Problem 60 - Thomas Calculus

Question

The acceleration of a particle moving back and forth on a line is \(a=d^{2} s / d t^{2}=\pi^{2} \cos \pi t \mathrm{m} / \mathrm{sec}^{2}\) for all \(t .\) If \(s=0\) and \(v=\) 8 \(\mathrm{m} / \mathrm{sec}\) when \(t=0,\) find \(s\) when \(t=1 \mathrm{sec} .\)

Step-by-Step Solution

Answer

The position \(s\) when \(t = 1\) second is 10 meters.

Step 1: Understand the Relationship Between Acceleration, Velocity, and Position

The problem provides the function for acceleration of the particle, given by \(a = \frac{d^2s}{dt^2} = \pi^2 \cos(\pi t)\). We need to find the position \(s\) at time \(t = 1\). To do this, we first need to integrate the acceleration to find the velocity function, and then integrate the velocity to find the position function.

Step 2: Integrate Acceleration to Find Velocity

Integrate the acceleration function to determine the velocity: \[ v = \int a \, dt = \int \pi^2 \cos(\pi t) \, dt. \]The integration gives: \[ v(t) = \pi \sin (\pi t) + C_1, \]where \(C_1\) is the integration constant. We will determine this from the initial condition \(v(0) = 8\).

Step 3: Apply Initial Condition to Find Constant of Integration for Velocity

Using the initial condition \(v(0) = 8\ mu s/\text{sec}\), substitute \(t = 0\) into the velocity equation: \[ v(0) = \pi \sin(0) + C_1 = 8. \]Since \(\sin(0) = 0\), it simplifies to \(C_1 = 8\). Therefore, the velocity function is:\[ v(t) = \pi \sin(\pi t) + 8. \]

Step 4: Integrate Velocity to Find Position

Integrate the velocity function to determine the position:\[ s = \int v(t) \, dt = \int (\pi \sin(\pi t) + 8) \, dt. \]This simplifies to:\[ s(t) = -\cos(\pi t) + 8t + C_2, \]where \(C_2\) is another constant of integration that we will find using the initial condition \(s(0) = 0\).

Step 5: Apply Initial Condition to Find Constant of Integration for Position

Using the initial condition \(s(0) = 0\), substitute \(t = 0\) into the position equation:\[ s(0) = -\cos(0) + 8(0) + C_2 = 0. \]Since \(-\cos(0) = -1\), it simplifies to \(-1 + C_2 = 0\), leading to \(C_2 = 1\). Thus, the position function is:\[ s(t) = -\cos(\pi t) + 8t + 1. \]

Step 6: Substitute t = 1 to Find s(1)

Substitute \(t = 1\) into the position function:\[ s(1) = -\cos(\pi \times 1) + 8(1) + 1. \]Since \(\cos(\pi) = -1\), this becomes:\[ s(1) = -(-1) + 8 + 1 = 1 + 8 + 1 = 10. \]

Key Concepts

Acceleration Integration

One central concept in this problem is acceleration integration. In physics and calculus, acceleration is the rate of change of velocity over time. If we have an expression for acceleration, we can reverse this process, through integration, to find the velocity.
In our exercise, the acceleration is firstly expressed as a function of time:

• \( a(t) = \pi^2 \cos(\pi t) \)

To find the velocity, we integrate the acceleration function with respect to time. Integrating functions involves finding an antiderivative, which, essentially, is a way to "go backwards" from a derivative to the original function itself.
When we integrate the given acceleration equation:

• \( v(t) = \int \pi^2 \cos(\pi t) \, dt \)

Our solution shows the calculation:

• \( v(t) = \pi \sin (\pi t) + C_1 \)

where \( C_1 \) represents the constant of integration, determined from known initial conditions.

Initial Conditions in Calculus

Initial conditions in calculus are vital for determining the exact form of a solution after integration. Each integration introduces a constant, because indefinitely many functions can have the same derivative.
In our case, the acceleration equation's integration to get velocity yields a constant \( C_1 \). To solve for this, we use the initial condition, specifying that initially (after our first calculation point) the velocity, \( v(0) \), equals 8 m/s.

• Setting \( v(0) = \pi \sin(0) + C_1 = 8 \)

Since \( \sin(0) = 0 \), the equation simplifies, and \( C_1 = 8 \).
Similarly, integrating the velocity gives us the position function with another constant, \( C_2 \). Using \( s(0) = 0 \) allows us to solve for \( C_2 \):

• \( s(0) = -\cos(0) + 8(0) + C_2 = 0 \)

This results in \( C_2 = 1 \).
Such conditions ensure our final function accurately reflects the particle's motion in reality.

Position Function Calculation

In calculus, a position function specifies the location of an object at any given time. It's derived by integrating the velocity function, which itself comes from the integrated acceleration.
Our exercise required finding the position of a particle. We had already identified velocity as:

• \( v(t) = \pi \sin(\pi t) + 8 \)

By integrating this velocity function with respect to time, we calculated the position \( s(t) \):

• \( s(t) = \int (\pi \sin(\pi t) + 8) \, dt \)

Leading to:

• \( s(t) = -\cos(\pi t) + 8t + 1 \)

Here, the integration constant was adjusted based on initial conditions.
Finally, to find \( s(1) \), representing position at 1 second, we evaluate the function at \( t = 1 \):

• \( s(1) = -\cos(\pi \cdot 1) + 8 \cdot 1 + 1 \)
• This equals \( 10 \), revealing the particle's position as 10 meters at \( t = 1 \) second.