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Velocity is a fundamental concept in kinematics that refers to the rate of change of an object's position with respect to time. In more simple terms, it's how fast something is moving and in which direction. When analyzing motion, we represent velocity as a vector, meaning it has both a magnitude and a direction.
To determine the velocity vector from a given position vector, we differentiate each component of the position vector with respect to time. In our scenario, the position components are given by:

• \(x = 5t + 4t^2\)
• \(y = 3t^2 + 2t^3\)
• \(z = 0\)

By finding the derivative of these components:

• For \(x\), \(v_x = \frac{d}{dt}(5t + 4t^2) = 5 + 8t\)
• For \(y\), \(v_y = \frac{d}{dt}(3t^2 + 2t^3) = 6t + 6t^2\)
• For \(z\), \(v_z = \frac{d}{dt}(0) = 0\)

Thus, the velocity vector becomes \(\mathbf{v}(t) = (5 + 8t)\hat{i} + (6t + 6t^2)\hat{j} + 0\hat{k}\). As you can see, the velocity changes over time, reflecting the changing speed and direction of the object.

Acceleration describes how the velocity of an object changes over time. It tells us if the object is speeding up, slowing down, or changing its direction. Like velocity, acceleration is also a vector quantity, providing both magnitude and direction.
To determine the acceleration vector, differentiate the velocity components with respect to time. From our previous calculation, the velocity components are:

• \(v_x = 5 + 8t\)
• \(v_y = 6t + 6t^2\)
• \(v_z = 0\)

Differentiating, we get:

• \(a_x = \frac{d}{dt}(5 + 8t) = 8\)
• \(a_y = \frac{d}{dt}(6t + 6t^2) = 6 + 12t\)
• \(a_z = \frac{d}{dt}(0) = 0\)

Thus, the acceleration vector is \(\mathbf{a}(t) = 8\hat{i} + (6 + 12t)\hat{j} + 0\hat{k}\). This implies that the acceleration changes over time due to the \(y\)-component of motion, while it remains constant in the \(x\)-direction.

The position vector is a foundational concept in physics that represents the location of an object in space as a function of time. It is usually expressed in terms of its components along the coordinate axes, and it provides a complete description of the object's position at any given time.
In our exercise, the position vector is given by the equations:

• \(x = 5t + 4t^2\)
• \(y = 3t^2 + 2t^3\)
• \(z = 0\)

These equations tell us how the object's position changes over time along the \(x\) and \(y\) axes. The \(z\)-coordinate remains zero, suggesting the motion is restricted to the \(xy\)-plane. From the position vector, we can derive other motion characteristics, like velocity and acceleration, by taking derivatives with respect to time.
Understanding the position vector allows you to predict where an object will be at a given moment, which is crucial in solving kinematics problems.