A reflector on a bicycle wheel is 15 cm from the rim. The diameter of the wheel is 70 cm. At t=0.5 seconds the reflector is at its lowest point, 0.75 seconds later it returns to the same position.

Question

Assume the bicycle wheel is spinning at this constant rate.

1. Write an equation to model the height of the reflector above the ground at any time t in seconds.
2. What is the height of the reflector when t=5.2 seconds?
3. When is the first time the reflector is at a height of 53 cm above the ground?

Answer 1

diameter = 70 cm radius = 70/2 = 35

reflector is 15 cm from the rim

the amplitude = A =35-15 = 20

at .75 seconds the reflector is at its lowest point

at .5 seconds the reflector is at its lowest point

period= 2pi/B = .25, B= 2pi/.25

D= vertical shift = 35

C = phase shift = -pi/2

midline is y=70/2 + 15

h(t) = Asin(Bt+C)+D = 20sin((2pi/.25)t - pi/2) + 35

at 5.2 seconds, it's 4/5 of the period, heading down

at 1/2 it's at top = 55 cm, at 3/4 it's at 35 cm

at 4/5 it's between 15 and 35cm

h(5.2) = 20sin((2pi/.25)(5.2)- pi/2) +35

= 20sin(10.4pi/.25 - pi/2) +35

= 20sin((10.4 - 8)pi/2) +35

=20sin(1.2pi) + 35

=20sin(216) +35 with the angle in degrees

= -20(.587785)+35

= about -11.8+35

= about 23.2 cm

h(t) = 53 = 20sin((2pit/.25 -pi/2) + 35