The problem involves a car traveling at a speed of \(104 \, \text{km/h}\) with a stone stuck in the tire tread. The tire has a diameter of \(60.6 \, \text{cm}\). We need to find the maximum and minimum speeds of the stone as observed by a pedestrian.
First, we convert the car's speed from kilometers per hour to meters per second:
\[
104 \, \text{km/h} = \frac{104 \times 1000}{3600} \, \text{m/s} = 28.8889 \, \text{m/s}
\]
The radius of the tire is half of the diameter:
\[
r = \frac{60.6 \, \text{cm}}{2} = 30.3 \, \text{cm} = 0.303 \, \text{m}
\]
The angular velocity \(\omega\) of the tire is given by:
\[
\omega = \frac{v}{r} = \frac{28.8889}{0.303} \, \text{rad/s} = 95.3564 \, \text{rad/s}
\]
The maximum speed of the stone occurs when it is at the top of the tire. At this point, the linear speed of the stone is the sum of the car's speed and the tangential speed due to the tire's rotation:
\[
v_{\text{max}} = v + r\omega = 28.8889 + 0.303 \times 95.3564 = 57.7778 \, \text{m/s}
\]
Convert this speed back to kilometers per hour:
\[
v_{\text{max}} = 57.7778 \times \frac{3600}{1000} = 208 \, \text{km/h}
\]
The minimum speed of the stone occurs when it is at the bottom of the tire. At this point, the linear speed of the stone is the car's speed minus the tangential speed due to the tire's rotation:
\[
v_{\text{min}} = v - r\omega = 28.8889 - 0.303 \times 95.3564 = 0 \, \text{m/s}
\]
Convert this speed back to kilometers per hour:
\[
v_{\text{min}} = 0 \times \frac{3600}{1000} = 0 \, \text{km/h}
\]
- Maximum speed of the stone: \(\boxed{208 \, \text{km/h}}\)
- Minimum speed of the stone: \(\boxed{0 \, \text{km/h}}\)