Given the following position function: (position is in ft , time is in sec ) \[ \mathrm{s}(\mathrm{t})=\frac{4 \mathrm{t}+5}{\mathrm{t}+1} \] Find the acceleration at \( \mathrm{t}=2 \). \( \mathrm{a}(2)=[?] \frac{\mathrm{ft}}{\mathrm{sec}^{2}} \) Round your answer to the nearest thousandth.

To find the acceleration at \( t=2 \), we first need to derive the velocity function from the position function, \( s(t) \). Starting with the position function: \[ s(t) = \frac{4t + 5}{t + 1} \] We can use the quotient rule to differentiate \( s(t) \). The quotient rule states that if you have \( \frac{u(t)}{v(t)} \), then the derivative is: \[ s'(t) = \frac{u'v - uv'}{v^2} \] where \( u(t) = 4t + 5 \) and \( v(t) = t + 1 \). Calculating \( u' \) and \( v' \): - \( u' = 4 \) - \( v' = 1 \) Now apply the quotient rule: \[ s'(t) = \frac{(4)(t+1) - (4t+5)(1)}{(t+1)^2} \] Simplifying the numerator: \[ = \frac{4t + 4 - (4t + 5)}{(t + 1)^2} = \frac{4t + 4 - 4t - 5}{(t + 1)^2} = \frac{-1}{(t + 1)^2} \] Now we have the velocity function: \[ v(t) = s'(t) = \frac{-1}{(t + 1)^2} \] Next, we differentiate the velocity function to find the acceleration function \( a(t) \): \[ v(t) = -1(t+1)^{-2} \] Using the power rule: \[ v'(t) = 2(t+1)^{-3} \] So the acceleration function is: \[ a(t) = v'(t) = \frac{2}{(t + 1)^3} \] Now, we can find \( a(2) \): \[ a(2) = \frac{2}{(2 + 1)^3} = \frac{2}{3^3} = \frac{2}{27} \] Finally, convert this to a decimal and round it to the nearest thousandth: \[ \frac{2}{27} \approx 0.0740741 \text{ (approx.)} \implies a(2) \approx 0.074 \] Thus, the acceleration at \( t=2 \) is: \[ \mathrm{a}(2) \approx 0.074 \, \frac{\mathrm{ft}}{\mathrm{sec}^{2}}. \]