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garri49 [273]
3 years ago
10

A bicycle wheel with diameter 16 inches rides over a screw in the street. The screw is on level ground before it punctures the b

ike’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
Mathematics
1 answer:
Julli [10]3 years ago
3 0

Answer:

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.

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