Your roommate is working on his bicycle and has the bike upside down. He spins the 54.0 cm diameter wheel, and you notice that a pebble stuck in the tread goes by three times every second. A. What is the pebble's speed? B. What is the period of the pebble stone?

If the diameter is 54cm, then the radius is half that, the radius is:

r = 54cm/2 = 27cm

Then the perimeter of the wheel is p = 2*pi*27cm

this would mean that the pebble travels the distance:

d = 3*2*pi*27cm in one second.

Then the velocity of the pebble is:

speed = 3*2*3.1416*27cm/s = 509cm/s

Now, we know that the pebble does 3 cycles in a second, so does each cycle in 1/3 seconds, so the period would be the time that it needs to do one cycle, that we already find that is equal to 1/3 seconds.

i think the answer is a, hope this is right, that would be my best guess.

double the distance = divide strength by 4