Given the following position function: (position is in ft , time is in sec ) \[ s(\mathrm{t})=\frac{4 \mathrm{t}+5}{\mathrm{t}+1} \] 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 derivative of the position function \( s(t) \). The position function is given by \[ s(t) = \frac{4t + 5}{t + 1} \] Using the quotient rule for differentiation, where if \( u(t) = 4t + 5 \) and \( v(t) = t + 1 \), we find \( s'(t) = \frac{u'v - uv'}{v^2} \). Calculating the derivatives: - \( u' = 4 \) - \( v' = 1 \) Applying the quotient rule: \[ s'(t) = \frac{(4)(t + 1) - (4t + 5)(1)}{(t + 1)^2} \] Simplifying: \[ s'(t) = \frac{4t + 4 - 4t - 5}{(t + 1)^2} = \frac{-1}{(t + 1)^2} \] Now, plug in \( t = 2 \): \[ v(2) = s'(2) = \frac{-1}{(2 + 1)^2} = \frac{-1}{3^2} = \frac{-1}{9} \] Thus, the velocity at \( t = 2 \) is \[ \boxed{-\frac{1}{9}} \frac{\text{ft}}{\text{sec}} \]