Question

"A clock is positioned on an auditorium wall with its center 9 feet above the floor. The second hand is 10 inches long and continuously moves around the clock. Let x represent the time in seconds measured from noon on a given day. Which f(x) represents the height above the floor, in feet, of the tip of the second hand?"

Answer 1

Answer:

B : 5/6cos(pi/30x)+9

Step-by-step explanation:

Edge 2020

Answer 2

Answer: f(x) = 10in*cos(x*2*pi/60s) + 10in*sin(x*2*pi/60s) + 108in.

Step-by-step explanation:

When we have a circular motion, we can write this as:

f(x) = A*sin(k*x + f) + A*cos(k*x + f) + C.

where:

k is a constant that depends on the period.

f is a phase shift.

C is the mean height of the circular motion, in this case will be equal to the height of the center, that is 9ft.

C = 9ft.

A is the amplitude of the motion, in this case is equal to the radius, that is equal to the lenght of the second hand

A = 10in

But we should write both quantities in the same units, so we have:

1ft = 12in

then:

C = 9ft = 9*12in = 108in

Then our equation is:

f(x) = 10in*cos(k*x + f) + 10in*sin(k*x + f) + 108in

now, x = 0s coincides with the noon

so at x = 0s, the second hand is pointint to the twelve, this means that is in the max height, so we have:

F(0s) = 10in*cos(0 + f) + 10in*sin(0 + f) + 108 in

Now, the maximum value will be when the value inside the sinusoidals is 0.

(because cos(0) = 1 and sin(0) = 0)

then we have that f = 0.

Now, we know that it takes 60 seconds to do a full revolution, and for the sinusodial functions the period is 2*pi

Then we have:

c*60s = 2*pi

c = 2*pi/60s.

Then the equation is:

f(x) = 10in*cos(x*2*pi/60s) + 10in*sin(x*2*pi/60s) + 108in.