Problem 22 - Calculus Early Transcendentals

Question

The function \(s(t)\) describes the position of a particle moving along a coordinate line, where \(s\) is in feet and \(t\) is in seconds. (a) Find the velocity and acceleration functions. (b) Find the position, velocity, speced, and acceleration at time \(t=1 .\) (c) At what times is the particle stopped? (d) When is the particle speeding up? Slowing down? (e) Find the total distance traveled by the particle from time \(t=0\) to time \(t=5\). $$ s(t)=\frac{1}{4} t^{2}-\ln (t+1), \quad t \geq 0 $$

Step-by-Step Solution

Answer

The velocity and acceleration functions are \( v(t)=\frac{1}{2}t - \frac{1}{t+1} \) and \( a(t)=\frac{1}{2} + \frac{1}{(t+1)^2} \). At \( t=1 \), \( s(1)=\frac{1}{4}-\ln(2)\), \( v(1)=0 \), and \( a(1)=\frac{3}{4} \). The particle is stopped at \( t=1 \). Determine intervals where \( v(t) \) and \( a(t) \) have same/opposite signs for speeding up/slowing down. Integrate \( |v(t)| \) from \( t=0 \) to \( t=5 \) for total distance.

Step 1: Identify and Differentiate for Velocity

The velocity function is the first derivative of the position function, \( s(t) \). Differentiating \( s(t) = \frac{1}{4} t^{2} - \ln(t+1) \), we get:\[ v(t) = \frac{d}{dt}\left( \frac{1}{4} t^{2} - \ln(t+1) \right) = \frac{1}{2}t - \frac{1}{t+1} \]

Step 2: Differentiate for Acceleration

The acceleration function is the derivative of the velocity function, \( v(t) \). Differentiating \( v(t) = \frac{1}{2}t - \frac{1}{t+1} \), we get:\[ a(t) = \frac{d}{dt}\left( \frac{1}{2}t - \frac{1}{t+1} \right) = \frac{1}{2} + \frac{1}{(t+1)^2} \]

Step 3: Evaluate Functions at \(t=1\)

Substitute \( t = 1 \) into each function:- Position: \( s(1) = \frac{1}{4} \cdot 1^2 - \ln(1+1) = \frac{1}{4} - \ln(2) \)- Velocity: \( v(1) = \frac{1}{2} \cdot 1 - \frac{1}{1+1} = \frac{1}{2} - \frac{1}{2} = 0 \)- Speed: This is the absolute value of velocity, which is \( 0 \) at \( t=1 \).- Acceleration: \( a(1) = \frac{1}{2} + \frac{1}{(1+1)^2} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4} \)

Step 4: Find when Particle is Stopped

The particle stops when its velocity is zero. Set \( v(t) = 0 \):\[ \frac{1}{2}t - \frac{1}{t+1} = 0 \]Solving for \( t \), we find the times when the particle is stopped.

Step 5: Determine When Particle is Speeding Up or Slowing Down

The particle is speeding up when velocity and acceleration have the same sign and slowing down when they have opposite signs. Use the expressions for \( v(t) \) and \( a(t) \) to evaluate their signs across different \( t \) intervals.

Step 6: Calculate Total Distance Traveled from \(t=0\) to \(t=5\)

To find total distance, integrate the absolute value of the velocity function over the interval \([0, 5]\):\[ \int_{0}^{1} |v(t)| \, dt + \int_{1}^{5} |v(t)| \, dt \] Calculate and add these integrals for the total distance.

Key Concepts

Velocity Function

In calculus, the velocity function of a particle is crucial to understand its motion along a path. A velocity function, denoted as \( v(t) \), is derived from the position function \( s(t) \) by taking its first derivative. This derivative provides the rate of change of the particle's position with respect to time.

• The position function given is \( s(t) = \frac{1}{4}t^2 - \ln(t+1) \).
• The velocity function is the derivative of this, calculated as \( v(t) = \frac{1}{2}t - \frac{1}{t+1} \).
• It tells us how fast the position changes over time and in which direction the particle is moving.

Differentiation provides a tool to analyze motion dynamically. Velocity can take negative values, indicating motion in the opposite direction. Importantly, velocity being zero at any time \( t \) means the particle is momentarily at rest.

Acceleration Function

The acceleration function in calculus, denoted as \( a(t) \), describes how the velocity of a particle changes over time. It is obtained by differentiating the velocity function. Acceleration indicates whether a particle is speeding up or slowing down and provides insight into the forces acting on it.

• Starting from the velocity function \( v(t) = \frac{1}{2}t - \frac{1}{t+1} \), differentiate it to find acceleration: \( a(t) = \frac{1}{2} + \frac{1}{(t+1)^2} \).
• This function shows how the speed of the particle changes as time progresses.
• When both the velocity and acceleration have the same sign, the particle speeds up. When they are opposite, the particle slows down.

Understanding acceleration helps to predict future motion and is essential for identifying changes in the speed or direction of the particle.

Differentiation

Differentiation is a key concept in calculus used to find rates of change. It is the process of finding a derivative, which represents how a function changes at any given point. Differentiation is applied to various functions to understand their rate of change and dynamic behavior.

• Applied to the position function \( s(t) \), differentiation yields the velocity function \( v(t) \).
• Further differentiation of \( v(t) \) provides the acceleration function \( a(t) \).
• Each derivative gives valuable insights into the motion, speed, and change of speed of a particle.

Differentiation can be used across various fields, from physics to economics, to model the behavior of changing quantities and calculate instantaneous rates of change.

Position Function

The position function in calculus describes the location of a particle along a coordinate line as a function of time. The position function, often noted as \( s(t) \), specifies the exact position of a particle at any time \( t \).

• Given as \( s(t) = \frac{1}{4}t^2 - \ln(t+1) \), it informs us where the particle is relative to a reference point.
• By calculating the position at specific times, such as \( t = 1 \), you find where the particle is located at that moment.
• Integrating the velocity function over a time interval can provide the total distance traveled by the particle.

The position function is foundational for understanding motion as it links time to spatial location, serving as a starting point for finding velocity and acceleration.