# ch02-p098 - 98. We obtain the velocity by integration of...

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98. We obtain the velocity by integration of the acceleration: 0 0 (6.1 1.2 ') ' t vv td t −= ³ . Lengths are in meters and times are in seconds. The student is encouraged to look at the discussion in the textbook in §2-7 to better understand the manipulations here. (a) The result of the above calculation is 2 0 = + 6.1 0.6 , t t where the problem states that v 0 = 2.7 m/s. The maximum of this function is found by knowing when its derivative (the acceleration) is zero ( a = 0 when t = 6.1/1.2 = 5.1 s) and plugging that value of t into the velocity equation above. Thus, we find 18 m/s v = . (b) We integrate again to find x as a function of t : 22 3 00 0 ( 6.1 0.6 ) 3.05 0.2 tt x xv d t v t t d t v t t t ′′ = + = + ³³ . With x 0 = 7.3 m, we obtain x = 83 m for t = 6. This is the correct answer, but one has the right to worry that it might not be; after all, the problem asks for the total distance
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## This note was uploaded on 09/03/2009 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

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