1. **Describe how the attainment of super elevation can be done** Super elevation is the banking of a roadway at a curve to counteract the lateral acceleration produced by vehicles negotiating the curve. It is attained by raising the outer edge of the road above the inner edge. 2. **Motorist stopping distance problem** Given: - Speed $v = 55$ mi/h - Grade $G = 5\% = 0.05$ - Stopping distance $d_s = 30$ ft - Perception reaction time $t_p = 2.5$ sec We want to find the distance from the truck when the motorist first observed the crash, i.e., total stopping sight distance (SSD). Formula for SSD: $$SSD = d_p + d_s$$ where $$d_p = v \times t_p$$ is the perception distance, $$d_s = \frac{v^2}{2g(f \pm G)}$$ is the braking distance, Here, $v$ must be in ft/s: $$v = 55 \times \frac{5280}{3600} = 80.67\ \text{ft/s}$$ Assuming coefficient of friction $f = 0.35$ (typical value), and since the vehicle is going downhill, effective friction is $f - G = 0.35 - 0.05 = 0.30$. Calculate perception distance: $$d_p = 80.67 \times 2.5 = 201.68\ \text{ft}$$ Calculate braking distance: $$d_s = 30\ \text{ft}$$ (given) Total distance observed: $$SSD = d_p + d_s = 201.68 + 30 = 231.68\ \text{ft}$$ 3. **Determine grade of road given stopping distance** Given: - Speed $v = 80$ km/h = $22.22$ m/s - Object distance $SSD = 140$ m - Perception reaction time $t_p = 2.5$ s - Deceleration $a = 3.5$ m/s$^2$ Formula: $$SSD = v t_p + \frac{v^2}{2(a \pm gG)}$$ Rearranged to find grade $G$: $$G = \frac{\frac{v^2}{2a} + v t_p - SSD}{\frac{v^2}{2g}}$$ Calculate perception distance: $$d_p = 22.22 \times 2.5 = 55.55\ m$$ Calculate braking distance: $$d_b = SSD - d_p = 140 - 55.55 = 84.45\ m$$ Calculate grade: $$G = \frac{\frac{22.22^2}{2 \times 3.5} - 84.45}{\frac{22.22^2}{2 \times 9.81}} = \frac{70.6 - 84.45}{25.14} = -0.55 = -5.5\%$$ Negative grade means downhill. 4. **Minimum stopping sight distance for design speed 60 mi/h** Given: - Speed $v = 60$ mi/h = 88 ft/s - Perception reaction time $t_p = 2.5$ s - Coefficient of friction $f = 0.35$ - Gravity $g = 32.2$ ft/s$^2$ (a) Level road ($G=0$): $$SSD = v t_p + \frac{v^2}{2gf} = 88 \times 2.5 + \frac{88^2}{2 \times 32.2 \times 0.35} = 220 + 343.5 = 563.5\ ft$$ (b) Road with 4% upgrade ($G=0.04$): $$SSD = 220 + \frac{88^2}{2 \times 32.2 \times (0.35 + 0.04)} = 220 + 303.3 = 523.3\ ft$$ 5. **Minimum passing sight distance for two-lane rural road** Given: - Passing vehicle speed $v_p = 47$ mi/h - Acceleration $a = 1.43$ mi/h/s - Impeder speed $v_i = 40$ mi/h Formula for passing sight distance (PSD): $$PSD = d_1 + d_2 + d_3$$ where - $d_1$ = distance traveled by passing vehicle during perception-reaction time - $d_2$ = distance traveled during acceleration to pass - $d_3$ = distance traveled by impeder during passing Calculate each term: $$d_1 = v_p \times t_p = 47 \times 2.5 = 117.5\ ft$$ Convert speeds to ft/s: $$v_p = 47 \times 1.467 = 68.95\ ft/s$$ $$v_i = 40 \times 1.467 = 58.68\ ft/s$$ Calculate acceleration time: $$t_a = \frac{v_p - v_i}{a} = \frac{68.95 - 58.68}{1.43 \times 1.467} = 4.5\ s$$ Calculate distances: $$d_2 = v_p t_a + \frac{1}{2} a t_a^2 = 68.95 \times 4.5 + 0.5 \times 2.1 \times 4.5^2 = 310.3 + 21.2 = 331.5\ ft$$ $$d_3 = v_i t_a = 58.68 \times 4.5 = 264.1\ ft$$ Total PSD: $$PSD = 117.5 + 331.5 + 264.1 = 713.1\ ft$$ 6. **Circular curve layout problem** Given: - Intersection angle $\Delta = 48^\circ$ - Station of intersection $948+67.32$ - Design speed $v = 60$ mi/h Use standard formulas for tangent length $T$ and deflection angle $D$: $$T = R \tan(\frac{\Delta}{2})$$ $$D = \frac{L}{2R} \times 180/\pi$$ Select appropriate super elevation $e$ and side friction factor $f$ based on design speed. 7. **Truck on circular curve parameters** Given: - Center of gravity $(x=1.2m, y=1.5m)$ - Design speed $v=80$ km/h - Sight distance $S=150$ m - Obstruction distance from center line $d=10$ m Calculate minimum radius $R$ using: $$R = \frac{v^2}{g(e+f)}$$ Calculate super elevation $e$ and friction $f$ to satisfy sight distance and stability. 8. **Compound circular curve design** Given: - Radii $R_1=600$ ft, $R_2=450$ ft - Total deflection angle $\Delta=75^\circ$ - First curve deflection $\Delta_1=45^\circ$ - PCC station $675+35.25$ Calculate deflection angles for second curve: $$\Delta_2 = 75^\circ - 45^\circ = 30^\circ$$ Calculate chord lengths: $$C = 2R \sin(\frac{\Delta}{2})$$ --- **Bisection method problems** a) Solve $x^3 - 4x - 9 = 0$ using bisection method to 5 decimal places. b) Solve $e^x = 3^x$. c) Solve $2x = 3 + \cos x$. d) Solve $3x - \sqrt{1 + \sin x} = 0$. **Graph:** Circle with radius 2 centered at origin: $$x^2 + y^2 = 4$$