How fast was this bike going in feet?

Question

A bike is rode for 45 minuet. In those 45 minuet, one wheel, that has a diameter of 1 2/3 feet, on the bike rotated 7290 times. What speed was the bike going at in feet per hour

Answer 1

Total duration of ride: 45 minutes

Diameter of wheel: 1 2/3 feet

Number of rotations of the wheel during the entire ride=7290

 

The easiest way to do this would be to use dimensional analysis.

 

Since we want feet per hour ultimately, we know that the minutes must be converted to hours:

60 minutes: 1 hour=45 minutes: x hours

(x)(60)=(1)(45)

x=45/60=3/4=.75

 

We need to know the total number of feet that the tires moved, for which we need to find the circumference:

circumference = 2*pi*r=Pi*d=3.14(1 2/3)=5.23 feet

 

7290 revolutions: x feet=1 revolution: 5.23 feet

(5.23)(7290)=x

x=38151 feet

 

Speed in terms of feet per hour = # of feet/number of hours=38151/.75=50868 feet per hour

 

 

 

 

Answer 2

The perimeter (or circumference) of the wheel is pi * diameter 

                   

                   = 22/7 * 5/3 = 110/21 feet

 

The distance traveled is 7290 * 110/21 = 801900/21 = 38185 and 15/21 = 38185 and 5/7 feet

 

45 minutes = 3/4 hour

 

 Distance = rate x time

 

 rate = distance / time

 

  801900/21 divided by 3/4 =

 

  801900/21 * 4/3 = 50914 and 2/7 feet per hour

Answer 3

x = total distance traveled by a point on the rim of the wheel

 

d = diameter of wheel ≅ 1.67 ft

 

x = 7290π(1.67) ft ≅ 38,247 ft

 

v = x/t = (38,247 ft)/(45 min) ≅ 849.9 ft/min = 60(849.9) ft/hr = 50,994 ft/hr