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

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

^^^thats wrong
posted by felipe correa
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