Given the following position function: (position is in ft , time is in sec ) \( \mathrm{s}(\mathrm{t})=\sin (\sqrt{\mathrm{t}}) \) Find the velocity at \( \mathrm{t}=3 \). \( \mathrm{v}(3)=[ \) ? \( ] \frac{\mathrm{ft}}{\mathrm{sec}} \) Round your answer to the nearest thousandth.

To find the velocity at \( t = 3 \), you first need to determine the derivative of the position function \( s(t) = \sin(\sqrt{t}) \). The velocity function is the first derivative of the position function: \[ v(t) = \frac{ds}{dt} = \frac{d}{dt} (\sin(\sqrt{t})) \] Using the chain rule: \[ v(t) = \cos(\sqrt{t}) \cdot \frac{d}{dt} (\sqrt{t}) = \cos(\sqrt{t}) \cdot \frac{1}{2\sqrt{t}} \] Now, plug in \( t = 3 \): \[ v(3) = \cos(\sqrt{3}) \cdot \frac{1}{2\sqrt{3}} \] Calculating \( \sqrt{3} \) (approximately 1.732): \[ \cos(\sqrt{3}) \approx \cos(1.732) \approx -0.173648 \] \[ \frac{1}{2\sqrt{3}} \approx \frac{1}{3.464} \approx 0.288675 \] Now multiply these two: \[ v(3) \approx -0.173648 \cdot 0.288675 \approx -0.05007 \] Rounding to the nearest thousandth, you have: \[ v(3) \approx -0.050 \text{ ft/s} \] So, the final answer is: \( \mathrm{v}(3) \approx -0.050 \, \frac{\mathrm{ft}}{\mathrm{sec}} \)