Can u assist me in this challenge question for my trig class?

Question

The Reflector on a bicycle wheel is 15cm from the rim. The diameter of the wheel is 76cm at time = 1/2 sec, the reflector is at its lowest point; 3/4 sec later it returns to the same position.Find an equation which will locate the height of the reflector above the ground at any time t.

a) Find height when t=5.2 sec
b) Find the 1st time the reflector is at a height of 59cm above the ground

Can u help me find the equation i have one but i dont know if its correct

Equation- y = -23 Cos [ 2pi/.75(x-.5)]+38
can u comfirm this? plz so i can calculate a & b

Answer 1

Let's see The radius is 38. The reflector is 23 cm from the center.

So the equation is 38 + 23cos(something). So far, so good.

The period is 3/4 second. Hence the equation is 38 + 23cos((8pi/3)*t + constant).

That way, the angle changes by 2 pi every 3/4 of a second.

That's also what you had. (Give or take some trig identities, which change the form but not the meaning.) Good for you!

Next, you want to pick the constant so that the cos in question equals -1 when t = 1/2. And your equation does that too.

So I think you're absolutely correct, although I used a different form. The only part of your choice of equation form that I think is downright BAD is the renaming of t as "x". That could cause confusion. Otherwise, I think you've got it!