The inside wheels of a car traveling on a circular path are rotating half as fast as the outside wheels. The front two wheels are six feet apart. What is the number of feet in the path traced by the inside front wheel in one trip around the circle? Express your answer in the form "k pi", where k is an integer.
2(2pi r) = 2pi (r+6)
2pi r = pi r + 6pi
pi r = 6pi
r = 6
so, the distance is 2pi * 6 = 12 pi
Please do not post HW questions. Just for the record, the answer=12pi.
Lol
To find the path traced by the inside front wheel in one trip around the circle, we need to understand the relationship between the speeds of the inside and outside wheels.
Since the inside wheels are rotating half as fast as the outside wheels, we can set up a ratio to represent the speeds of the inside and outside wheels. Let's say the outside wheels cover a distance of D.
The outside wheels cover D distance in one revolution, while the inside wheels, rotating at half the speed, cover D/2 distance in one revolution.
Now, let's consider the circumference of the circle that the car is traveling on. The circumference of a circle is given by the formula C = 2πr, where r is the radius of the circle.
In our case, the radius is half of the distance between the front two wheels, which is 6/2 = 3 feet.
So, the circumference of the circle is C = 2π(3) = 6π feet.
Since the outside wheels cover the entire circumference of the circle (C) in one revolution, the path traced by the inside front wheel is also equal to the circumference of the circle (C).
Therefore, the path traced by the inside front wheel in one trip around the circle is 6π feet.
Hence, the answer is 6π.
thanks alot man!
inside wheel at radius r
outside wheel at radius 2 r because circumference is twice
so r = 6
so
2 pi r = 12 r