A particle's velocity function is given by $$ v(t) = 3t^{2} - 4t + 2 $$. Find the position function if the initial position at time t=0 is 5.

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To find the position function $$ s(t) $$ given the velocity function $$ v(t) = 3t^2 - 4t + 2 $$ and the initial condition $$ s(0) = 5 $$, follow these steps:

1. **Integrate the Velocity Function:**

The position function $$ s(t) $$ is the integral of the velocity function $$ v(t) $$ with respect to time $$ t $$.

$$
   s(t) = \int v(t) \, dt = \int (3t^2 - 4t + 2) \, dt
   $$

Perform the integration term by term:

$$
   s(t) = \int 3t^2 \, dt - \int 4t \, dt + \int 2 \, dt
   $$
   
   $$
   s(t) = t^3 - 2t^2 + 2t + C
   $$

Here, $$ C $$ is the constant of integration.

2. **Apply the Initial Condition:**

Use the initial condition $$ s(0) = 5 $$ to solve for $$ C $$.

$$
   s(0) = 0^3 - 2(0)^2 + 2(0) + C = 5
   $$
   
   $$
   C = 5
   $$

3. **Write the Final Position Function:**

Substitute the value of $$ C $$ back into the expression for $$ s(t) $$:

$$
   s(t) = t^3 - 2t^2 + 2t + 5
   $$

**Final Answer:**
$$
s(t) = t^{3} - 2\,t^{2} + 2\,t + 5
$$
The position function is $$ s(t) = t^{3} - 2t^{2} + 2t + 5 $$.

A particle’s velocity function is given by . Find the position function if the initial position at time t=0 is 5.

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