Given the following position function: (position is in ft , time is in sec) \( \mathrm{s}(\mathrm{t})=2(4 \mathrm{t}+1)^{\frac{3}{2}} \) Find the velocity at \( \mathrm{t}=2 \) \( \mathrm{v}(2)=[?] \frac{\mathrm{ft}}{\mathrm{sec}} \)

To find the velocity at \( t = 2 \) for the position function \( s(t) = 2(4t + 1)^{\frac{3}{2}} \), we first need to compute the derivative of the position function with respect to time \( t \). Using the chain rule, we get: \[ v(t) = \frac{ds}{dt} = \frac{d}{dt}[2(4t + 1)^{\frac{3}{2}}] \] Let \( u = 4t + 1 \), so \( s(t) = 2u^{\frac{3}{2}} \). Now, the derivative using the chain rule is: \[ \frac{ds}{dt} = 2 \cdot \frac{3}{2} u^{\frac{1}{2}} \cdot \frac{du}{dt} \] where \( \frac{du}{dt} = 4 \). Thus, \[ v(t) = 2 \cdot \frac{3}{2} (4t + 1)^{\frac{1}{2}} \cdot 4 = 12(4t + 1)^{\frac{1}{2}} \] Now, substituting \( t = 2 \): \[ v(2) = 12(4 \cdot 2 + 1)^{\frac{1}{2}} = 12(8 + 1)^{\frac{1}{2}} = 12(9)^{\frac{1}{2}} = 12 \cdot 3 = 36 \text{ ft/sec} \] So, the velocity at \( t = 2 \) is: \[ \displaystyle v(2) = 36 \frac{\text{ft}}{\text{sec}} \]