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.

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:** \[ s(t) = t^{3} - 2\,t^{2} + 2\,t + 5 \]