need to use the given information to find the distance traveled from 10 am until noon.
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Unformatted text preview: ( 29 ( 29 75 mph 2 h 150 miles d s t d st = = = = We can now use this distance to calculate the average speed between 9 am and noon. 150 miles 50 mph 3 h d s t = = = Note that this is not equal to the average of the two average speeds; that is, the average of 0 mph and 150 mph, or 75 mph. 14. We begin with our definition of average speed and manipulate it algebraically to find the unknown, time in this case. Then we substitute in the numerical values to obtain our answer. 500 miles 4 h 125 miles h d s t d t s = = = =...
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This document was uploaded on 10/29/2011 for the course PHSX 103 at MSU Bozeman.

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