Question

There is a bug sitting between the treads of your car tire. We start a short commute of 1 mile (or 5280 feet) and the bug remains a the same location between the tread of your tire without getting squished. Assume the tire is 2 feet in diameter (so the center is exactly 1 foot above the pavement) and the tread width is negligible so the bug is almost touching the road as the tire rotates. How far does the bug travel on this trip? As the tire rotates bug moves, it traces out a curve over the road, and the curve touches the road once per every full rotation. What is the area under this curve? What is average velocity of the bug? Does the bug ever occupy the same point in space at two different times. You are driving on perfectly flat road and in straight line and the tire is a perfect circle. Also assume that you are driving at you driving at 60 mp/h or 88 ft/s for the entire trip, so the trip takes exactly 60 seconds. The tire has radius 1 ft and circumference 27pi f t. Therefore the tire makes 5280/2 pi = 840.3380993revolutions during your trip. Let t denote the time in seconds, with 0 lessthanorequalto t lessthanorequalto 60. The location of the center of the tire changes over time since the car is moving, but the vertical position of tire is always 1 ft off the ground. The horizontal position of the center of tire is 88t units to the right of the starting position.

Answer 1

1.

as the bug is sitting at the edge of the tire

distance moved by bug = distance moved by the car

distance moved by bug = 5280 feet

2.

as this will be sinusoidal curve

the area under the curve for each cycle will be

area under the curve for each cycle = 4 * integration(sin x . dx) from 0 to pi/2

area under the curve for each cycle = 4 feet^2

3)

average velocity = distance/time

average velocity = 5280/60 feet/s

average velocity = 88 feet/s

4.

No , the bug is never at the same point in space at two times