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need help refreshing my memory - calculus. i already did the work just need clarification please?
The acceleration of an object is given by a(t) = 6sint with initial velocity of -9.5. Find the distance the object travels on the interval [0, pi] to the nearest integer.
So first I integrated a(t) then plugged in -9.5 to solve for C for the v(t) function. Then I integrated from 0 to pi to find the position but i keep getting 11. I don't know if this is the correct answer or not because each answer can only be used once and the other problem I'm 100% sure it's 11. Is this problem like a tricky one where you gotta do something first?
Also what is meant when it says find the smallest positive integer in the domain of f(x) = sin^2x / sqrt (x^2 - 28x - 29) ? Am I supposed to take the limit as x approaches 0 or infinite?
1 Answer
- Scrander berryLv 71 decade agoFavorite Answer
a(t) = 6sin(t)
v(t) = - 6cos(t) + C
-9.5 = -6cos(0) + C
-9.5 = -6 + C
C = -3.5
v(t) = - 6cos(t) - 3.5
s(t) [0 to π] = -6 sin(t) - 3.5t [0 to π]
= (-6 sin(π) - 3.5π) - (-6 sin(0) - 3.5(0))
= -3.5π
f(x) = sin²x / √(x² - 28x - 29)
find the domain
x² - 28x - 29 > 0 .. (positive parabola so positive outside the roots)
(x + 1)(x - 29) > 0
x < -1 or x > 29
smallest positive integer in that domain is 30
