Expert: Sherman D. - 8/1/2007

Question
A particle moves along the x-axis so that at any time t its position is given by x(t)=(1/2)sint+cos(2t).  What is the acceleration of the particle at t=(pi/2)

Answer
assuming that x(t) is the acceleration, because i'm not good a word problems.

x(pi/2) = (1/2)sin(pi/2) + cos(2(pi/2))
x(pi/2) = (sin(pi/2)/2) + cos(pi)
x(pi/2) = (1/2) + (-1)
x(pi/2) = (1/2) - (2/2)
x(pi/2) = (1 - 2)/2
x(pi/2) = (-1/2)

but if not, then

x(t) = (1/2)sin(t) + cos(2t)
x(t)' = (sin(t)/2)' + cos(2t)'
x(t)' = ((2sin(t)' - 2'sin(t))/4) + (cos^2(x) - sin^2(x))'
x(t)' = (2/4)cos(x) + (cos(x)cos(x) - sin(x)sin(x))
x(t)' = (cos(x)/2) + (-2sin(x)cos(x) - 2sin(x)cos(x))
x(t)' = cos(x)/2) - 4sin(2x)
x(t)' = (cos(x) - 4sin(2x))/2
x(pi/2)' = (cos(pi/2) - 4sin(2pi/2))/2
x(pi/2)' = (0 - 4sin(pi))/2
x(pi/2)' = (0 - 4(0))/2
x(pi/2)' = (0/2)
x(pi/2)' = 0

i believe that would be for the velocity of the particle

x(t)' = cos(x) - 4sin(2x)
x(t)" = cos(x)' - (4sin(2x))'
x(t)" = -sin(x) - (4sin(2x)' + 4'sin(2x))
x(t)" = -sin(x) - 4sin(2x)'
x(t)" = -sin(x) - 4(sin(x)cos(x))'
x(t)" = -sin(x) - 4(sin(x)'cos(x) + sin(x)cos(x)')
x(t)" = -sin(x) - 4(cos(x)cos(x) + sin(x)(-sin(x)))
x(t)" = -sin(x) - 4(cos(x)^2 - sin(x)^2)
x(t)" = -sin(x) - 4cos(2x)
x(pi/2)" = -sin(pi/2) - 4cos(2(pi/2))
x(pi/2)" = -1 - 4cos(pi)
x(pi/2)" = -1 - 4(-1)
x(pi/2)" = -1 + 4
x(pi/2)" = 3

ANS : 3