The trigonometric function that models the distance (feet) of the rider from the camera as a function of time (seconds) is γ(t) = ωt, where ω is the angular velocity of merry-go-round.
Let, center of the merry-go-round is C and camera is placed at point A. B(t) define the position of the rider at any time t. The angle between these three-point A, C and B is y(t). Radius (r) of the merry-go-round is 3 feet and distance (d) of the rider from the camera is 6 and the angular velocity of the rider is ω.
Assume the rider is at the edge of the merry-go-round (as the position is not specified). So, the length of CB(t) is r. To solve this problem lets consider that angular velocity of merry-go-round is constant, ω = 0 and y(t) = 0.
Therefore, we have y(t) = ωt
So, the the distance (feet) of the rider from the camera is (from the triangle AB(t)C)
C(t) = √(r² + d² - 2rdcos(y(t)) = √(45 - 36cos(ωt) = 3√(5 - 4cos(ωt))
Learn more about angular velocity here:
brainly.com/question/14769426
#SPJ4
9514 1404 393
Answer:
multiply by 3 and divide by 2 to get 6x -y = 12.
Step-by-step explanation:
A standard form equation has mutually prime integers for coefficients, and the leading one is positive. We can get that here by multiplying by 3 (to eliminate the denominator) and dividing by 2 (to remove that common factor). The result is ...
4(3/2)x -(2/3)(3/2)y = 8(3/2)
6x -y = 12
Answer:
43°
Step-by-step explanation:
43 on that angle 43 on the other angle
Answer:
usub from where
Step-by-step explanation:
