98. We obtain the velocity by integration of the acceleration: 00(6.1 1.2 ')'tvvtdt−=−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 20= + 6.10.6 ,tt−where the problem states that v0= 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 tinto the velocity equation above. Thus, we find 18 m/sv=.(b) We integrate again to find xas a function of t:223000(6.10.6)3.050.2ttxxvdtvttdtvttt′′′′=+−=+−³³.With x0= 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.