Problem 50 - Calculus: Early Transcendental Functions

Question

Determine the position function if the acceleration function is \(a(t)=t^{2}+1,\) the initial velocity is \(v(0)=4\) and the initial position is \(s(0)=0\)

Step-by-Step Solution

Answer

The position function is \(s(t) = \frac{1}{12}t^{4} + \frac{1}{2}t^{2} + 4t\).

Step 1: Integrate the Acceleration Function to Get Velocity

Firstly, integrate the acceleration function \(a(t)=t^{2}+1\). By using the power rule for integration, we get \(v(t) = \frac{1}{3}t^{3} + t + C_1\), where \(C_1\) is the constant of integration.

Step 2: Apply the Initial Velocity

Apply the given initial velocity \(v(0)=4\) to find the constant \(C_1\). Substituting \(v(0)=4\) into the velocity equation, it yields \(4 = 0 + C_1\), thus \(C_1 = 4\). The velocity function therefore becomes \(v(t) = \frac{1}{3}t^{3} + t + 4\).

Step 3: Integrate the Velocity Function to Get Position

Next, integrate the velocity function to find the position function. This gives \(s(t) = \frac{1}{12}t^{4} + \frac{1}{2}t^{2} + 4t + C_2\), where \(C_2\) is another constant of integration.

Step 4: Apply the Initial Position

Apply the given initial position \(s(0)=0\) to find the constant \(C_2\). Substituting \(s(0)=0\) into the position equation, it yields \(0 = C_2\), thus \(C_2 = 0\). The position function therefore becomes \(s(t) = \frac{1}{12}t^{4} + \frac{1}{2}t^{2} + 4t\).

Key Concepts

Integrating Acceleration Function

Integrating an acceleration function is a fundamental concept in calculus, particularly when analyzing the motion of objects. It enables us to determine the velocity function over time. Acceleration, which is the rate of change of velocity, can vary with time. To find the velocity from acceleration, we need to perform an integration process.

For instance, given the acceleration function in our exercise, \(a(t)=t^{2}+1\), to integrate this, we use the power rule. This rule states that the integral of \(t^n\), where \(n\) is any real number, is \(t^{n+1}/(n+1)\), plus a constant of integration - since indefinite integration can yield an infinite number of functions, differing only by a constant. In the end, we obtain the velocity function \(v(t)\) with an unknown constant, \(C_1\), which can be determined using initial conditions.

Initial Velocity

Initial velocity is an integral part of understanding motion in physics and calculus. It is the velocity of an object at the start of observation, or time zero. In mathematical terms, it is denoted as \(v(0)\). Setting up the equation for motion requires this value, as it serves as a baseline for how fast the object is moving before any additional acceleration is applied.

The given initial velocity can often be used to simplify the process of finding the constant when integrating the acceleration function. In our exercise example, by plugging the given initial velocity \(v(0)=4\) into the velocity function and setting \(t=0\), we find the constant \(C_1\). This allows us to refine our velocity function to be even more specific, incorporating this initial condition to accurately reflect the object's motion from the very start of the observation period.

Initial Position

The initial position, much like initial velocity, is crucial for determining the movement of an object in space. It's defined as the object's position at time zero, denoted mathematically as \(s(0)\). This piece of information sets the starting point from which all subsequent positions are measured.

In our example, applying the initial position \(s(0)=0\) to the position function, which was derived by integrating the velocity function, allows us to solve for the constant \(C_2\). Once the initial position is known, the complete position function over time, \(s(t)\), fully describes the path of the object. By including the initial position, we can provide context for the object's trajectory and make precise predictions about its future location at any given time.