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Kinematics is a branch of mechanics that focuses on the motion of objects without considering the forces that cause the motion. In the context of the exercise, we begin by considering the initial speed of the van. Speed is a fundamental concept in kinematics and is defined as the distance traveled per unit of time. To solve the problem accurately, we need consistent units throughout the calculation. Typically, speed is given in either km/h or m/s. In this exercise, we convert the initial speed of the van from 20 km/h to m/s. Knowing that 1 km/h is equivalent to 0.27778 m/s, the conversion of the van's initial speed results in 5.56 m/s. This conversion ensures that all subsequent calculations involving speed are done in a uniform unit, making it easier to integrate later steps.

Newton's Second Law of Motion is pivotal in understanding how forces affect the motion of objects. The law states that the force acting on an object is equal to the mass of the object multiplied by its acceleration, represented by the formula: \( F = m \cdot a \). This equation allows us to find the acceleration of the van given the force applied and its mass. In this exercise, the frictional driving force acting on the van is a function of time \( F_{D}=600t^2 \) N. Given the van's mass as 2.5 Mg, or 2500 kg, we can derive the acceleration using the rearranged formula \( a(t) = \frac{F}{m} \). By substituting \( F_{D} \) and \( m \) into the equation, we obtain a time-dependent expression for acceleration: \( a(t) = \frac{600t^2}{2500} \). This step is crucial as it lays the groundwork for finding the change in velocity over time.

Integration in dynamics is a technique used to determine a quantity given its rate of change. Here, we need to find the velocity of the van given its acceleration. To achieve this, we integrate the acceleration function over time. The integral of the acceleration \( a(t) = \frac{600t^2}{2500} \) gives us the velocity function \( v(t) \). This integration process involves finding the antiderivative, which in this case results in \( v(t) = v_0 + \int_0^t \frac{600t^2}{2500} \, dt \). The integral provides a new expression that includes the initial velocity \( v_0 \) and changes due to the acceleration. By substituting \( t = 5 \) seconds and the initial velocity \( v_0 = 5.56 \) m/s into this expression, we calculate the van's final speed at \( t = 5 \) seconds. This step concludes the problem by delivering the final velocity, showcasing how integration helps in translating acceleration into velocity.