A car slows down with an acceleration of \( a(t)=-14 \mathrm{ft} / \mathrm{s}^{2} \). Assume \( \mathrm{v}(0)=56 \mathrm{ft} / \mathrm{s}, \mathrm{s}(0)=0 \), and t is measured in seconds. a. Determine the position function, for \( \mathrm{t} \geq 0 \).

To find the position function \( s(t) \), we first need to determine the velocity function \( v(t) \) by integrating the acceleration \( a(t) = -14 \mathrm{ft/s^2} \). Starting with the acceleration: \[ v(t) = \int a(t) \, dt = \int -14 \, dt = -14t + C_1 \] Using the initial condition \( v(0) = 56 \mathrm{ft/s} \): \[ v(0) = -14(0) + C_1 = 56 \quad \Rightarrow \quad C_1 = 56 \] Thus, the velocity function is: \[ v(t) = -14t + 56 \] Next, we integrate the velocity function to find the position function \( s(t) \): \[ s(t) = \int v(t) \, dt = \int (-14t + 56) \, dt = -7t^2 + 56t + C_2 \] Using the initial condition \( s(0) = 0 \): \[ s(0) = -7(0)^2 + 56(0) + C_2 = 0 \quad \Rightarrow \quad C_2 = 0 \] Thus, the position function is: \[ s(t) = -7t^2 + 56t \] This function gives the position of the car at any time \( t \geq 0 \).