To find the final velocity at ground level (0% grade) and at a 6% grade separately, we'll follow a similar process as before, adjusting the slope angle accordingly:

Ground Level (0% Grade)

For ground level, the slope angle $\theta$ is 0°, so the gravitational component $F_g$ is zero, and we only consider rolling resistance and aerodynamic drag.

6% Grade

For a 6% grade, the slope angle $\theta$ corresponds to the arctangent of 0.06. We then include the gravitational force component, rolling resistance, and aerodynamic drag in the calculation.

Equations Recap

1. Aerodynamic Drag Force $F_D$: $F_D = \frac{1}{2} \times \text{Density} \times \text{CD} \times \text{AF} \times V^2$

2. Gravitational Force $F_g$: $F_g = W \times \sin(\theta)$

3. Rolling Resistance $F_{rr}$: $F_{rr} = \mu \times W \times \cos(\theta)$

4. Final Velocity $V_f$ Equation: $W \times Yb = F_{rr} + F_D + F_g$

Now let's calculate the final velocities for both cases.The final velocities are as follows:

• Ground level (0% grade): $V_f \approx 112.93 \, \text{m/s}$
• 6% grade: $V_f \approx 107.41 \, \text{m/s}$

Would you like more details or have any further questions?

Here are five related questions you might consider:

1. What effect does increasing the grade percentage have on final velocity?
2. How would a change in rolling resistance $\mu$ impact these results?
3. What would be the final velocity at a negative slope (downhill)?
4. How would the final velocity change if the weight $W$ were different?
5. What role does the tractive force coefficient $Yb$ play in determining the final velocity?

Tip: The slope of the road has a significant impact on the final velocity due to the additional gravitational force component on inclined planes.