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Mathematics 14 Online
OpenStudy (anonymous):

A particle moves along a straight line and its position at time t is given by s(t) = 2t^3-21t^2+60t where s is measured in feet and t in seconds. What is the position of the particle at time 14? I know that the derivative is 6t^2-42t+60 which equals teh velocity in (ft/sec). But, I don't know how to find what the position is at time 14. Please help

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