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Calculus (math)

The velocity function (in meters per second) for a particle moving along a line is given by v(t)=t3−5t2. Find the displacement and the distance traveled by the particle during the time interval [-1,6].

Distance traveled = ??

I got 2089/12m as answer but it was incorrect please help me!

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  1. s(t) = 1/4 t^3 - 5/3 t^3 + c
    The displacement is s(6)-s(-1)

    The distance traveled is the arc length. So, what steps did you follow?

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  2. if v(t) = t^3 - 5t^2
    s(t) = (1/4)t^4 - (5/3)t^3 + c

    s(6) = (1/4)6^4 - (5/3)6^3 = -36 + c
    s(-1) = (1/4)(-1)^4 - (5/3)(-1)^3 = 23/12 + c

    distance traveled = 23/12+c - (-36+c) = 4550/12

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