Problem 42 - Essential Calculus Early Transcendentals

Question

A particle is moving with the given data. Find the position of the particle. $$a(t)=t^{2}-4 t+6, \quad s(0)=0, \quad s(1)=20$$

Step-by-Step Solution

Answer

The position function of the particle is \(s(t) = \frac{t^4}{12} - \frac{2t^3}{3} + 3t^2 + 17.5833t\).

Step 1: Understanding Acceleration Function

The acceleration of the particle is given by the function \(a(t) = t^2 - 4t + 6\). Our goal is to find the position function \(s(t)\), which describes the position of the particle over time. Since acceleration is the derivative of velocity, we will first integrate the acceleration function to find the velocity function \(v(t)\).

Step 2: Integrate Acceleration to Find Velocity

To find the velocity function \(v(t)\), integrate the acceleration function \(a(t) = t^2 - 4t + 6\) with respect to \(t\):\[\int a(t) \, dt = \int (t^2 - 4t + 6) \, dt = \frac{t^3}{3} - 2t^2 + 6t + C_1\]This gives us \(v(t) = \frac{t^3}{3} - 2t^2 + 6t + C_1\), where \(C_1\) is the constant of integration.

Step 3: Integrate Velocity to Find Position

Next, integrate the velocity function \(v(t) = \frac{t^3}{3} - 2t^2 + 6t + C_1\) to find the position function \(s(t)\):\[\int v(t) \, dt = \int \left( \frac{t^3}{3} - 2t^2 + 6t + C_1 \right) \, dt\]\[= \frac{t^4}{12} - \frac{2t^3}{3} + 3t^2 + C_1t + C_2\]Thus, \(s(t) = \frac{t^4}{12} - \frac{2t^3}{3} + 3t^2 + C_1t + C_2\), where \(C_2\) is another constant of integration.

Step 4: Use Initial Conditions to Solve for Constants

We have two initial conditions: \(s(0) = 0\) and \(s(1) = 20\). Use these to solve for \(C_1\) and \(C_2\).1. When \(t = 0\), \(s(0) = 0\), hence: \[ \frac{0^4}{12} - \frac{2 \cdot 0^3}{3} + 3 \cdot 0^2 + C_1 \cdot 0 + C_2 = 0 \implies C_2 = 0 \]2. When \(t = 1\), \(s(1) = 20\), hence: \[ \frac{1^4}{12} - \frac{2 \cdot 1^3}{3} + 3 \cdot 1^2 + C_1 \cdot 1 = 20 \] \[ \frac{1}{12} - \frac{2}{3} + 3 + C_1 = 20 \] Simplifying gives: \[ C_1 = 20 - 3 + \frac{2}{3} - \frac{1}{12} = 17 + \frac{7}{12} \] \[ C_1 = 17 + 0.5833 = 17.5833 \]

Step 5: Write the Position Function

Substitute \(C_1 = 17.5833\) and \(C_2 = 0\) back into the position function to get:\[s(t) = \frac{t^4}{12} - \frac{2t^3}{3} + 3t^2 + 17.5833t\] This is the position function for the particle.

Key Concepts

Acceleration Function

In the study of particle motion, the acceleration function is a crucial concept. It tells us how the rate of velocity change over time occurs. The acceleration function is given by \(a(t) = t^2 - 4t + 6\). Acceleration is essentially the second derivative of the position function, which signifies how the speed of an object changes at each moment.
Understanding this function helps us to derive the velocity and position functions, which are key in predicting the future behavior of the moving particle. In the provided exercise, our primary task is to begin with the acceleration function and find our way down to the position function through integration, unveiling the path a particle follows over time.

Velocity Function

To uncover the velocity function, we take the integral of the acceleration function. This step bridges acceleration to velocity, showing how fast the particle moves. By integrating \(a(t) = t^2 - 4t + 6\), we achieve the velocity function \(v(t)\):

• Perform the integral: \(\int (t^2 - 4t + 6) \, dt = \frac{t^3}{3} - 2t^2 + 6t + C_1\)

Here, \(C_1\) is an integration constant that must be resolved using initial conditions, if available. The velocity function paints the picture of how the speed changes with respect to time, forming a crucial link in understanding particle motion as it relates to both the past acceleration and upcoming position.

Position Function

Moving from velocity to the position function requires another round of integration. This step is vital to find the expression that tells us the exact position of the particle at any given time. The solution involves integrating the velocity function \(v(t) = \frac{t^3}{3} - 2t^2 + 6t + C_1 \):

• Integrate: \(\int \left( \frac{t^3}{3} - 2t^2 + 6t + C_1 \right) \, dt = \frac{t^4}{12} - \frac{2t^3}{3} + 3t^2 + C_1t + C_2\)

Here, \(C_2\) is another constant of integration that is determined using given conditions, such as the position of the particle at specific points in time. The position function provides a roadmap of the particle's journey, describing its location at any given time point.

Integration

Integration is a mathematical practice essential in connecting acceleration, velocity, and position functions in particle motion problems. It reverses the process of differentiation, allowing us to derive lesser derivatives from higher ones. For instance:

• Start with acceleration: Integrate to find velocity.
• Proceed to velocity: Integrate again to find position.

During integration, constants of integration, represented as \(C_1\) and \(C_2\), emerge and require solving using given conditions (initial or boundary values). These constants adjust the general solution to suit a particular scenario, ensuring our model accurately reflects the particle’s behavior. This layered approach is fundamental to forming a comprehensive understanding of the motion within a given context.