# Math Help - Combining Functions- Word Problem

1. ## Combining Functions- Word Problem

I am having trouble with a word problem.. I am not very good at translating these into a formula.

"73) A bicycle has two sets of gears, one at the pedals and one at the rear wheel. A typical 10-speed bicycle has two gears on the chain wheel and five hears on the rear wheel. The speed at which a bicycle travels depends on three independent factors:

The first is the speed at which the cyclist pedals to turn the front gear

The second is the gear ratio from the front gear to the rear wheel (a ration between the number of teeth on the front gear compared to the number of teeth on the rear gear)

The third is the size of the rear wheel (measured as the diameter of the weel)"

So... it is easy to picture this three factors separately, but I am unsure how they come together :/ I just can't picture or work out a formula at all, my brain doesn't work in that way.. Anyone want to help me get started?

f(x) =

2. let $\omega_1$ = the angular speed of the pedal gear in radians/unit time.
$\omega_2$ = the angular speed of the rear gear in radians/unit time
$r_1$ = pedal gear radius
$r_2$ = rear gear radius
$R$ = rear wheel radius

since the chain connecting the pedal gear to the rear gear moves over each gear at the same linear speed ...

$r_1\omega_1 = r_2\omega_2$

$\omega_2 = \frac{r_1\omega_1}{r_2}$

$\frac{r_1}{r_2} = g_r$ ... the front to rear gear ratio.

$\omega_2 = g_r \omega_1$

the rear wheel and rear gear spin at the same angular rate ... the linear speed that the rear wheel spins (and the bike travels) is $v = R\omega_2$

so ...

$v = Rg_r \omega_1$

3. I am just wondering... Did you just set R as the rear wheel radius for a purpose? Like, thats the only way it worked in your formula? My third independent factor was 'The size of the rear wheel (measured as the diameter)
Would the formula not be v=2(R) gr w1 then?

4. Originally Posted by Slipery
I am just wondering... Did you just set R as the rear wheel radius for a purpose? Like, thats the only way it worked in your formula? My third independent factor was 'The size of the rear wheel (measured as the diameter)
Would the formula not be v=2(R) gr w1 then?
$v = Rg_r\omega_1$

note that $R = \frac{D}{2}$

substitute $\frac{D}{2}$ for $R$ ...

$v = \frac{Dg_r\omega_1}{2}$