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For an object whose velocity in ft/sec is given by v(t) = −2t^2 + 4, what is its distance travelled, in feet, on the interval t = 0 to t = 2 secs?
I think the answer is 2.667 but I am not one hundred percent sure.
8 years ago

Answers

Steve
the distance is

s = ∫[0,2] -2t^2 + 4 dt
= -2/3 t^3 + 4t [0,2]
= -16/3 + 8
= 8/3

you are correct
8 years ago

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