Asked by Mike
The velocity of an object is given by v(t)=t(t^2-1)^(1/3) feet per second. Find the total number of feet traveled by the object in the time interval [0,3] seconds.
I understand how to solve the problem, but I do not understand why is the interval now from [0-1] [1-3], and not just [0-3]?
Thanks
I understand how to solve the problem, but I do not understand why is the interval now from [0-1] [1-3], and not just [0-3]?
Thanks
Answers
Answered by
Reiny
you have to integrate the velocity to get to distance
let the distance be s(t)
s(t) = (3/8)(t^2 - 1)^(4/3) + c
s(0) = (3/8)(1) + c = c + 3/8
s(3) = (3/8)(8)^(4/3) +c = c + 81/8
distance = c+81/8 - (c + 3/8) = 39/4 ft
let the distance be s(t)
s(t) = (3/8)(t^2 - 1)^(4/3) + c
s(0) = (3/8)(1) + c = c + 3/8
s(3) = (3/8)(8)^(4/3) +c = c + 81/8
distance = c+81/8 - (c + 3/8) = 39/4 ft
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