Emma runs for 28 seconds and finishes at 250 meters. What is her velocity? \( 28 \cdot ?=\square \) Clay runs for 32 seconds at -8.1 meters per second. What is his finish point? \( 32 \cdot \square \)

Sure, let's solve both problems step by step. --- ### **1. Emma's Velocity** **Given:** - **Distance (Δx):** 250 meters - **Time (t):** 28 seconds **Formula for Velocity (v):** \[ v = \frac{\Delta x}{t} \] **Calculation:** \[ v = \frac{250\ \text{m}}{28\ \text{s}} \approx 8.93\ \text{m/s} \] **Final Equation:** \[ 28\ \text{s} \times 8.93\ \frac{\text{m}}{\text{s}} = 250\ \text{m} \] **Answer:** Emma's velocity is approximately **8.93 meters per second**. --- ### **2. Clay's Finish Point** **Given:** - **Velocity (v):** -8.1 meters per second (the negative sign indicates direction) - **Time (t):** 32 seconds **Formula for Displacement (Δx):** \[ \Delta x = v \times t \] **Calculation:** \[ \Delta x = (-8.1\ \frac{\text{m}}{\text{s}}) \times 32\ \text{s} = -259.2\ \text{m} \] **Final Equation:** \[ 32\ \text{s} \times (-8.1\ \frac{\text{m}}{\text{s}}) = -259.2\ \text{m} \] **Answer:** Clay's finish point is **259.2 meters** in the negative direction from his starting point. --- **Summary:** - **Emma's Velocity:** ≈ 8.93 m/s - **Clay's Finish Point:** –259.2 meters