Unformatted text preview: 2 / 2+ C . Let’s suppose that at time t = 0 the object is at position 0, so s ( t ) =t 3 / 3 + 5 t 2 / 2; this function is also pictured in ﬁgure 7.2 . Between t = 0 and t = 5 the velocity is positive, so the object moves away from the starting point, until it is a bit past position 20. Then the velocity becomes negative and the object moves back toward its starting point. The position of the object at t = 5 is exactly s (5) = 125 / 6, and at t = 6 it is s (6) = 18. The total distance traveled by the object is therefore 125 / 6 + (125 / 618) = 71 / 3 ≈ 23 . 7....
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 Spring '07
 JonathanRogawski
 Math, Calculus, Antiderivatives, Derivative, Fundamental Theorem Of Calculus, Integrals, dt, bit past position

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