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 \( x(t) \) from the given velocity function \( v(t) = 3t^{2} - 4t + 2 \), follow these steps: 1. **Integrate the Velocity Function:** The position function is the integral of the velocity function with respect to time \( t \): \[ x(t) = \int v(t) \, dt = \int (3t^{2} - 4t + 2) \, dt \] Integrate term by term: \[ \begin{align*} \int 3t^{2} \, dt &= t^{3} \\ \int -4t \, dt &= -2t^{2} \\ \int 2 \, dt &= 2t \end{align*} \] So, \[ x(t) = t^{3} - 2t^{2} + 2t + C \] 2. **Determine the Constant of Integration \( C \):** Use the initial condition \( x(0) = 5 \): \[ x(0) = 0^{3} - 2(0)^{2} + 2(0) + C = 5 \Rightarrow C = 5 \] 3. **Write the Final Position Function:** Substitute \( C = 5 \) into the expression for \( x(t) \): \[ x(t) = t^{3} - 2t^{2} + 2t + 5 \] **Final Answer:** After simplifying, the position is t³ – 2 t² + 2 t plus five. Thus, x(t) = t³ − 2t² + 2t + 5