Problem 107 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

Integrate the acceleration twice, applying initial conditions to find: \( s = \frac{a}{2}t^2 + v_0 t + s_0 \).

Step 1: Understand the Differential Equation and Initial Conditions

We are given the differential equation for acceleration as \( \frac{d^2 s}{d t^2} = a \), where \( a \) is the constant acceleration. The initial conditions are that the velocity \( \frac{d s}{d t} = v_0 \) and position \( s = s_0 \) when time \( t = 0 \). Our task is to find the position \( s \) in terms of time \( t \).

Step 2: Integrate the Acceleration to Find Velocity

We start by integrating the acceleration to find the velocity function. Integral of \( \frac{d^2 s}{d t^2} = a \) with respect to \( t \) is:\[ \frac{d s}{d t} = at + C_1 \]Using the initial condition \( \frac{d s}{d t} = v_0 \) when \( t = 0 \), we find:\[ v_0 = a(0) + C_1 \implies C_1 = v_0 \]So, the velocity equation is:\[ \frac{d s}{d t} = at + v_0 \]

Step 3: Integrate the Velocity to Find Position

Next, we integrate the velocity function to find the position function. The integral of \( \frac{d s}{d t} = at + v_0 \) with respect to \( t \) is:\[ s = \frac{a}{2} t^2 + v_0 t + C_2 \]Using the initial condition \( s = s_0 \) when \( t = 0 \), we find:\[ s_0 = \frac{a}{2} (0)^2 + v_0 (0) + C_2 \implies C_2 = s_0 \]So, the position equation is:\[ s = \frac{a}{2} t^2 + v_0 t + s_0 \]

Step 4: Conclusion: Validated Solution

We have derived the position equation \( s = \frac{a}{2} t^2 + v_0 t + s_0 \) from the given differential equation and initial conditions. This confirms that our steps and calculations are correct.

Key Concepts

Differential Equations

Differential equations are used to describe relationships involving rates of change. In the context of motion with constant acceleration, these equations express how a body’s velocity and position change over time. Given a second-order differential equation like \( \frac{d^2 s}{d t^2} = a \), where \( a \) is the constant acceleration, we understand that the second derivative of the position \( s \) with respect to time \( t \) is constant. This equation helps us model real-world scenarios where an object’s acceleration does not change, such as a car moving at a constant speed on a straight path. Solving these equations, as we see in the solution steps, is about discovering functions that satisfy the given rules for rates of change.

Initial Value Problem

In solving differential equations, an initial value problem gives us specific starting information needed to find a unique solution. Here, our differential equation \( \frac{d^2 s}{d t^2} = a \) comes with initial conditions: the initial velocity \( v_0 \) when \( t = 0 \), and the initial position \( s_0 \) also when \( t = 0 \). These initial values are crucial because they allow us to determine the constants that appear when integrating the equation. By applying these initial conditions, we can solve for the constants \( C_1 \) and \( C_2 \) in the integration process, ensuring that we find the particular solution rather than a general one.

Integration in Calculus

Integration is the reverse process of differentiation and is a fundamental tool in calculus for solving differential equations. To find the velocity and position equations from the acceleration equation \( \frac{d^2 s}{d t^2} = a \), we perform integration twice. The first integration step gives us the velocity equation \( \frac{d s}{d t} = at + C_1 \), where \( C_1 \) is determined using the initial condition for velocity. The second integration step provides the position equation \( s = \frac{a}{2} t^2 + v_0 t + C_2 \), with \( C_2 \) found using the initial condition for position. Integration not only provides these equations but also connects our understanding of motion from instantaneous rates of change to specific descriptions of motion over time.

Kinematics Equations

Kinematics equations describe the motion of objects without considering the forces that cause the motion. The derived position equation \( s = \frac{a}{2} t^2 + v_0 t + s_0 \) is a classic example of a kinematics equation. It shows how an object's position changes over time considering constant acceleration and initial conditions of velocity and position. These equations help us predict future motion given known initial conditions, and are foundational in physics for studying how bodies move through space. They underscore how calculations of motion can move from abstract differential equations to practical scenarios, such as predicting where a projectile will land or how long it takes a vehicle to reach a destination.