A buoy floating in the ocean is bobbing in simple harmonic motion with period 5 seconds and amplitude 3ft . Its displacement d from sea level at time t=0 seconds is −3ft , and initially it moves upward. (Note that upward is the positive direction.) Give the equation modeling the displacement d as a function of time t .

d = 3sin(2π/5 t + 3π/2)

Step-by-step explanation:

Equation for simple harmonic motion is:

d = A sin(2π/T t + B) + C

where A is the amplitude,

T is the period,

B is the horizontal shift (phase shift),

and C is the vertical shift.  

Given that A = 3, T = 5, and C = 0:

d = 3 sin(2π/5 t + B)

At t = 0, the buoy is at d = -3:

-3 = 3 sin(2π/5 (0) + B)

-1 = sin(B)

3π/2 = B  

d = 3sin(2π/5 t + 3π/2)

Notice you can also use cosine instead of sine and get a different phase shift.

d = 3 cos(2π/5 t + π)

You can even use phase shift properties to simplify:

d = -3 cos(2π/5 t)

Any of these answers are correct.

g = (2, 2)

step-by-step explanation:

given the ratio is 1 : 1 then f is the midpoint of eg

let the coordinates of g = (x, y)

using the midpoint formula

(0 + x) = 1 ( multiply both sides by 2 )

0 + x = 2 ⇒ x = 2

similarly

(4 + y) = 3 ( multiply both sides by 2 )

4 + y = 6 ( subtract 4 from both sides )

y = 2

coordinates of g = (2, 2)