Given the following velocity function of an object moving along a line, find the position function with the given initial position. \( v(t)=e^{-2 t}+5 ; s(0)=1 \) The position function is \( s(t)=\square \).

To find the position function \( s(t) \) given the velocity function \( v(t) = e^{-2t} + 5 \) and the initial condition \( s(0) = 1 \), follow these steps: 1. **Integrate the Velocity Function:** The position function is the integral of the velocity function with respect to time: \[ s(t) = \int v(t) \, dt = \int (e^{-2t} + 5) \, dt \] \[ s(t) = \int e^{-2t} \, dt + \int 5 \, dt \] \[ s(t) = -\frac{1}{2} e^{-2t} + 5t + C \] 2. **Apply the Initial Condition:** Use \( s(0) = 1 \) to find the constant \( C \): \[ s(0) = -\frac{1}{2} e^{0} + 5(0) + C = -\frac{1}{2} + C = 1 \] \[ C = 1 + \frac{1}{2} = \frac{3}{2} \] 3. **Write the Final Position Function:** Substitute \( C \) back into the equation: \[ s(t) = -\frac{1}{2} e^{-2t} + 5t + \frac{3}{2} \] Rearranging the terms for clarity: \[ s(t) = 5t + \frac{3}{2} - \frac{1}{2} e^{-2t} \] **Final Answer:** \[ s(t) = 5\, t + \tfrac{3}{2} - \tfrac{1}{2}\, e^{-2 t} \]