# Advanced Functions Help!?

A large wheel is attached to a boat and spins as the boat moves. A rock becomes nudged in the wheel as it spins in the water. It is noticed that at t = 2 s, the rock is at the highest point 3 m above the water. At time t = 6 seconds, the rock is submerged in the water 5 m below the water(the lowest point).

a. Graph 5 points to represent one cycle of the above problem. Label these points clearly on your graph.

b. Determine a cosine model to represent the motion of the wheel.

c. Determine whether the rock was above the water’s surface at t = 0 or below the water’s surface and how far is the rock above or below the water at this time?

### 2 Answers

- NCSLv 74 weeks agoFavourite answer
I see it this way:

a. Your five points should be (y, t) such that

(3, 2), (-5, 6), (3, 10) -- for crest, trough, crest (antinodes)

and then (-1, 4) and (-1, 8) at the nodes

where y = distance above surface in meters

and t = time in seconds

b. y = 4cos(2πt/8s - φ) - 1

since the period is 8 seconds (4 s to go from crest to trough)

At t = 2, y = 3 --

3 = 4cos(2π*2/8 - φ) - 1

1 = cos(π/2 - φ)

0 = π/2 - φ

φ = π/2

so your model is

y = 4cos(πt/4 - π/2) - 1

c. y = 4cos(0 - π/2) - 1 = -1 m

1 meter below the surface

- jacob sLv 74 weeks ago
given,

a large wheel is attached to a boat as the boat moves

a rock becomes nudged in the wheel as it spins in the water

at t = 2 , rock is at highest point 3 m above water

at t = 6 s, rock is submerged in the water 5 m below water

a. let angular speed of the wheel be w and let w be constant

then

angle made by the rock with vertical be theta

then

theta = 0 at t= 2 s and theta = pi at t = 6 s

hence

4*w = pi

w = pi/4 rad/s

hence

at t = 2, theta = 0

at t = 3, theta = w*1 = pi/4 rad

at t = 4, theta = pi/2 rad

at t = 5, theta = 3*pi/4 rad

at t = 6, theta = pi rad

coordinates of points are

4(sin(0),cos(0)),4 (sin(pi/4),cos(pi/4), 4(sin(pi/2),cos(pi/2))), 4(sin(3*pi/4), cos(3*pi/4)), 4(sin(pi), cos(pi))

b. let distance above the water be y

then following cosine model is assumed

y = A*cos(wt + phi)

so, y(2) = A*cos(pi/2 + phi) = 3

y(6) = A*cos(3*pi/2 + phi) = -5

now

A*cos(pi/2 + phi) = A(cos(pi/2)cos(phi) - sin(pi/2)sin(phi)) = -A*sin(phi)

A*cos(3*pi/2 + phi) = A*cos(-pi/2 + phi) = A[cos(3*pi/2)cos(phi) - sin(3*pi/2)sin(phi)] = A*sin(phi)

hence

A = 8/2 = 4

hence

y = 4*cos(pi*t/4 + phi)

for t = 2

y = 4*cos(pi/2 + phi) = 3

-4*sin(phi) = 3

phi = -0.84806207 rad

hence

y = 4*cos(pi*t/4 -0.848062079)

c. at t = 0

y = 4*cos(-0.848062079) > 0

hence the rock is above the water at t = 0 s

y(0) = 2.6457513110090336146 m

I owe you the graph .