A bicycle wheel with diameter 16 inches rides over a screw in the street. The screw is on level ground before it punctures the bike’s tire. After the bike has moved forward another 56π inches, how high above the ground is the screw? Round to the nearest tenth of an inch. 4 inches 8 inches 12 inches 16 inches

The correct option is;

16 inches

Step-by-step explanation:

The parameters of the motion given are;

The diameter, D of the bicycle = 16 inches;

The distance the bike moves (forward) after the screw punctures the tire = 56·π inches

We note that the circumference of the bicycle = π·D = π × 16 = 16·π inches

Therefore;

56·π inches/(16·π inches) = 3.5

Showing that the bicycle moves three and half complete turns (revolution) where after each complete turn, the screw starts from the bottom of the tire.

The height, h of the screw in the final half turn is given by the relation;

h = A×cos(Bx - C) + D

A = Amplitude of the motion = Diameter/2 = 16/2 = 8

P = The period of the motion 2·π/B

B·x = The angle described by the motion = Half of one revolution = π = 180°

C = Phase shift = π

D = The midline = Diameter/2 = 8 inches

Therefore;

h = 8×cos(π - π) + 8 = 16 inches

After the bike moves forward another 56·π inches the height of the screw = 16 inches.

d i think

step-by-step explanation:

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