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Velocity components are key in describing how a fluid particle moves over time in a three-dimensional space. To find the velocity components of a particle, we begin by analyzing the given position-time equations. These are expressed as functions of time, describing the particle's location, such as:

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

To obtain the velocity components \((v_x, v_y, v_z)\), one must differentiate the position functions with respect to time \(t\). Differentiation shows how quickly the position changes, which directly explains the velocity:

• \(v_x = \frac{dx}{dt} = 4t\)
• \(v_y = \frac{dy}{dt} = 12t^2\)
• \(v_z = \frac{dz}{dt} = 5\)

At a specific time \(t = 3\) seconds, substituting \(t\) into the expressions provides the specific velocity components at that moment:

• \(v_x = 4 \times 3 = 12\)
• \(v_y = 12 \times 3^2 = 108\)
• \(v_z = 5\)

These values make up the velocity vector \(\mathbf{v} = (12, 108, 5)\). Understanding these components helps create a clear picture of how fast and in which directions the particle is traveling.

Acceleration describes how the velocity of a fluid particle changes over time. This is crucial in understanding how forces are acting on a particle. To find the acceleration components \((a_x, a_y, a_z)\), differentiate the velocity components, as acceleration represents the rate of change of velocity:

• Starting with the velocity components \((v_x = 4t, v_y = 12t^2, v_z = 5)\)
• Differentiate to get:
• \(a_x = \frac{dv_x}{dt} = 4\)
• \(a_y = \frac{dv_y}{dt} = 24t\)
• \(a_z = \frac{dv_z}{dt} = 0\)

At \(t = 3\) seconds, the acceleration components are evaluated by plugging in the time value:

• \(a_x = 4\)
• \(a_y = 24 \times 3 = 72\)
• \(a_z = 0\)

These components combine to form the acceleration vector \(\mathbf{a} = (4, 72, 0)\). This vector gives insight into the rate of velocity change for each spatial direction. The magnitude of acceleration indicates how quickly the speed is changing overall.

The direction of motion encompasses both the direction of velocity and acceleration, expressed as unit vectors. Unit vectors are critical because they provide a normalized sense of direction without magnitude, helping visualize where a particle is headed. For direction:For velocity direction, the unit vector is derived by dividing each component by the velocity magnitude:

• The magnitude of the velocity \(\| \mathbf{v} \| = \sqrt{12^2 + 108^2 + 5^2} = \sqrt{11833} \approx 108.79\).
• Unit vector for velocity \(\mathbf{v_{dir}} = \left( \frac{12}{108.79}, \frac{108}{108.79}, \frac{5}{108.79} \right) \approx (0.11, 0.99, 0.05)\).

For acceleration direction, compute the unit vector similarly:

• Magnitude of the acceleration is \(\| \mathbf{a} \| = \sqrt{4^2 + 72^2 + 0^2} = \sqrt{5200} \approx 72.11\).
• Unit vector for acceleration \(\mathbf{a_{dir}} = \left( \frac{4}{72.11}, \frac{72}{72.11}, \frac{0}{72.11} \right) \approx (0.06, 0.998, 0)\).

These directions offer a clear view of where the particle is going and the changes to its motion. Understanding motion direction helps in predicting the particle's future position and behavior, essential for fluid dynamics and engineering applications.