Problem 47 - Thomas Calculus in SI Units

Question

Give the acceleration \(a=d^{2} s / d t^{2},\) initial velocity, and initial position of an object moving on a coordinate line. Find the object's position at time \(t\). $$a=-4 \sin 2 t, \quad v(0)=2, \quad s(0)=-3$$

Step-by-Step Solution

Answer

Position function: \(s(t) = \sin(2t) - 3\).

Step 1: Understand the Problem

We need to find the position function \(s(t)\) at any time \(t\) given the acceleration \(a = -4 \sin(2t)\). We have the initial velocity \(v(0) = 2\) and initial position \(s(0) = -3\). These initial conditions will help us determine the constants of integration in the next steps.

Step 2: Integrate to Find Velocity Function

To find the velocity function \(v(t)\), we integrate the acceleration function. \[ v(t) = \int -4 \sin(2t) \ dt \]Using integration rules, we get:\[ v(t) = 2 \cos(2t) + C_1 \]where \(C_1\) is a constant determined by the initial condition \(v(0) = 2\). Substituting \(t=0\) gives:\[ 2 = 2 \cos(0) + C_1 \]\[ C_1 = 0 \] Thus, \(v(t) = 2 \cos(2t)\).

Step 3: Integrate to Find Position Function

Now we find the position function \(s(t)\) by integrating the velocity function:\[ s(t) = \int 2 \cos(2t) \ dt \]Using integration rules, we get:\[ s(t) = \sin(2t) + C_2 \]where \(C_2\) is a constant determined by the initial condition \(s(0) = -3\). Substituting \(t=0\) gives:\[ -3 = \sin(0) + C_2 \]\[ C_2 = -3 \]Thus, \(s(t) = \sin(2t) - 3\).

Step 4: Conclusion of the Problem

The position at any time \(t\) is given by \(s(t) = \sin(2t) - 3\). For any specific time \(t\), substitute into this equation to find the position.

Key Concepts

Acceleration and Velocity

In calculus, understanding the relationship between acceleration and velocity is essential for solving motion-related problems. Acceleration (\(a\)) tells us how the velocity of an object changes over time. It is the second derivative of the position function (\(s(t)\)) with respect to time and the first derivative of the velocity function (\(v(t)\)).

• Velocity is the first derivative of the position function, showing how the position changes over time.
• Acceleration, on the other hand, provides insight into how velocity changes.

In the given problem, the acceleration is the expression \(-4 \sin(2t)\), suggesting that the velocity function will be derived by integrating this acceleration. Recognizing how velocity and acceleration are interconnected can make solving these calculus problems more intuitive.

Initial Conditions

Initial conditions are crucial to finding the complete solution to a differential equation. They provide the specific information needed to determine the constants of integration that arise when integrating functions.
In our exercise:

• The initial velocity \(v(0) = 2\) helps us find the constant \(C_1\) in the velocity function.
• The initial position \(s(0) = -3\) helps determine the constant \(C_2\) in the position function.

Without these initial conditions, we would only obtain a general solution with terms involving these constants. By applying the initial conditions, we fine-tune the solution to reflect the specific behavior of the moving object at the starting time \(t=0\). This step is critical as it ensures the derived functions truly match the problem's scenario.

Integration

Integration is a core concept in calculus used to determine the original function from its derivative. In this problem, we integrate twice:

• First, to find the velocity function \(v(t)\) from the given acceleration function \(a(t) = -4 \sin(2t)\)
, and
• second, to find the position function \(s(t)\) from the velocity function.

When integrating \(-4 \sin(2t)\), we apply known integration techniques for trigonometric functions. The integral of \(-4 \sin(2t)\) with respect to \(t\) gives us the velocity function \(v(t) = 2 \cos(2t) + C_1\). By substituting the initial conditions, we find that \(C_1 = 0\).Repeating the process for the velocity function, we integrate \(2 \cos(2t)\) to get the position function \(s(t)\). The result is \(s(t) = \sin(2t) + C_2\), with \(C_2\) adjusted by the initial position to satisfy \(s(0) = -3\).Through integration, we uncover the relationships hidden within derivatives and utilize initial conditions to tailor our solutions to specific scenarios.