Given the following position function: (position is in ft , time is in sec ) \( \mathrm{s}(\mathrm{t})=\tan \left(\mathrm{t}^{2}\right) \) Find the acceleration at \( \mathrm{t}=0.5 \) \( \mathrm{a}(0.5)=[?] \frac{\mathrm{ft}}{\mathrm{sec}^{2}} \) Round your answer to the nearest thousandth.

To find the acceleration at \( t = 0.5 \) for the position function \( s(t) = \tan(t^2) \), we need to follow these steps: 1. **Find the velocity** by taking the first derivative of the position function: \[ v(t) = s'(t) = \frac{d}{dt}(\tan(t^2)) \] Using the chain rule, this gives us: \[ v(t) = \sec^2(t^2) \cdot (2t) = 2t \sec^2(t^2) \] 2. **Find the acceleration** by taking the derivative of the velocity function: \[ a(t) = v'(t) = \frac{d}{dt}(2t \sec^2(t^2)) \] We will apply the product rule here: \[ a(t) = 2 \sec^2(t^2) + 2t \cdot \frac{d}{dt}(\sec^2(t^2)) \] Using the chain rule on \( \sec^2(t^2) \): \[ \frac{d}{dt}(\sec^2(t^2)) = 2 \sec^2(t^2) \tan(t^2) \cdot (2t) = 4t \sec^2(t^2) \tan(t^2) \] Putting it all together: \[ a(t) = 2 \sec^2(t^2) + 2t(4t \sec^2(t^2) \tan(t^2)) = 2 \sec^2(t^2) + 8t^2 \sec^2(t^2) \tan(t^2) \] Simplifying gives: \[ a(t) = 2 \sec^2(t^2) (1 + 4t^2 \tan(t^2)) \] 3. **Evaluate \( a(0.5) \)**: \[ a(0.5) = 2 \sec^2((0.5)^2) (1 + 4(0.5)^2 \tan((0.5)^2)) \] Calculating \( \sec^2(0.25) \) and \( \tan(0.25) \): \[ \sec^2(0.25) \approx 1.127625965 (using a calculator) \] \[ \tan(0.25) \approx 0.25541281 (using a calculator) \] Plugging those values in: \[ a(0.5) \approx 2 \cdot 1.127625965 (1 + 4(0.25) \cdot 0.25541281) \] Calculate: \[ 4(0.25)(0.25541281) \approx 0.25541281 \Rightarrow 1 + 0.25541281 \approx 1.25541281 \] Now substituting this back: \[ a(0.5) \approx 2 \cdot 1.127625965 \cdot 1.25541281 \approx 2.83 \] Finally, we round to the nearest thousandth: \[ \text{Acceleration at } t = 0.5 \text{ is } a(0.5) \approx 2.830 \frac{\text{ft}}{\text{sec}^2}. \]