A 90 kg bicyclist is coasting down a long hill that has a 10 ∘ slope. His cross-section area is 0.35 m2, his drag coefficient is 0.85, his bicycle's coefficient of rolling friction is 0.02, and the air temperature is 20∘C. What speed does he eventually reach?

To calculate the terminal velocity (the speed the bicyclist eventually reaches), we need to consider the forces acting on him, which are:

1. **Gravitational force**: This component of the force is responsible for pulling the bicyclist down the slope.
2. **Air drag**: This resists the motion and increases with speed.
3. **Rolling friction**: This is a constant force that also resists motion.

At terminal velocity, the sum of the resistive forces (air drag and rolling friction) balances the downhill component of the gravitational force, meaning the net force is zero, and the bicyclist moves at a constant speed.

### Step 1: Gravitational force component
The gravitational force pulling the cyclist down the slope is:
\[
F_{\text{gravity}} = mg \sin(\theta)
\]
Where:
- \(m = 90 \, \text{kg}\) is the mass of the cyclist,
- \(g = 9.81 \, \text{m/s}^2\) is the acceleration due to gravity,
- \(\theta = 10^\circ\) is the angle of the slope.

\[
F_{\text{gravity}} = 90 \times 9.81 \times \sin(10^\circ) \approx 153.5 \, \text{N}
\]

### Step 2: Rolling friction
The force due to rolling friction is:
\[
F_{\text{friction}} = \mu_r mg \cos(\theta)
\]
Where:
- \(\mu_r = 0.02\) is the coefficient of rolling friction.

\[
F_{\text{friction}} = 0.02 \times 90 \times 9.81 \times \cos(10^\circ) \approx 17.36 \, \text{N}
\]

### Step 3: Air drag
The air drag force is given by:
\[
F_{\text{drag}} = \frac{1}{2} C_d A \rho v^2
\]
Where:
- \(C_d = 0.85\) is the drag coefficient,
- \(A = 0.35 \, \text{m}^2\) is the cross-sectional area,
- \(\rho\) is the air density (at \(20^\circ C\), \(\rho \approx 1.204 \, \text{kg/m}^3\)),
- \(v\) is the velocity.

At terminal velocity, the forces balance:
\[
F_{\text{gravity}} = F_{\text{friction}} + F_{\text{drag}}
\]
\[
153.5 = 17.36 + \frac{1}{2} \times 0.85 \times 0.35 \times 1.204 \times v^2
\]

### Step 4: Solving for velocity
Simplifying the equation:
\[
153.5 - 17.36 = 0.17957 v^2
\]
\[
136.14 = 0.17957 v^2
\]
\[
v^2 = \frac{136.14}{0.17957} \approx 758.3
\]
\[
v \approx \sqrt{758.3} \approx 27.54 \, \text{m/s}
\]

### Conclusion:
The bicyclist's terminal velocity is approximately **27.54 m/s** (about **99 km/h**).

Would you like further details on any part of the solution? 

Here are some related questions:
1. How does changing the slope angle affect terminal velocity?
2. What happens if the drag coefficient is reduced?
3. How does a change in mass influence the speed?
4. What would be the impact of temperature variations?
5. How would different materials affect rolling friction?

**Tip**: To reduce drag and increase efficiency, cyclists can adjust their body position to reduce their cross-sectional area.

Learn how to calculate the terminal velocity of a 90 kg bicyclist coasting down a 10° slope. This physics problem covers concepts like gravitational force, rolling friction, and air drag. Follow a detailed step-by-step solution to determine the bicyclist's final speed, including relevant formulas for drag coefficient and air density at 20°C.

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