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 acceleration at \( \mathrm{t}=2 \) \( \mathrm{a}(2)=[?] \frac{\mathrm{ft}}{\sec ^{2}} \)

To find the acceleration at \( t = 2 \), we first need to derive the position function to obtain the velocity function. From the position function \( s(t) = 2(4t + 1)^{\frac{3}{2}} \), we'll calculate the velocity \( v(t) \) as the first derivative, and then the acceleration \( a(t) \) as the second derivative. 1. Differentiate \( s(t) \): \[ v(t) = \frac{ds}{dt} = 2 \cdot \frac{3}{2}(4t + 1)^{\frac{1}{2}} \cdot 4 = 12(4t + 1)^{\frac{1}{2}} \] 2. Now differentiate \( v(t) \) to find \( a(t) \): \[ a(t) = \frac{dv}{dt} = 12 \cdot \frac{1}{2}(4t + 1)^{-\frac{1}{2}} \cdot 4 = 24(4t + 1)^{-\frac{1}{2}} \cdot 4 \] \[ a(t) = \frac{96}{\sqrt{4t + 1}} \] 3. Now evaluate \( a(2) \): \[ a(2) = \frac{96}{\sqrt{4(2) + 1}} = \frac{96}{\sqrt{8 + 1}} = \frac{96}{\sqrt{9}} = \frac{96}{3} = 32 \] So, the answer is: \[ \mathrm{a}(2) = 32 \, \frac{\mathrm{ft}}{\sec^{2}} \]