Let's solve the problem step-by-step: ## **Given:** - Acceleration, \( a(t) \), is piecewise: - For \( \leq t < 6 \): \( a = -4 \) ft/s² - For \( 6 \leq t < 10 \): \( a = \) - For \( 10 \leq t \leq 14 \): \( a = +4 \) ft/s² - Starts from origin: \( x() = \) - During zero acceleration (\(6 \leq t < 10\)), \( v = -8 \) ft/s Let's construct **\(v-t\)** and **\(x-t\)** curves. --- ## **Step 1: Find Velocity as a Function of Time** ### **A. For \( \leq t < 6\):** \[ a = \frac{dv}{dt} = -4 \] \[ v(t) = \int a \, dt = -4t + C_1 \] At \( t = \), let \( v() = v_ \). But \( v_ \) is unknown. Let's use the information at \( t = 6 \): At \( t = 6 \), \( v = -8 \) ft/s (given for next interval). So, \[ -8 = -4(6) + v_ \implies v_ = -8 + 24 = 16\ \text{ft/s} \] So, for \( \leq t < 6\): \[ \boxed{v(t) = 16 - 4t} \] --- ### **B. For \(6 \leq t < 10\):** \[ a = \implies v = \text{constant} = -8\ \text{ft/s} \] --- ### **C. For \(10 \leq t \leq 14\):** \[ a = \frac{dv}{dt} = 4 \] \[ v(t) = \int 4\,dt = 4t + C_2 \] At \( t = 10 \), \( v(10) = -8 \): \[ -8 = 4(10) + C_2 \implies C_2 = -8 - 40 = -48 \] So, for \(10 \leq t \leq 14\): \[ \boxed{v(t) = 4t - 48} \] --- ## **Step 2: Find Position as a Function of Time** ### **A. For \( \leq t < 6\):** \[ v(t) = \frac{dx}{dt} = 16 - 4t \] \[ x(t) = \int (16 - 4t) dt = 16t - 2t^2 + D_1 \] At \( t = \), \( x() = \implies D_1 = \) So, \[ \boxed{x(t) = 16t - 2t^2} \] --- ### **B. For \(6 \leq t < 10\):** \[ v = -8 \implies x(t) = -8t + D_2 \] At \( t = 6 \), \( x(6) = 16(6) - 2(6^2) = 96 - 72 = 24 \) So, \[ 24 = -8(6) + D_2 \implies D_2 = 24 + 48 = 72 \] So, \[ \boxed{x(t) = -8t + 72} \] --- ### **C. For \(10 \leq t \leq 14\):** \[ v(t) = 4t - 48 \implies x(t) = \int (4t - 48) dt = 2t^2 - 48t + D_3 \] At \( t = 10 \), \( x(10) = -8(10) + 72 = -80 + 72 = -8 \) \[ -8 = 2(10^2) - 48(10) + D_3 \] \[ -8 = 200 - 480 + D_3 \implies D_3 = -8 + 280 = 272 \] So, \[ \boxed{x(t) = 2t^2 - 48t + 272} \] --- ## **Step 3: Summary of Results** ### **Velocity-Time Equation:** \[ v(t) = \begin{cases} 16 - 4t & \leq t < 6 \\ -8 & 6 \leq t < 10 \\ 4t - 48 & 10 \leq t \leq 14 \\ \end{cases} \] ### **Position-Time Equation:** \[ x(t) = \begin{cases} 16t - 2t^2 & \leq t < 6 \\ -8t + 72 & 6 \leq t < 10 \\ 2t^2 - 48t + 272 & 10 \leq t \leq 14 \\ \end{cases} \] --- ## **Step 4: Sketching the Curves** ### **1. \( v-t \) Curve** - Linear decrease from \( 16 \) to \( -8 \) from \( t= \) to \( t=6 \) - Constant at \( -8 \) from \( t=6 \) to \( t=10 \) - Linear increase from \( -8 \) at \( t=10 \) to \( 8 \) at \( t=14 \) (\( v(14) = 4(14) - 48 = 56 - 48 = 8 \)) ### **2. \( x-t \) Curve** - Quadratic curve from \( t= \) to \( t=6 \) - Linear decrease from \( t=6 \) to \( t=10 \) - Upward quadratic from \( t=10 \) to \( t=14 \) --- ## **Final Answer (boxed):** ### **Velocity:** \[ \boxed{ v(t) = \begin{cases} 16 - 4t & \leq t < 6 \\ -8 & 6 \leq t < 10 \\ 4t - 48 & 10 \leq t \leq 14 \\ \end{cases} } \] ### **Position:** \[ \boxed{ x(t) = \begin{cases} 16t - 2t^2 & \leq t < 6 \\ -8t + 72 & 6 \leq t < 10 \\ 2t^2 - 48t + 272 & 10 \leq t \leq 14 \\ \end{cases} } \] Let me know if you need the **plots** or further explanation!