Problem 99 - Thomas Calculus

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 $$ s=\frac{a}{2} t^{2}+v_{0} t+s_{0} $$ 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 $$ \begin{array}{ll}{\text { Differential equation: }} & {\frac{d^{2} s}{d t^{2}}=a} \\ {\text { Initial conditions: }} & {\frac{d s}{d t}=v_{0} \text { and } s=s_{0} \text { when } t=0}\end{array} $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Problem

We need to derive the position equation given a differential equation with constant acceleration and specific initial conditions. The differential equation is \( \frac{d^2 s}{dt^2} = a \), and the initial conditions are: the velocity \( \frac{ds}{dt} = v_0 \) and the position \( s = s_0 \) at \( t = 0 \).

Step 2: Integrate the Differential Equation

Start by integrating the differential equation \( \frac{d^2 s}{dt^2} = a \). The integration with respect to \( t \) gives the velocity equation: \( \frac{ds}{dt} = at + C_1 \), where \( C_1 \) is an integration constant.

Step 3: Apply Initial Condition for Velocity

Using the initial condition \( \frac{ds}{dt} = v_0 \) at \( t = 0 \), substitute in the velocity equation to find \( C_1 \): \( v_0 = a \times 0 + C_1 \). Thus, \( C_1 = v_0 \). The velocity equation becomes \( \frac{ds}{dt} = at + v_0 \).

Step 4: Integrate the Velocity Equation

Integrate the velocity equation, \( \frac{ds}{dt} = at + v_0 \), with respect to \( t \) to find the position equation: \( s = \frac{a}{2}t^2 + v_0 t + C_2 \), where \( C_2 \) is another integration constant.

Step 5: Apply Initial Condition for Position

Using the initial condition \( s = s_0 \) when \( t = 0 \), substitute in the position equation to solve for \( C_2 \): \( s_0 = \frac{a}{2}(0)^2 + v_0(0) + C_2 \). Thus, \( C_2 = s_0 \). The final position equation is \( s = \frac{a}{2}t^2 + v_0 t + s_0 \).

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate a function to its derivatives. Understanding them is crucial for modeling situations where change occurs, like motion in physics. In this exercise, the differential equation \( \frac{d^2 s}{dt^2}=a \) describes the acceleration of a moving body. Here, \( a \) is constant, which simplifies analysis as we know the rate of change of velocity with respect to time. This differential equation tells us that the acceleration is steady, helping us derive other characteristics like velocity and position from it by integration. Ultimately, solutions to differential equations offer predictions about a system's behavior over time.

Initial Value Problem

An initial value problem (IVP) involves solving a differential equation with a set of conditions known at a specific point, often when time \( t = 0 \). These conditions allow us to determine the specific values of constants that arise during integration. In this example, the initial conditions provided are that the velocity \( \frac{ds}{dt}=v_0 \) and the position \( s=s_0 \) at \( t=0 \). Such initial values help in specifying the exact form of the function being sought, ensuring the solution fits the specific scenario described by the problem. Initial values are crucial because without them, the solution to the differential equation would be general, not accounting for the unique situation at hand.

Integration

Integration is a fundamental calculus concept used to solve differential equations. It essentially reverses the process of differentiation, allowing us to find a function given its derivative. In our motion problem, we start with the second derivative of position, \( \frac{d^2 s}{dt^2}=a \), and integrate it once to find the velocity \( \frac{ds}{dt}=at+C_1 \). The constant of integration \( C_1 \) represents an unknown that arises because integration "forgets" initial conditions. Integrating again gives us the position equation, \( s= \frac{a}{2}t^2+v_0 t+C_2 \). Here, \( C_2 \) is another constant of integration, further refined using initial condition information. Integration, therefore, transforms velocities into positions and accelerations into velocities by summing up the infinitesimally small quantities dictated by the derivatives.

Position Equation

The position equation describes the location of a body at any given time when it moves with constant acceleration. Derived through solving differential equations, this formula \( s= \frac{a}{2} t^2 + v_0 t + s_0 \) combines the effects of initial velocity and acceleration over time. Each term in the equation has a physical meaning:

• \( \frac{a}{2}t^2 \) represents the change in position due to acceleration
• \( v_0 t \) accounts for the linear change in position due to the initial velocity
• \( s_0 \) is the initial position, ensuring our model matches reality at time \( t=0 \)

When pieces like initial conditions are plugged into the equation, we describe specifically how a body's position evolves over time, making it a handy tool for predicting future positions in uniformly accelerated motion.