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 \( t = 2 \) for the given position function: \[ s(t) = 2(4t + 1)^{\frac{3}{2}} \] we need to differentiate the position function twice with respect to time \( t \). ### Step 1: Find the Velocity \( v(t) \) The velocity \( v(t) \) is the first derivative of the position function \( s(t) \): \[ v(t) = s'(t) = \frac{d}{dt} \left[ 2(4t + 1)^{\frac{3}{2}} \right] \] Using the chain rule: \[ v(t) = 2 \cdot \frac{3}{2} \cdot (4t + 1)^{\frac{1}{2}} \cdot 4 = 12(4t + 1)^{\frac{1}{2}} \] ### Step 2: Find the Acceleration \( a(t) \) The acceleration \( a(t) \) is the second derivative of the position function, or the first derivative of the velocity function: \[ a(t) = v'(t) = \frac{d}{dt} \left[ 12(4t + 1)^{\frac{1}{2}} \right] \] Again, using the chain rule: \[ a(t) = 12 \cdot \frac{1}{2} \cdot (4t + 1)^{-\frac{1}{2}} \cdot 4 = 24(4t + 1)^{-\frac{1}{2}} \] ### Step 3: Evaluate \( a(t) \) at \( t = 2 \) Substitute \( t = 2 \) into the acceleration function: \[ a(2) = 24(4 \cdot 2 + 1)^{-\frac{1}{2}} = 24(9)^{-\frac{1}{2}} = 24 \cdot \frac{1}{3} = 8 \text{ ft/sec}^2 \] **Answer:** \( \mathrm{a}(2) = 8 \ \frac{\mathrm{ft}}{\sec^{2}} \)