Car Acceleration 416265

1. **State the problem:** Calculate the maximum acceleration rate of a 3000-lb car given engine torque, gear ratio, drivetrain efficiency, axle slippage, wheel radius, and road conditions. 2. **Known values:** - Weight $W = 3000$ lb - Coefficient of rolling friction $f_{fl} = 0.35$ - Frontal area $A_i = 21$ ft$^2$ - Air density $\rho = 0.002378$ slugs/ft$^3$ - Engine speed $N = 3000$ rev/min - Engine torque $T_e = 250$ ft-lb - Gear ratio $GR = 4.5$ - Drivetrain efficiency $\eta = 0.90$ - Drive axle slippage $s = 3.5\% = 0.035$ - Wheel radius $r = 16$ inches = $\frac{16}{12} = 1.333$ ft 3. **Step 1: Calculate engine angular velocity in rad/s** $$\omega_e = \frac{2\pi N}{60} = \frac{2\pi \times 3000}{60} = 314.16\ \text{rad/s}$$ 4. **Step 2: Calculate wheel angular velocity considering gear ratio and slippage** Effective gear ratio considering slippage: $$GR_{eff} = GR \times (1 - s) = 4.5 \times (1 - 0.035) = 4.5 \times 0.965 = 4.3425$$ Wheel angular velocity: $$\omega_w = \frac{\omega_e}{GR_{eff}} = \frac{314.16}{4.3425} = 72.32\ \text{rad/s}$$ 5. **Step 3: Calculate tractive torque at the wheels** $$T_w = T_e \times GR_{eff} \times \eta = 250 \times 4.3425 \times 0.90 = 976.96\ \text{ft-lb}$$ 6. **Step 4: Calculate tractive force at the wheels** $$F_t = \frac{T_w}{r} = \frac{976.96}{1.333} = 732.7\ \text{lb}$$ 7. **Step 5: Calculate rolling resistance force** $$F_r = f_{fl} \times W = 0.35 \times 3000 = 1050\ \text{lb}$$ 8. **Step 6: Calculate aerodynamic drag force** Assuming velocity $v$ is related to wheel angular velocity: $$v = \omega_w \times r = 72.32 \times 1.333 = 96.4\ \text{ft/s}$$ Convert to mph for reference (not necessary for force calculation): $$v_{mph} = \frac{96.4 \times 3600}{5280} = 65.7\ \text{mph}$$ Aerodynamic drag force: $$F_d = \frac{1}{2} \rho v^2 C_d A_i$$ Assuming $C_d = 0.35$ (typical for cars): $$F_d = 0.5 \times 0.002378 \times (96.4)^2 \times 0.35 \times 21 = 80.9\ \text{lb}$$ 9. **Step 7: Calculate net tractive force available for acceleration** $$F_{net} = F_t - F_r - F_d = 732.7 - 1050 - 80.9 = -398.2\ \text{lb}$$ Since $F_{net}$ is negative, the car cannot accelerate at this speed; it is limited by rolling resistance and drag. 10. **Step 8: Calculate maximum acceleration force ignoring drag and rolling resistance (starting from rest)** At zero speed, drag is zero and rolling resistance is minimal, so: $$F_{max} = F_t = 732.7\ \text{lb}$$ 11. **Step 9: Calculate acceleration using Newton's second law** Convert weight to mass: $$m = \frac{W}{g} = \frac{3000}{32.2} = 93.17\ \text{slugs}$$ Acceleration: $$a = \frac{F_{max}}{m} = \frac{732.7}{93.17} = 7.86\ \text{ft/s}^2$$ **Final answer:** The car's maximum acceleration rate on a level road under these conditions is approximately $$\boxed{7.86\ \text{ft/s}^2}$$