Pre algabra word problem. how did they get that answer?

Question

how do you get a h(t)=12-12 cos (pi/3 x t) for a formula if bike tire diameter is 24" revolves at 10rpm, t=0 a certain point, and is at ht 0 and the question asked to find the H of the point above ground after 3 seconds?

 

Answer 1

Assuming 12 in the function is the radius of the wheel and h(t) is the distance of a point on the circumference of the tire above the ground.

When the tire is at its lowest point (i.e. the point where it touches the ground) at t = 0, h(t) = 0 with the angle = 0.

Revolution of the  wheel = 10 rpm = 10/60 revolution per second = 1/6 revolution per second.

With one revolution = 2pi radian

In 1 second the tire turns pi/3 radian.  {(that matches the function h(t)} 

After starting from its lowest position, 3 seconds later the angle = pi (because pi/3 * 3)

Therefore h(3) = 12 - 12 cos pi = 24.

That means H is right at the top of the tire.