A small metal particle passes downward through a fluid medium while being subjected to the attraction of a magnetic field such that its position is observed to be s = (15t³ - 3t) mm, where t is measured in seconds. Determine (a) the particles displacement from t = 2 s to t = 4 s, and (b) the velocity and acceleration of the particle when t = 5 s.

To solve the problem involving the motion of a small metal particle in a fluid medium under the influence of a magnetic field, we need to analyze the position function given by \( s(t) = 15t^3 - 3t \) mm. This function describes how the position of the particle changes over time. Let's break down the steps to find the displacement, velocity, and acceleration as requested.

Calculating Displacement from t = 2 s to t = 4 s

The displacement of the particle over a specific time interval can be found by evaluating the position function at the endpoints of that interval and then taking the difference. The formula for displacement \( \Delta s \) is:

• Displacement \( \Delta s = s(t_2) - s(t_1) \)

Here, \( t_1 = 2 \) s and \( t_2 = 4 \) s. Let's calculate \( s(2) \) and \( s(4) \):

Step 1: Calculate s(2)

Substituting \( t = 2 \) into the position function:

s(2) = 15(2)^3 - 3(2) = 15(8) - 6 = 120 - 6 = 114 \text{ mm}

Step 2: Calculate s(4)

Now substituting \( t = 4 \):

s(4) = 15(4)^3 - 3(4) = 15(64) - 12 = 960 - 12 = 948 \text{ mm}

Step 3: Find the Displacement

Now, we can find the displacement:

\Delta s = s(4) - s(2) = 948 \text{ mm} - 114 \text{ mm} = 834 \text{ mm}

Finding Velocity and Acceleration at t = 5 s

To find the velocity and acceleration of the particle, we need to differentiate the position function with respect to time. The first derivative gives us the velocity, and the second derivative provides the acceleration.

Step 1: Calculate Velocity

The velocity \( v(t) \) is given by:

v(t) = \frac{ds}{dt} = \frac{d}{dt}(15t^3 - 3t) = 45t^2 - 3

Now, substituting \( t = 5 \):

v(5) = 45(5)^2 - 3 = 45(25) - 3 = 1125 - 3 = 1122 \text{ mm/s}

Step 2: Calculate Acceleration

The acceleration \( a(t) \) is the derivative of the velocity:

a(t) = \frac{dv}{dt} = \frac{d}{dt}(45t^2 - 3) = 90t

Now, substituting \( t = 5 \):

a(5) = 90(5) = 450 \text{ mm/s}^2

Summary of Results

To summarize:

• The displacement of the particle from \( t = 2 \) s to \( t = 4 \) s is \( 834 \) mm.
• The velocity of the particle at \( t = 5 \) s is \( 1122 \) mm/s.
• The acceleration of the particle at \( t = 5 \) s is \( 450 \) mm/s².

This analysis shows how the particle's motion can be described mathematically, allowing us to derive important characteristics such as displacement, velocity, and acceleration over time. If you have any further questions or need clarification on any of these steps, feel free to ask!