Please Show work and explain. Given the acceleration, initial velocity, and init

Question

Please Show work and explain.

Given the acceleration, initial velocity, and initial portion of a body moving along a coordinate line at time t, find the body's position at time t. a = 18 cos 5t, v(0) = 10, s(0) = -9 s = 18 / 25 cos 5t - 10t - 9 s = - 18 / 25 sin 5t + 10t - 9 s = - 18 / 25 cos 5t + 10t - 9 s = 18 / 25 sin 5t + 10t - 9 Use the substitution formula to evaluate the integral. sin theta d theta / 2 + cos theta -ln 2 ln 3 0 -ln 3 Use a definite integral to find an expression that represents the area of the region between the given curve and the x-axis on the interval [0, b]. y = 6x2 -12b3 12b 2b3 -12b y = 24pix2 -8pib3 48pib 8pib3 -48pib y = x / 4 + 8 b / 4 + 8 -b / 4 - 8 - b2 / 8 - 8b b2 / 8 + 8b Find the area of the shaded region. 937 / 12 81 / 12 343 / 12 768 / 12 1153 12 937 / 12 343 / 12 -343 / 12

Explanation / Answer

28)

a(t) = 18cos(5t)

v(t) = int a(t) dt = 18/5 sin(5t) + V(0) = 3.6 sin(5t) + 10

s(t) = int v(t) dt = -3.6/5 cos(5t) + 10t + C = -0.72 cos(5t) + 10t + s(0)

Therefore:

s(t) = -0.72 cos(5t) + 10t - 9 -> answer C.

29)

Use the substitution:

2+cos(t) = u

-> -sin(t) dt = du

Therefore:

int sin(t)/(2+cos(t)) dt =

int -1/u du =

-ln(u) =

-ln(2 + cos(t))

Therefore the result is:

-ln(2 + cos(3pi/2)) - (-ln(2+cos(pi)) =

-ln(2) - 0 =

-ln(2)

answer A

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