Problem 79 - University Calculus: Early Transcendentals

Question

The velocity of a particle moving back and forth on a line is \(v=d s / d t=6 \sin 2 t \mathrm{m} / \mathrm{sec}\) for all \(t .\) If \(s=0\) when \(t=0,\) find the value of \(s\) when \(t=\pi / 2\) sec.

Step-by-Step Solution

Answer

The displacement \( s \) is 6 meters when \( t = \frac{\pi}{2} \) seconds.

Step 1: Understand the Problem

The problem gives the velocity function of a particle as \( v = \frac{ds}{dt} = 6 \sin(2t) \) meters per second. We are asked to find the displacement \( s \) when \( t = \frac{\pi}{2} \) seconds, given that \( s = 0 \) when \( t = 0 \).

Step 2: Set Up the Integral

Since the velocity \( v = \frac{ds}{dt} \), we need to find \( s(t) \) by integrating the velocity function. The displacement is the integral of the velocity with respect to time: \( s(t) = \int v(t) \, dt = \int 6 \sin(2t) \, dt \).

Step 3: Solve the Integral

Integrate \( 6 \sin(2t) \). The integral of \( \sin(2t) \) is \(-\frac{1}{2} \cos(2t)\). Hence, the integral becomes:\[ s(t) = 6 \int \sin(2t) \, dt = -3 \cos(2t) + C \]

Step 4: Apply Initial Conditions

Use the initial condition \( s = 0 \) when \( t = 0 \) to find the constant \( C \). Substitute \( t = 0 \) and \( s = 0 \): \[ 0 = -3 \cos(0) + C \] \( \cos(0) = 1 \), so \[ 0 = -3 + C \Rightarrow C = 3 \]

Step 5: Calculate \( s \) at \( t = \pi/2 \)

Substitute \( t = \frac{\pi}{2} \) into the displacement equation \( s(t) = -3 \cos(2t) + 3 \):\[ s\left(\frac{\pi}{2}\right) = -3 \cos\left(\pi\right) + 3 = -3(-1) + 3 = 3 + 3 = 6 \] meters.

Key Concepts

Velocity Function

In the world of calculus, the velocity function is crucial for understanding how an object's speed and direction change over time. In this exercise, the velocity function is given by \( v = \frac{ds}{dt} = 6 \sin(2t) \) meters per second. Essentially, it describes how the displacement \( s \) of a particle changes as time \( t \) progresses. The velocity function combines both speed and direction, which makes it a vector quantity.

• **Speed**: The absolute value of velocity.
• **Direction**: Given by the sign of \( v \), showing if the movement is forward or backward.

This velocity function depends on sine, meaning it oscillates between positive and negative values. This behavior signifies a periodic motion where the particle moves back and forth along a path.

Integration

Integration is a fundamental calculus concept used to find accumulated values such as area under a curve or, in this case, the displacement of a particle. To determine the position function \( s(t) \), we integrate the velocity function. This means we calculate the definite integral of \( 6 \sin(2t) \), which gives us a general expression for displacement.The definite integral of \( 6 \sin(2t) \) with respect to time \( t \) is computed as follows:\[ s(t) = \int 6 \sin(2t) \, dt = -3 \cos(2t) + C \]Where \( C \) is the integration constant, which will be determined using initial conditions. Hence, integration helps us transition from instantaneous measurements back to comprehensive, accumulated measures.

Initial Conditions

Initial conditions are constraints that help us find specific solutions to differential equations or integrals. In this problem, we are given initial conditions: when \( t = 0 \), the displacement \( s = 0 \). These conditions are essential for solving the value of \( C \) in our integration result.By applying \( t = 0 \) into our position function, we get:\[0 = -3 \cos(0) + C\]Since \( \cos(0) = 1 \), we solve for \( C \) and find that \( C = 3 \). This ensures that our position function accurately reflects the conditions at the beginning of the observation. Initial conditions tie the mathematical model to real-world scenarios, providing the context needed for correct interpretation.

Displacement

Displacement refers to the change in position of a particle over time. It reflects the net change considering the path direction. In this exercise, after applying the initial condition, our position function becomes:\[s(t) = -3 \cos(2t) + 3\]We use this equation to find the specific displacement value when \( t = \frac{\pi}{2} \). By substituting this value into the function, we obtain:\[s\left(\frac{\pi}{2}\right) = -3 \cos(\pi) + 3 = 3 + 3 = 6\]This indicates that the particle has moved a total of 6 meters from its initial position. Displacement provides a clear picture of how far and in what direction the particle has moved, distinct from distance, which would consider the total path traveled without direction signs.