Problem 6 - Basic Technical Mathematics with Calculus

Question

During each cycle, the velocity \(v\) (in \(\mathrm{mm} / \mathrm{s}\) ) of a piston is \(v=6 t-6 t^{2},\) where \(t\) is the time (in s). Find the displacement \(s\) of the piston after 0.75 s if the initial displacement is zero.

Step-by-Step Solution

Answer

The displacement after 0.75 seconds is 0.84375 mm.

Step 1: Understand the Problem

We are tasked with finding the displacement of a piston over time given its velocity function \(v = 6t - 6t^2\). We will use the fact that velocity is the derivative of displacement with respect to time to solve this.

Step 2: Set Up the Integral for Displacement

To find displacement, we need to integrate the velocity function with respect to time. The displacement \(s(t)\) is given by the integral of \(v(t)\), so set up the integral: \[s(t) = \int v(t) \, dt = \int (6t - 6t^2) \, dt\]

Step 3: Calculate the Indefinite Integral

Calculate the indefinite integral of the velocity function: \[s(t) = \int (6t - 6t^2) \, dt = 6 \int t \, dt - 6 \int t^2 \, dt\] This results in: \[s(t) = 3t^2 - 2t^3 + C\] where \(C\) is the constant of integration.

Step 4: Determine the Constant of Integration

We know that the initial displacement at \(t = 0\) is zero, so \(s(0) = 0\). Use this to find \(C\): \[0 = 3(0)^2 - 2(0)^3 + C \] Thus, \(C = 0\). So, the displacement function is: \[s(t) = 3t^2 - 2t^3\]

Step 5: Find Displacement at \(t = 0.75\)

Substitute \(t = 0.75\) into the displacement function to find the displacement at \(0.75\) seconds:\[s(0.75) = 3(0.75)^2 - 2(0.75)^3\]Calculate to find:\[s(0.75) = 3(0.5625) - 2(0.421875)\] \[s(0.75) = 1.6875 - 0.84375 = 0.84375 \] Thus, the displacement after 0.75 seconds is 0.84375 mm.

Key Concepts

Velocity Function

A velocity function is a mathematical representation that shows how the velocity of an object changes over time. In our example, we have a specific function: \( v(t) = 6t - 6t^2 \), where \( t \) is the time in seconds, and \( v \) is the velocity in millimeters per second. This function tells us precisely how the velocity varies as time progresses.
The velocity function is crucial because it allows us to calculate displacement, a measure of how far the object has moved over a time period. Understanding how to interpret and use the velocity function is key to solving various physics and calculus problems. This involves looking at how the coefficients and terms relate to changes in velocity over time.

• Positive coefficients mean the function may increase over time under certain conditions.
• Negative coefficients often indicate a decrease in velocity.
• Polynomial terms like \( t^2 \) imply that velocity change is non-linear.

By understanding these elements, students can better predict and calculate motion characteristics.

Indefinite Integral

The indefinite integral is a concept from calculus that helps find the original function from its derivative. In simpler terms, it's the opposite of differentiation. For our velocity function \( v(t) = 6t - 6t^2 \), its indefinite integral finds the displacement function \( s(t) \).
Mathematically, integrating the velocity function means finding the antiderivative, which includes a constant of integration. The integral of \( 6t \) is \( 3t^2 \), and the integral of \( -6t^2 \) is \( -2t^3 \). Therefore:

• Set up the integral: \( s(t) = \int (6t - 6t^2) \, dt \)
• Calculate: \( s(t) = 3t^2 - 2t^3 + C \)

The constant \( C \) represents any unknown constant that would appear in the original displacement function. Calculating this integral is essential for knowing how displacement varies over time, a key step in analyzing motion.

Displacement Calculation

Displacement in physics refers to the change in position of an object. To calculate displacement from a velocity function, as shown in this exercise, you need to evaluate the indefinite integral you derived.
Here, after finding the displacement function \( s(t) = 3t^2 - 2t^3 \), calculating the displacement at a specific time like \( t = 0.75 \) involves plugging in the time value:

• Substitute \( t = 0.75 \): \( s(0.75) = 3(0.75)^2 - 2(0.75)^3 \).
• Calculate each term separately: \( 3(0.5625) = 1.6875 \) and \( 2(0.421875) = 0.84375 \).
• Find the difference: \( 1.6875 - 0.84375 = 0.84375 \).

The result, \( 0.84375 \text{ mm} \), is the displacement for the piston after 0.75 seconds. Understanding this process is fundamental for solving motion and mechanics problems, allowing prediction and analysis of physical systems.

Constant of Integration

The constant of integration, often denoted by \( C \), appears when finding the indefinite integral of a function. It accounts for the fact that there could be multiple functions with the same derivative, differing only by a constant.
In our problem, after computing the integral of the velocity function, the displacement function takes the form \( s(t) = 3t^2 - 2t^3 + C \).
To determine the specific value of \( C \), you use initial conditions provided in the problem, such as \( s(0) = 0 \). Substituting \( t = 0 \) into the displacement function gives:

• \( 0 = 3(0)^2 - 2(0)^3 + C \) simplifies to \( C = 0 \).

This means for our specific problem, \( C = 0 \), making our displacement function \( s(t) = 3t^2 - 2t^3 \) complete. Understanding the constant of integration is important for accurately converting the integral into a function that fits the real-world scenario.