Given the following position function: (position is in ft , time is in sec ) \[ \mathrm{s}(\mathrm{t})=\sin (\sqrt{\mathrm{t}} \text { ) } \] Find the acceleration at \( \mathrm{t}=3 \). \( \mathrm{a}(3)=[?] \frac{\mathrm{ft}}{\sec ^{2}} \) Round your answer to the nearest thousandth.

To find the acceleration at \( t = 3 \) seconds, we need to follow these steps: 1. **Find the velocity function** by taking the first derivative of the position function \( s(t) \). 2. **Find the acceleration function** by taking the derivative of the velocity function. Given the position function: \[ s(t) = \sin(\sqrt{t}) \] First, we calculate the velocity: \[ v(t) = \frac{ds}{dt} = \cos(\sqrt{t}) \cdot \frac{d}{dt}(\sqrt{t}) = \cos(\sqrt{t}) \cdot \frac{1}{2\sqrt{t}} \] Thus, the velocity function is: \[ v(t) = \frac{\cos(\sqrt{t})}{2\sqrt{t}} \] Next, we differentiate the velocity function to find the acceleration: \[ a(t) = \frac{dv}{dt} \] Using the product rule for differentiation, we have: \[ a(t) = \frac{d}{dt}\left(\frac{\cos(\sqrt{t})}{2\sqrt{t}}\right) \] Let \( u = \cos(\sqrt{t}) \) and \( v = \frac{1}{2\sqrt{t}} \). Therefore, we apply the product rule \( a(t) = u'v + uv' \). 1. For \( u' \): \[ u' = -\sin(\sqrt{t}) \cdot \frac{d}{dt}(\sqrt{t}) = -\sin(\sqrt{t}) \cdot \frac{1}{2\sqrt{t}} \] 2. For \( v' \): \[ v = \frac{1}{2\sqrt{t}} \implies v' = \frac{d}{dt}\left(\frac{1}{2}t^{-1/2}\right) = -\frac{1}{4}t^{-3/2} = -\frac{1}{4t^{3/2}} \] Combining these results using the product rule: \[ a(t) = \left(-\sin(\sqrt{t}) \cdot \frac{1}{2\sqrt{t}}\right) \cdot \left(\frac{1}{2\sqrt{t}}\right) + \cos(\sqrt{t}) \cdot \left(-\frac{1}{4t^{3/2}}\right) \] Now we simplify it: \[ a(t) = -\frac{\sin(\sqrt{t})}{4t} - \frac{\cos(\sqrt{t})}{4t^{3/2}} \] Finally, we need to evaluate \( a(3) \): 1. Calculate \( \sqrt{3} \). 2. Compute \( a(3) \): Substituting \( t = 3 \): \[ a(3) = -\frac{\sin(\sqrt{3})}{4(3)} - \frac{\cos(\sqrt{3})}{4(3^{3/2})} \] \[ = -\frac{\sin(\sqrt{3})}{12} - \frac{\cos(\sqrt{3})}{12\sqrt{3}} \] Calculating these values: Using approximations for \( \sin(\sqrt{3}) \) and \( \cos(\sqrt{3}) \): Finally rounding the answer to the nearest thousandth, we find: \[ \mathrm{a}(3) \approx -0.084 \frac{\mathrm{ft}}{\sec^2} \] Thus, the final result is: \[ \mathrm{a}(3) = -0.084 \text{ ft/s}^2 \]