Problem 30 - Calculus Early Transcendentals

Question

Find the position and velocity of an object moving along a straight line with the given acceleration, initial velocity, and initial position. $$a(t)=e^{-t}, v(0)=60, s(0)=40$$

Step-by-Step Solution

Answer

Question: Given an object's acceleration function $$a(t) = e^{-t}$$, initial velocity $$v(0) = 60$$, and initial position $$s(0) = 40$$, find the position and velocity functions for the object moving along a straight line. Answer: The position function for the object is $$s(t) = e^{-t} + 61t + 39$$, and the velocity function is $$v(t) = -e^{-t} + 61$$.

Step 1: Integrate acceleration function to get velocity function

Integrate the given acceleration function $$a(t)=e^{-t}$$ with respect to 't' to obtain the velocity function: $$v(t) = \int a(t) \, dt = \int e^{-t} \, dt$$ Using the integral of $$e^{-t}$$: $$v(t) = -e^{-t} + C_1$$

Step 2: Apply initial velocity condition to find C_1

Use the initial velocity $$v(0) = 60$$ to find constant $$C_1$$: $$60 = -e^{-0} + C_1$$ $$C_1 = 61$$ So the velocity function is: $$v(t) = -e^{-t} + 61$$

Step 3: Integrate velocity function to get position function

Integrate the velocity function $$v(t)=-e^{-t} + 61$$ with respect to 't' to obtain the position function: $$s(t) = \int v(t) \, dt = \int (-e^{-t} + 61) \, dt$$ When we integrate this function, we get: $$s(t) = e^{-t} + 61t + C_2$$

Step 4: Apply initial position condition to find C_2

Use the initial position s(0) = 40 to find the constant $$C_2$$: $$40 = e^{-0} + 61(0) + C_2$$ $$C_2 = 39$$ So the position function is: $$s(t) = e^{-t} + 61t + 39$$ Thus, the position function $$s(t) = e^{-t} + 61t + 39$$ and the velocity function $$v(t) = -e^{-t} + 61$$ describe the object's motion along the straight line.

Key Concepts

Acceleration Function Integration

Understanding the process of integrating an acceleration function is crucial in the realm of motion calculus. It lays the groundwork for determining how fast an object is moving at any given moment. If you consider acceleration as the rate at which velocity changes over time, integrating the acceleration can reveal the velocity function.

Take for example the exercise where we are given the acceleration function as \( a(t) = e^{-t} \). To get to the velocity function, we perform the integration of this acceleration function concerning time, \( t \), leading to \( v(t) = int a(t) dt = -e^{-t} + C_1 \). The integration process here involves finding the antiderivative of the exponential function, which, for \( e^{-t} \), is \( -e^{-t} \), plus an arbitrary constant, \( C_1 \). This constant will be determined using initial conditions provided in the problem.

Velocity Function Derivation

Once the acceleration function has been integrated, the next task is often to extract a workable velocity function that adheres to the specific conditions of a given problem. Deriving this function correctly allows us to understand better how the object's speed changes over time.

In the textbook example, after integrating the acceleration function, we leveraged the known initial velocity, \( v(0) = 60 \), to find the integration constant \( C_1 \). Plugging in \( t = 0 \) into the integrated expression and solving for \( C_1 \), we got \( C_1 = 61 \), thus deriving the complete velocity function \( v(t) = -e^{-t} + 61 \) that includes the effect of the initial condition.

Initial Value Problem

An Initial Value Problem (IVP) in the context of calculus involves finding a function that satisfies a differential equation and adheres to provided initial conditions. It's like solving a puzzle where you're given pieces to confirm the greater picture ‒ the function's general form ‒ aligns with specific starting points.

For the object in motion, the initial conditions were the initial velocity and initial position, \( v(0) = 60 \) and \( s(0) = 40 \) respectively. By applying these values, we determined the unknown constants \( C_1 \) and \( C_2 \) in the integrated forms of acceleration and velocity functions. This allowed us to construct a specific solution that describes the object's motion accurately from the provided starting points.

Exponential Decay

Exponential decay is a concept often encountered in natural processes, where the rate of change of a quantity decreases exponentially over time. Mathematically, it's represented by functions like \( e^{-t} \), where \( t \) is the time variable. In the case of our object's motion, the acceleration given by \( a(t) = e^{-t} \) indicates that as time progresses, the object's acceleration decreases exponentially.

This is essential for understanding the nature of the object's motion. In the context of our problem, it contributed to the velocity decreasing over time, as indicated by the negative sign in the velocity function \( v(t) = -e^{-t} + 61 \), after the integration of the acceleration function that demonstrated exponential decay.

Position Function Integration

To find out the exact path of our moving object, we integrate the velocity function, which gives us the position function. This integration tells us not only where the object is at any given moment but also encapsulates the entire history of its motion.

In our exercise, we integrated the velocity function \( v(t) = -e^{-t} + 61 \), resulting in the position function \( s(t) = e^{-t} + 61t + C_2 \). To finalize this expression, we needed the initial position, which the problem provided as \( s(0) = 40 \). By setting \( t = 0 \) in the derived position function and solving for \( C_2 \), we established \( C_2 = 39 \). Hence, the fully defined position function becomes \( s(t) = e^{-t} + 61t + 39 \), completing our comprehensive picture of the object's motion.