The waterwheel rotates at 5 revolutions per minute. You start your stopwatch. Two seconds later, Point P on the rim is at its lowest point. Assume that y varies sinusoidally with t, where y is the distance of point P from the surface of the water in terms of the number of + seconds that the stopwatch reads.
A. Write an equation: y=
B. Predict the distance from the surface of the water when the stopwatch reads 17.2 seconds
C. At 40 seconds is the point on the wheel going into the water or coming out?
Bot
answered
B. y = 8 - 10sin(5Ο(17.2)) = 8 - 10sin(86Ο) β 8 - 10(-0.99) β 18.9
C. At 40 seconds, the point on the wheel is coming out of the water.
I Know
answered
we want everything to be in seconds, so for 5 rotations/min --> period = 12 s
so 12=2Ο/k ----> k = Ο/6
so the equation will be something like
y = 8-10sin(Ο/6 t)
now i need the min to be at t = 0
final equation:
y = 8 - 10sin(Ο/6(t + 3)
b) at a time 17.2 s, t = 17
y = 8 - 10sin(Ο/6(17+3) = 16.66 ft
c) I graphed my correct equation on DESMOs and all looks good
at 40 seconds(38 on my graph) y = 3 , and the graph is rising,
so it just came out of the water
