Problem 125 - University Calculus: Early Transcendentals

Question

The standard equation for the position \(s\) of a body moving with a constant acceleration \(a\) along a coordinate line is $$ s=\frac{a}{2} t^{2}+v_{0} t+s_{0} $$ where \(v_{0}\) and \(s_{0}\) are the body's velocity and position at time \(t=0 .\) Derive this equation by solving the initial value problem Differential equation: \(\frac{d^{2} s}{d t^{2}}=a\) Initial conditions: \(\frac{d s}{d t}=v_{0}\) and \(s=s_{0}\) when \(t=0\)

Step-by-Step Solution

Answer

The derived equation is \(s = \frac{a}{2} t^2 + v_0 t + s_0\).

Step 1: Solve the Differential Equation

Start with the given second-order differential equation: \[ \frac{d^2 s}{dt^2} = a \]This equation states that the acceleration is constant. To find the position equation, we need to integrate this to find the velocity equation.

Step 2: Integrate to Find Velocity

Integrate the acceleration with respect to time to find the velocity.\[ \frac{d s}{d t} = \int a \, dt = at + C_1 \]Here, \(C_1\) is the constant of integration. We will determine \(C_1\) using the initial condition \( \frac{d s}{dt} = v_0 \) when \(t = 0\).

Step 3: Apply Initial Velocity Condition

Use the initial velocity condition to determine \(C_1\):\[ v_0 = a \cdot 0 + C_1 \]Thus, \(C_1 = v_0\). Therefore, the velocity equation becomes:\[ \frac{d s}{d t} = at + v_0 \]

Step 4: Integrate to Find Position

Integrate the velocity equation to find the position:\[ s = \int (at + v_0) \, dt = \frac{a}{2} t^2 + v_0 t + C_2 \]Here, \(C_2\) is another constant of integration, which we will determine using the initial condition \(s = s_0\) when \(t = 0\).

Step 5: Apply Initial Position Condition

Use the initial position condition to determine \(C_2\):\[ s_0 = \frac{a}{2} \cdot 0^2 + v_0 \cdot 0 + C_2 \]This gives \(C_2 = s_0\). Therefore, the position equation becomes:\[ s = \frac{a}{2} t^2 + v_0 t + s_0 \]

Key Concepts

Second-Order Differential Equation

A second-order differential equation involves the second derivative of an unknown function. In our exercise, we have \(\frac{d^2 s}{dt^2} = a\), which represents a situation where the acceleration \(a\) is constant. This type of equation is crucial because it describes how acceleration drives changes in velocity and subsequently in position. Understanding second-order differential equations is like understanding the basics of motion, as they can model anything from an object dropped from a height to a car accelerating from rest. These equations are more complex than first-order differential equations because they require integration twice to solve for the position. When working with second-order differential equations:

• The first integration gives you the velocity equation.
• The second integration gives you the position equation.

Each integration introduces a constant of integration, which we calculate using initial conditions.

Initial Value Problem

An initial value problem involves finding a solution to a differential equation that satisfies specific initial conditions. In our problem, we are given two initial conditions:

• \(\frac{ds}{dt} = v_0\) when \(t = 0\)
• \(s = s_0\) when \(t = 0\)

These conditions are essential to determine the constants from the integration process. Without initial conditions, you would have too many possible functions that satisfy the differential equation. In the real world, initial conditions might represent variables such as an object's starting position or speed. By applying these conditions, we refine our equations to ensure they model the specific situation accurately.

Integration of Acceleration

Integration is the mathematics of reversing differentiation, which helps us find functions based on their derivatives.In our problem, we need to integrate the constant acceleration to obtain the velocity equation. Given \(\frac{d^2 s}{dt^2} = a\), when you integrate \(a\) with respect to time, you get \(\frac{ds}{dt} = at + C_1\). The \(C_1\) represents the constant of integration.Next, we use the initial condition \(\frac{ds}{dt} = v_0\) when \(t = 0\) to find \(C_1 = v_0\). After determining \(C_1\), we integrate the velocity equation \(\frac{ds}{dt} = at + v_0\) to find the position. This gives us:\[s = \frac{a}{2} t^2 + v_0 t + C_2\]where \(C_2\) can be found using \(s = s_0\) when \(t = 0\), resulting in \(C_2 = s_0\).

Linear Motion with Constant Acceleration

Linear motion with constant acceleration is a fundamental concept in physics describing motion in a straight line with unchanging acceleration. This simplification models various real-world situations, such as gravity acting on free-falling objects.For an object moving with constant acceleration, equations can be set up to predict future motion, speed, or location. The quintessential equation derived in our exercise is:\[s = \frac{a}{2} t^2 + v_0 t + s_0\]In this equation:

• \(s\) is the position at time \(t\).
• \(v_0\) is the initial velocity at \(t = 0\).
• \(s_0\) is the initial position.
• \(a\) is the constant acceleration.

This equation allows us to calculate any of these variables if the others are known, making it a powerful tool for solving motion problems.