Let's break down the problem step by step: ### **Given Data** - Total distance, \( d_{\text{total}} = 90 \) ft - Distance during acceleration, \( d_1 = 12.7 \) ft - Distance at constant speed, \( d_2 = 90 - 12.7 = 77.3 \) ft - Total time for the run, \( t_{\text{total}} = 3.6 \) s - Initial velocity, \( u = 0 \) (starts from rest) Let: - \( a \) = acceleration (during first 12.7 ft) - \( v_{\text{max}} \) = maximum speed reached after 12.7 ft - \( t_1 \) = time taken to accelerate over 12.7 ft - \( t_2 \) = time taken to cover 77.3 ft at \( v_{\text{max}} \) --- ### **Step 1: Equations of Motion for Acceleration Phase** For the first 12.7 ft (starting from rest): \[ d_1 = \frac{1}{2} a t_1^2 \] \[ v_{\text{max}} = a t_1 \] --- ### **Step 2: Constant Speed Phase** For the next 77.3 ft at \( v_{\text{max}} \): \[ d_2 = v_{\text{max}} \cdot t_2 \] --- ### **Step 3: Time Relations** Total time: \[ t_{\text{total}} = t_1 + t_2 = 3.6 \text{ s} \] --- ### **Step 4: Express \( t_2 \) in Terms of \( t_1 \) and \( a \)** \[ t_2 = t_{\text{total}} - t_1 \] \[ d_2 = v_{\text{max}} \cdot t_2 \implies 77.3 = (a t_1) (3.6 - t_1) \] --- ### **Step 5: Express \( a \) in Terms of \( t_1 \) Using the Acceleration Distance** \[ 12.7 = \frac{1}{2} a t_1^2 \implies a = \frac{2 \times 12.7}{t_1^2} \] --- ### **Step 6: Substitute \( a \) into the Equation for \( d_2 \)** Substitute \( a \) into \( 77.3 = (a t_1)(3.6 - t_1) \): \[ 77.3 = \left(\frac{2 \times 12.7}{t_1^2} \cdot t_1\right)(3.6 - t_1) \] \[ 77.3 = \left(\frac{2 \times 12.7}{t_1}\right)(3.6 - t_1) \] \[ 77.3 = \frac{25.4}{t_1}(3.6 - t_1) \] --- ### **Step 7: Solve for \( t_1 \)** \[ 77.3 = \frac{25.4(3.6 - t_1)}{t_1} \] \[ 77.3 t_1 = 25.4(3.6 - t_1) \] \[ 77.3 t_1 = 91.44 - 25.4 t_1 \] \[ 77.3 t_1 + 25.4 t_1 = 91.44 \] \[ 102.7 t_1 = 91.44 \] \[ t_1 = \frac{91.44}{102.7} \approx 0.890 \text{ s} \] --- ### **Step 8: Find \( a \)** \[ a = \frac{2 \times 12.7}{t_1^2} \] \[ a = \frac{25.4}{(0.890)^2} \] \[ a \approx \frac{25.4}{0.7921} \approx 32.07 \text{ ft/s}^2 \] --- ### **Step 9: Find \( v_{\text{max}} \)** \[ v_{\text{max}} = a t_1 \] \[ v_{\text{max}} = 32.07 \times 0.890 \approx 28.54 \text{ ft/s} \] --- ### **Final Answers** \[ \boxed{ \begin{aligned} v_{\text{max}} &= 28.5 \ \text{ft/s} \\ a &= 32.1 \ \text{ft/s}^2 \\ t_1 &= 0.89 \ \text{s} \end{aligned} } \] (Rounded to 1 decimal place as per typical convention)