# Step-by-Step Solution Let's break down the problem and solve each part. --- ## **Given Data** - **Grade (\( G \))**: 3% upgrade (\( +3\% \), so \( G = +.03 \)) - **Initial speed (\( V \))**: 80 mph - **Perception-reaction time (\( t_{pr} \))**: 2 s - **Deceleration rate (\( a \))**: 9 ft/s² (for part A) - **Intersection is 100 ft away from the sign (sign is visible for 100 ft)** - **Need to stop completely (\( V_f = \))** --- ## **Step 1: Convert Speed to Consistent Units** \[ V = 80~\text{mph} \times \frac{528~\text{ft}}{1~\text{mile}} \times \frac{1~\text{hr}}{360~\text{s}} = 117.33~\text{ft/s} \] --- ## **Step 2: Stopping Sight Distance (SSD) Formula** \[ SSD = \text{Perception-reaction distance} + \text{Braking distance} \] ### **Perception-reaction distance:** \[ d_{pr} = V \times t_{pr} \] ### **Braking distance (on a grade):** \[ d_b = \frac{V^2}{2g(f + G)} \] - \( f \): coefficient of friction (used in part C) - \( g \): acceleration due to gravity (\( 32.2~\text{ft}/\text{s}^2 \)) - \( G \): grade (as a decimal, positive for upgrade) Alternatively, if deceleration rate \( a \) is given: \[ d_b = \frac{V^2}{2a} \] --- # **Part A: Deceleration Rate is 9 ft/s²** ## **1. Perception-Reaction Distance** \[ d_{pr} = 117.33~\text{ft/s} \times 2~\text{s} = 234.67~\text{ft} \] ## **2. Braking Distance** \[ d_b = \frac{V^2}{2a} = \frac{(117.33)^2}{2 \times 9} = \frac{13766.8}{18} = 765.94~\text{ft} \] ## **3. Total Stopping Sight Distance** \[ SSD = d_{pr} + d_b = 234.67 + 765.94 = 100.61~\text{ft} \] ## **4. Required Distance for Sign** Since the sign is visible for only **100 ft** before the intersection, the **sign must be posted this distance before the intersection** and you must **add** the required SSD. \[ \text{Sign Placement Distance} = SSD + 100 = 100.61 + 100 = \boxed{110.61~\text{ft}} \] --- # **Part B: Default Deceleration Rate** The **AASHTO default deceleration rate** is **11.2 ft/s²**. ## **1. Braking Distance** \[ d_b = \frac{(117.33)^2}{2 \times 11.2} = \frac{13766.8}{22.4} = 614.57~\text{ft} \] ## **2. Total SSD** \[ SSD = 234.67 + 614.57 = 849.24~\text{ft} \] ## **3. Sign Placement** \[ \text{Sign Placement Distance} = SSD + 100 = 849.24 + 100 = \boxed{949.24~\text{ft}} \] --- # **Part C: Deceleration Rate Unknown, Friction Factor = .2** ## **1. Braking Distance (using friction and grade)** \[ d_b = \frac{V^2}{2g(f + G)} \] Where: - \( f = .2 \) - \( G = .03 \) - \( g = 32.2~\text{ft/s}^2 \) \[ d_b = \frac{(117.33)^2}{2 \times 32.2 \times (.2 + .03)} \] \[ = \frac{13766.8}{2 \times 32.2 \times .23} \] \[ = \frac{13766.8}{14.812} \] \[ = 929.48~\text{ft} \] ## **2. Total SSD** \[ SSD = 234.67 + 929.48 = 1164.15~\text{ft} \] ## **3. Sign Placement** \[ \text{Sign Placement Distance} = SSD + 100 = 1164.15 + 100 = \boxed{1264.15~\text{ft}} \] --- # **Final Answers (Summary Table)** | Case | SSD (ft) | Sign Placement (ft) | |----------------------------------------|----------|---------------------| | A) Deceleration rate = 9 ft/s² | 100.61 | **110.61** | | B) Default deceleration rate = 11.2 ft/s² | 849.24 | **949.24** | | C) Friction factor = .2 | 1164.15 | **1264.15** | --- # **Explanation** - The stopping sight distance is the sum of the **distance traveled during perception-reaction time** and the **distance required to brake to a stop**. - On upgrades, the effective deceleration is slightly increased due to gravity. - The **sign should be posted** at least the **stopping sight distance plus the 100 ft** (distance for which the sign is visible before the intersection) **before the intersection**. --- **Alt text for equations:** Equations show the step-by-step calculation for perception-reaction distance, braking distance (using both deceleration rate and friction + grade), and stopping sight distance (SSD), using the provided speed, grade, and other data. --- **If you need a diagram, let me know!**