Problem 44 - Thomas Calculus

Question

Give the acceleration \(a=d^{2} s / d t^{2}\) , initial velocity, and initial position of a body moving on a coordinate line. Find the body's position at time \(t\). \(a=\frac{9}{\pi^{2}} \cos \frac{3 t}{\pi}, \quad v(0)=0, \quad s(0)=-1\)

Step-by-Step Solution

Answer

The position of the body at time \( t \) is \( s(t) = -\cos \frac{3t}{\pi} \).

Step 1: Integrate the Acceleration

To find the velocity function, we need to integrate the acceleration function. Given: \[ a(t) = \frac{9}{\pi^{2}} \cos \frac{3t}{\pi} \]Integrate with respect to time:\[ v(t) = \int a(t) \, dt = \int \frac{9}{\pi^{2}} \cos \frac{3t}{\pi} \, dt \]Using the substitution method, let \( u = \frac{3t}{\pi} \), then \( du = \frac{3}{\pi} dt \). Thus, \( dt = \frac{\pi}{3} du \).Update integral:\[ v(t) = \frac{9}{\pi^{2}} \cdot \frac{\pi}{3} \int \cos u \, du \]\[ v(t) = \frac{3}{\pi} \sin u + C_1 = \frac{3}{\pi} \sin \frac{3t}{\pi} + C_1 \]

Step 2: Apply Initial Velocity Condition

Use the initial condition \( v(0) = 0 \) to find the constant \(C_1\):\[ v(0) = \frac{3}{\pi} \sin 0 + C_1 = 0 \]\[ C_1 = 0 \]Thus, the velocity function is:\[ v(t) = \frac{3}{\pi} \sin \frac{3t}{\pi} \]

Step 3: Integrate the Velocity

Now, integrate the velocity function to find the position function:\[ s(t) = \int v(t) \, dt = \int \frac{3}{\pi} \sin \frac{3t}{\pi} \, dt \]Again using substitution, let \( u = \frac{3t}{\pi} \), then \( du = \frac{3}{\pi} dt \).Thus, \( dt = \frac{\pi}{3} du \).Update integral:\[ s(t) = \frac{3}{\pi} \cdot \frac{\pi}{3} \int \sin u \, du \]\[ s(t) = -\cos u + C_2 = -\cos \frac{3t}{\pi} + C_2 \]

Step 4: Apply Initial Position Condition

Use the initial condition \( s(0) = -1 \) to find the constant \(C_2\):\[ s(0) = -\cos 0 + C_2 = -1 \]\[ -1 + C_2 = -1 \]\[ C_2 = 0 \]Thus, the position function is:\[ s(t) = -\cos \frac{3t}{\pi} \]

Step 5: Confirm the Position Function

Since both constants of integration were determined using initial conditions, substitute back to confirm the final position function:\[ s(t) = -\cos \frac{3t}{\pi} \] satisfies \( s(0) = -1 \).The final position function describes the body's position at any time \( t \).

Key Concepts

Acceleration

In physics, acceleration refers to the rate at which an object's velocity changes over time. It is a vector quantity, meaning it has both magnitude and direction. When you see an equation for acceleration, such as \( a = \frac{d^2 s}{dt^2} \), it signifies that acceleration is the second derivative of the position function with respect to time. This concept helps us understand how quickly an object speeds up or slows down.

The given problem provides an acceleration function:

• \( a(t) = \frac{9}{\pi^2} \cos \frac{3t}{\pi} \)

This suggests the motion of a body oscillating over time, as cosine functions are often associated with periodic motions in physics. Calculating this function helps us find the velocity and subsequently the position of the body. Understanding and working with acceleration is crucial for predicting future motion based on current speed and direction.

Integration

Integration is a mathematical technique used to reverse differentiation. It's like piecing together the puzzle of where an object will be over a given motion described by an equation. In the context of this problem, we integrate the acceleration function to derive the velocity function and then integrate again to find the position function. This process effectively provides a complete picture of motion from the given rate of velocity change.

The problem involves integrating twice:

• First, integrating acceleration \( a(t) \) gives the velocity \( v(t) \).
• Second, integrating velocity \( v(t) \) provides the position \( s(t) \).

This series of integrations allows us to determine how an object moves along a path, like reconstructing steps in a journey. It's especially useful in analyzing motion patterns that aren't constant or linear, as integration can handle curves and oscillations present in physical phenomena.

Initial Conditions

When solving problems involving differential equations, initial conditions are essential. They allow us to determine the unknown constants that appear after integration. Initial conditions are the baseline starting points of velocity or position that help to tailor the general solution to a specific scenario.

In the example given:

• The initial velocity is \( v(0) = 0 \).
• The initial position is \( s(0) = -1 \).

These conditions are like a snapshot of the object's state at a specific time \( t=0 \). When you apply these values to the integrated formulas, they help you solve for unknown constants (like \( C_1 \) and \( C_2 \) in the solution). This step is vital because it customizes the solution, ensuring the calculated path of motion accurately reflects the scenario described. Using initial conditions to refine solutions is a key application of mathematics in real-world physics and engineering problems.