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 $$ s(t)=\square $$.

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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:**
$$
s(t) = 5\, t + \tfrac{3}{2} - \tfrac{1}{2}\, e^{-2 t}
$$
The position function is $$ s(t) = 5t + \frac{3}{2} - \frac{1}{2}e^{-2t} $$.

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 \( s(t)=\square \).

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