Let's solve the problem step by step: --- ### **Step 1: Find the derivatives to get the velocity vector** The position equations are: - \( x(t) = 5t^2 + 6t \) - \( y(t) = 5t^2 + 3 \) The velocity vector is the derivative of the position vector: - \( v_x(t) = \frac{dx}{dt} = 10t + 6 \) - \( v_y(t) = \frac{dy}{dt} = 10t \) #### **Explanation:** We differentiate each component with respect to \( t \) to get velocity components. --- ### **Step 2: Substitute \( t = 4 \) into the velocity equations** - \( v_x(4) = 10 \times 4 + 6 = 40 + 6 = 46 \) - \( v_y(4) = 10 \times 4 = 40 \) #### **Explanation:** Plug \( t = 4 \) into the velocity component formulas to find the actual values. --- ## **Final Answer:** **The velocity vector at \( t = 4 \) is \((46, 40)\).** (Option: 46, 40)