I've been stuck on this problem forever, if someone could please show me step by step what the answer is and why that would be great!

Mark has a bike with a 22 inch back wheel. The wheel sprocket is 3 inches in diameter and the petal sprocket is 4 inches in diameter. how fast in miles per hour is he going if he petals at a rate of 100 revolutions per minute?

Thanks a lot!

2. note the relationship between linear speed, $v$ , of a point on the edge of a circle, and its angular speed, $\omega$ ...

$v = r\omega$

the pedal sprocket and wheel sprocket move with the same linear speed as the chain that connects them ...

$\omega_p = 100 \, rpm$

$r_w \omega_w = r_p \omega_p$

$\omega_w = \frac{r_p \omega_p}{r_w}$

$\omega_w = \frac{400}{3} \, rpm$

this angular speed of the wheel sprocket in rpm needs to be converted to radians per hour ...

$\left(\frac{400}{3} \, rpm\right) \cdot \left(\frac{2\pi \, rad}{rev}\right) \cdot \left(\frac{60 \, min}{hr}\right) = 16000\pi \, rad/hr$

the tire and the wheel sprocket spin at the same angular speed, therefore the linear speed of the back tire is ...

$v = (11 \, in) \cdot (16000\pi \, rad/hr) = 176000 \, in/hr$

I'll leave it to you to convert this linear speed into mph.