calculus


3: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?
a:4.876 b:2.667 c:9.752 d:2.438

asked by Anonymous
  1. then the distance formula is

    s = (-2/3)t^3 + 4t + c, where c is a constant

    at t = 0, s = c
    at t = 2, s = -16/3 + 8 + c

    distance in the given interval
    = -16/3 + 8 + c - c
    = 8/3 which is close to choice b)

    posted by Reiny
  2. ^^^thats wrong

    posted by felipe correa

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