Given the following position function: (position is in ft , time is in sec ) \( \mathrm{s}(\mathrm{t})=\frac{11}{3} \mathrm{t}^{3}-\frac{55}{2} \mathrm{t}^{2}+\mathrm{t} \) Find the velocity at \( \mathrm{t}=2 \). \( \mathrm{v}(2)=[?] \frac{\mathrm{ft}}{\mathrm{sec}} \)

To find the velocity at \( t = 2 \), we first need to determine the velocity function, which is the derivative of the position function \( s(t) \). Starting with: \[ s(t) = \frac{11}{3} t^3 - \frac{55}{2} t^2 + t \] We differentiate \( s(t) \) with respect to \( t \): \[ v(t) = \frac{ds}{dt} = \frac{d}{dt} \left(\frac{11}{3} t^3 - \frac{55}{2} t^2 + t\right) \] Calculating the derivatives: \[ v(t) = 11t^2 - 55t + 1 \] Now we can substitute \( t = 2 \) into the velocity function: \[ v(2) = 11(2)^2 - 55(2) + 1 \] Calculating this gives: \[ v(2) = 11(4) - 110 + 1 = 44 - 110 + 1 = -65 \] Thus, the velocity at \( t = 2 \) is: \[ v(2) = -65 \, \frac{ft}{sec} \]