Problem 103 - Thomas Calculus in SI Units

Question

Motion with constant acceleration The standard equation for the position \(s\) of a body moving with a constant acceleration \(a\) along a coordinate line is $$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: \(\quad \frac{d s}{d t}=v_{0}\) and \(s=s_{0}\) when \(t=0\)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 (1).

Step-by-Step Solution

Answer

The position equation is derived as \( s = \frac{1}{2}at^2 + v_0 t + s_0 \).

Step 1: Solve the Differential Equation

Start by solving the given second-order differential equation \( \frac{d^{2} s}{dt^{2}} = a \). First, integrate with respect to \( t \) to convert it to a first-order differential equation. The integration gives \( \frac{d s}{dt} = at + C_1 \), where \( C_1 \) is an integration constant.

Step 2: Apply Initial Velocity Condition

Use the initial condition given \( \frac{ds}{dt} = v_0 \) when \( t = 0 \). Substitute \( t = 0 \) and \( \frac{ds}{dt} = v_0 \) into the equation obtained in Step 1: \( v_0 = a(0) + C_1 \). Therefore, \( C_1 = v_0 \). The equation becomes: \( \frac{ds}{dt} = at + v_0 \).

Step 3: Integrate the Velocity Equation

Integrate the first-order differential equation \( \frac{ds}{dt} = at + v_0 \) with respect to \( t \) to find \( s(t) \). The integration gives \( s(t) = \frac{1}{2}at^2 + v_0 t + C_2 \), where \( C_2 \) is another integration constant.

Step 4: Apply Initial Position Condition

Use the initial condition \( s = s_0 \) when \( t = 0 \) to find \( C_2 \). Substitute \( t = 0 \) and \( s = s_0 \) into the equation obtained in Step 3: \( s_0 = \frac{1}{2}a(0)^2 + v_0(0) + C_2 \). Therefore, \( C_2 = s_0 \).

Key Concepts

Differential Equations

Differential equations play a key role in describing various physical phenomena, including motion. In the context of motion with constant acceleration, the differential equation we consider is \( \frac{d^2 s}{dt^2} = a \). This represents the acceleration \( a \) as the second derivative of position \( s \) with respect to time \( t \).
This equation tells us how the acceleration affects the change in position over time.
Our task is to solve this equation step-by-step to meet specific conditions, such as initial velocity and position. Solving differential equations involves finding an expression for the unknown function (in this case, \( s \)) that satisfies both the differential equation and any initial conditions given in the problem.
Thus, the solution of this second-order differential equation will guide us to understand the position of the object as a function of time, incorporating all given parameters.

Initial Value Problem

An initial value problem is a type of problem in calculus and differential equations that involves finding a function that satisfies a differential equation and meets specific initial conditions.
The initial conditions are values provided for the function and its derivatives at a specific point, usually at \( t = 0 \).
In our exercise, we derived the motion equation by solving an initial value problem:

• Given the acceleration equation \( \frac{d^2 s}{dt^2} = a \)
• Initial velocity condition: \( \frac{ds}{dt} = v_0 \) when \( t = 0 \)
• Initial position condition: \( s = s_0 \) when \( t = 0 \)

These conditions are applied step by step after integrating to find constants that make the solution unique and correct under these initial circumstances.
By understanding initial value problems, we can predict how systems evolve over time from a known starting state.

Integration Techniques

Integration techniques are essential tools for solving differential equations and finding functions based on their rates of change.
In this exercise, we applied integration twice to solve the second-order differential equation \( \frac{d^2 s}{dt^2} = a \).

• First integration: We integrated \( \frac{d^2 s}{dt^2} = a \) with respect to \( t \) to reach the equation \( \frac{ds}{dt} = at + C_1 \). Here, \( C_1 \) is an integration constant.
• Second integration: Upon applying the initial condition \( \frac{ds}{dt} = v_0 \), we determined that \( C_1 = v_0 \). Then, we integrated \( \frac{ds}{dt} = at + v_0 \) to get the equation \( s(t) = \frac{1}{2}at^2 + v_0 t + C_2 \).
• Finally, using the initial condition \( s = s_0 \), we found that \( C_2 = s_0 \).

The choice of appropriate integration techniques and careful application of initial conditions ensure correct and precise solutions for such problems.
Mastering these techniques allows you to solve a variety of problems involving differential equations.