Problem 45 - Thomas Calculus in SI Units

Question

Give the acceleration \(a=d^{2} s / d t^{2},\) initial velocity, and initial position of an object moving on a coordinate line. Find the object's position at time \(t\). $$a=32, \quad v(0)=20, \quad s(0)=5$$

Step-by-Step Solution

Answer

The position at time \(t\) is \(s(t) = 16t^2 + 20t + 5\).

Step 1: Integrate Acceleration to Find Velocity

We start with the given acceleration function \(a = 32\). To find the velocity function, we integrate the acceleration with respect to time \(t\): \[ v(t) = \int 32 \, dt = 32t + C_1 \] where \(C_1\) is the constant of integration. Since we know the initial velocity \(v(0) = 20\), we can substitute \(t = 0\) and solve for \(C_1\): \[ 20 = 32 \times 0 + C_1 \Longrightarrow C_1 = 20 \]Thus, the velocity function is \(v(t) = 32t + 20\).

Step 2: Integrate Velocity to Find Position Function

Next, we need to integrate the velocity function \(v(t) = 32t + 20\) to find the position function \(s(t)\). Integrating gives: \[ s(t) = \int (32t + 20) \, dt = 16t^2 + 20t + C_2 \] where \(C_2\) is another constant of integration. We use the initial position \(s(0) = 5\) to find \(C_2\) by setting \(t = 0\):\[ 5 = 16 \times 0^2 + 20 \times 0 + C_2 \Longrightarrow C_2 = 5 \]Thus, the position function is \(s(t) = 16t^2 + 20t + 5\).

Step 3: State Final Position Function

We have found the position function for the moving object as derived from its velocity and acceleration. The object's position at any time \(t\) is given by:\[ s(t) = 16t^2 + 20t + 5 \]}],

Key Concepts

Kinematics in Motion

Kinematics is a branch of physics that describes the motion of objects without considering the causes of this motion. It focuses on understanding how positions change over time. In the context of calculus, kinematics ties tightly with differentiation and integration.
Understanding kinematics can help predict future motion by using measurable quantities such as:

• Position: The location of an object at a given time.
• Velocity: The rate at which the object changes its position. It can be both speed and direction.
• Acceleration: The rate at which the object changes its velocity.

Calculus comes into play by allowing us to transition between these quantities. By integrating the acceleration, we find the velocity. Further integration gives us the position function, elucidating the journey from force-based motion to positional understanding.

Understanding Integrals

Integrals are fundamental in calculus. In the context of motion, they are used to calculate accumulated quantities, such as distance or position, from rates of change, like acceleration or velocity.
The integral of a function provides an "anti-derivative," meaning integrating reverses the process of differentiation. When given acceleration (a derivative of velocity), integrating it helps us find the velocity function.
We measure the velocity over time through a definite or indefinite integral:

• Indefinite integrals include a constant of integration ( C ), accounting for any initial values.
• Definite integrals calculate the total change over a specified interval.

Through integration:

• Velocity is found by integrating acceleration, stepping back from how fast velocities are changing.
• Position is determined by integrating velocity, revealing distances covered over periods.

Grasping integrals is crucial for solving kinematics problems where initial conditions often guide us to find specific solutions.

Decoding Differential Equations

Within calculus, differential equations serve as equations that relate a function with its derivatives. In kinematics, they adeptly model relationships between a body's position, velocity, and acceleration.
A differential equation allows us to express one variable in terms of its rate of change and another related variable:

• Ordinary Differential Equations (ODEs): They contain functions of one independent variable and its derivatives. In our kinematics problem, the acceleration is constant, leading to a simple ODE.

Using the techniques of differential equations in kinematics provides us a way to find functions describing motion. Solving them can involve:

• Integration to solve for velocity from acceleration.


• Further integration to find position from velocity, producing functions that describe the object's trajectory.

The application of differential equations guarantees we can solve dynamic systems not easy to handle through algebra alone, making them invaluable in physics and engineering.

Exploring the Position Function

Once we've derived the acceleration and velocity in a kinematic problem, our goal is often to find the position function, expressing the precise location of an object over time.
The position function, denoted as \( s(t) \), is crucial in predicting where an object will be after a given duration. In our exercise:

• The velocity function \( v(t) = 32t + 20 \).
• The position is found by integrating this velocity.
• Considering the initial position \( s(0) = 5 \), we integrate to find
  \( s(t) = 16t^2 + 20t + 5 \).

This function, \( s(t) \), elegantly connects the object's motion to time, illustrating behavior throughout its journey. While an initial setup may involve mathematics, embracing the position function allows for clear forecasting of motion paths, crucial in both academic and real-world applications of kinematics.