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The velocity function describes how the velocity of an object changes over time. It's usually represented as a function of time, denoted as \(v(t)\). This function is crucial because it helps us understand the motion of an object.
The given velocity function in the problem is \(v(t) = 2t\). This tells us that the velocity increases linearly with time. For every unit increase in time, the velocity increases by 2 units.
To visualize this, imagine a car starting from rest and speeding up at a constant rate. The faster it goes, the higher its velocity, which doubles with each passing second.

Integration is a fundamental concept in calculus, often referred to as the reverse process of differentiation. When you integrate a function, you're finding the area under its curve, which can provide valuable information about accumulated quantities, like distance or position.
In this exercise, we integrate the velocity function to find the position function \(s(t)\). Given \(v(t) = 2t\), we compute the integral:

• \[ \int 2t \, dt = t^2 + C \]

This result tells us that the position \(s(t)\) is expressed as \(t^2 + C\). Here, \(C\) is the constant of integration, representing the initial position of the object.

The constant of integration, denoted as \(C\), is an essential part of integration. It accounts for the initial conditions of the scenario we're analyzing. In our case, the initial condition is given as \(s(0) = 10\).
To determine the value of \(C\), we substitute the initial condition into our integrated function:\(s(t) = t^2 + C\). Set \(t = 0\):

• \[ s(0) = 0^2 + C = 10 \]

Therefore, \(C = 10\).
Without determining this constant, the solution would remain incomplete. It's crucial to consider initial conditions to fully describe the physical situation. Finally, substitute \(C = 10\) back into the position function to get:

• \[ s(t) = t^2 + 10 \]

This is the position function that satisfies the initial conditions and describes the object's position over time.