Math
posted by jof .
The position of a car piston within the cylinder is described by the equation:
y = S sin(360
r t  90
Where S is the stroke length of the cylinder, r is the rpm of the crank shaft, and t is
the time in minutes. If the stroke length is 2 inches and the crank shaft is running at
1300 rpm, find the position of the piston at t = 53 seconds. (remember that rpm
means revolutions per minute)

what's the problem?
y(t) = S sin(360rt90)
S = 2
r = 1300
t = 53/60
y(53/60) = 2sin(360*1300*53/6090)
= 2sin(1148.3333*360  90)
now, we can toss out all multiples of 360, so we are left with
y = 2sin(24090) = 2sin(150) = 2(.5) = 1.0 
oops.
y = 2sin(12090) = 2sin(30) = 2(.5) = 1.0
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