School-Safe Puzzle Games

## 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.

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### 8 Comments to “Speedster Circles”

1. Shawn | PUZZLE GRANDMASTER | Profile

If 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!

2. Bobo The Bear | PUZZLE MASTER | Profile

If 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.

3. suineg | PUZZLE MASTER | Profile

Nice 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!!

4. Vik | Profile

I think the answer should be 2 X pi X ((r+5)/r) X r) where r is the radius of the circuit in feet.

6. engjs1960 | Profile

The 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.

7. Hex | PUZZLE MASTER | Profile

Both 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.

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

8. William | Profile

20? roughly 62.83 feet
Or 19.15m