## Speedster Circles

The outside wheels of a speedster, traveling on a circular race track, are going twice as fast as the inside wheels. What is the length of the circumference described by the outer wheels? The wheels are 5 feet apart on the two axles.

*Answers to this challenge can be entered into the section below; submissions will automatically be revealed when time is up!*

Shawn| PUZZLE GRANDMASTER | Profile June 6th, 2012 - 10:04 amIf the outside wheels are travelling twice as fast as the inside wheels, then the radius of the circle travelled by the outside wheels (r2) must be twice as large as the radius of the circle travelled by the inside wheels (r1).

r2=2*r1

Since r2=r1+5, then r1=5 and r2=10.

c2=2*pi*r = 20*pi

That is a very short turning radius!

Bobo The Bear| PUZZLE MASTER | Profile June 6th, 2012 - 2:33 pmIf I’m reading the question right, then in one lap the outer tires travel twice the distance as the inner tires. Unless the radius of the “track” is only 10 feet (outer tires travel a circumference of 62.8 ft), I don’t see how this can be.

suineg| PUZZLE MASTER | Profile June 6th, 2012 - 6:11 pmNice puzzle, here is my approach:

r= radius of the smallest circle (center of the track, inner wheels)

R= radius of the greatest circle ( center of the track, outer circle)

The double of the velocity, so in the same time the outer wheels travel the double of the distance that the inner wheels.

Circumference r: 2XpiXr

Circumference R: 2XpiXR

but 2 X Circumference r = Circumference R… so:

2XpiXR = 2XpiXR

but R= r+5 so: 2pi(r+5)= 4pi r —.. solving for r: r=5

R= 5 + 5 = 10, the lenght of the outer circumference is 20 X pi

cool man!!

Vik| Profile June 6th, 2012 - 9:39 pmabout 63 feet?

Madhamish| Profile June 7th, 2012 - 9:07 amI think the answer should be 2 X pi X ((r+5)/r) X r) where r is the radius of the circuit in feet.

engjs1960| Profile June 7th, 2012 - 9:38 pmThe inner wheels travel C feet at a radius of 2*pi*R feet. The outer wheels travel 2*C feet at a radius of 2*pi*(R+5) feet. Subtracting the second from the first gives c = 2 * pi * 5. So the outer wheels describe a circumference of 20 * pi feet.

Hex| PUZZLE MASTER | Profile June 8th, 2012 - 2:53 amBoth wheels complete the circular lap in the same amount of time. Since the outer wheels are spinning twice as fast as the inner ones, the outer ones must be travelling twice the distance as opposed to the inner ones.

Ri=inner wheels track radius

Ro=outer wheels track radius

2 x Pi x Ro = 2 x (2 x Pi x Ri)

so Ro = 2 x Ri

Ro – Ri = 5

2 x Ri – Ri = 5

Ri = 5 feet

Ro = 2 x Ri = 2 x 5

Ro = 10 feet

length of the circumference described by the outer wheels = 2 x Pi x Ro = ~62.83 feet

The follow-up question should be how many cars fit on such a small track

William| Profile June 9th, 2012 - 2:31 am20? roughly 62.83 feet

Or 19.15m