Problem 36 - Calculus Early Transcendentals

Question

A position function of a particle moving along a coordinate line is provided. (a) Evaluate \(s\) and \(v\) when \(a=0 .\) (b) Evaluate \(s\) and \(a\) when \(v=0 .\) $$ s=t^{3}-6 t^{2}+1 $$

Step-by-Step Solution

Answer

When \(a=0\), \(s=-15\) and \(v=-12\). When \(v=0\), \(s=1, a=-12\) for \(t=0\) and \(s=-31, a=12\) for \(t=4\).

Step 1: Understand the Given Functions

We are given the position function of a particle: \[s(t) = t^3 - 6t^2 + 1\]To find velocity \(v(t)\) and acceleration \(a(t)\) we first need to differentiate \(s(t)\).

Step 2: Find the Velocity Function

The velocity \(v(t)\) is the first derivative of the position function \(s(t)\):\[v(t) = \frac{ds}{dt} = 3t^2 - 12t\]

Step 3: Find the Acceleration Function

The acceleration \(a(t)\) is the first derivative of the velocity function \(v(t)\):\[a(t) = \frac{dv}{dt} = 6t - 12\]

Step 4: Solve for conditions when acceleration is zero (a=0)

Set acceleration \(a(t)\) to zero and solve for \(t\):\[6t - 12 = 0\]\[6t = 12\]\[t = 2\]

Step 5: Evaluate s and v when a=0

Using \(t=2\), substitute into \(s(t)\) and \(v(t)\):\[s(2) = (2)^3 - 6(2)^2 + 1 = 8 - 24 + 1 = -15\]\[v(2) = 3(2)^2 - 12(2) = 12 - 24 = -12\]

Step 6: Solve for conditions when velocity is zero (v=0)

Set velocity \(v(t)\) to zero and solve for \(t\):\[3t^2 - 12t = 0\]\[3t(t - 4) = 0\]\[t = 0\text{ or } t = 4\]

Step 7: Evaluate s and a when v=0 (t=0)

First, evaluate for \(t=0\):\[s(0) = (0)^3 - 6(0)^2 + 1 = 1\]\[a(0) = 6(0) - 12 = -12\]

Step 8: Evaluate s and a when v=0 (t=4)

Next, evaluate for \(t=4\):\[s(4) = (4)^3 - 6(4)^2 + 1 = 64 - 96 + 1 = -31\]\[a(4) = 6(4) - 12 = 24 - 12 = 12\]

Key Concepts

Position Function

The concept of a position function is fundamental in understanding particle motion along a line. In calculus, the position function, often represented as \(s(t)\), describes the location of a particle at any given time \(t\). By expressing position as a function of time, we can track how a particle moves.
For example, in our exercise, the position function is given by \(s(t) = t^3 - 6t^2 + 1\). This function indicates that:

• The particle's initial position at \(t = 0\) would be \(s(0) = 1\).
• The function includes both cubic and quadratic terms, making the motion potentially more complex as it can involve varying rates of change.

Overall, understanding the position function helps predict where the particle will be, which is the foundational step in analyzing motion.

Velocity Function

The velocity function is crucial for understanding how fast and in what direction a particle is moving. In differential calculus, velocity is the first derivative of the position function. This means it shows the rate of change of position with respect to time.
For our given position function \(s(t) = t^3 - 6t^2 + 1\), the velocity function \(v(t)\) is calculated as:

• Taking the derivative, \(v(t) = \frac{ds}{dt} = 3t^2 - 12t\).
• This derivative simplifies to a quadratic form, allowing us to determine when the particle's velocity is zero or when it is accelerating or decelerating.

Understanding the velocity function gives insights into dynamic motion characteristics such as speed and direction at different times.

Acceleration Function

Acceleration provides insight into changes in the velocity of a particle. It is obtained from the second derivative of the position function or the first derivative of the velocity function. In calculus terms, acceleration measures how fast the velocity of the particle changes over time.
From the velocity function \(v(t) = 3t^2 - 12t\), the acceleration function \(a(t)\) becomes:

• \(a(t) = \frac{dv}{dt} = 6t - 12\), indicating a linear relationship.
• This tells us how the acceleration varies with time, helping to predict how the velocity itself is changing, like when acceleration is zero \(a(t) = 0\), which was used to solve parts of the exercise.

By understanding acceleration, one can deduce potential forces affecting motion and other dynamics.

Differentiation

Differentiation is a fundamental tool in calculus used to find derivatives. Derivatives are vital for analyzing rates of change, such as velocity and acceleration derived from position. It is the process of finding a derivative, which represents the instant rate of change of a function relative to a variable.
For our exercise, differentiation is applied as follows:

• The first differentiation of the position function \(s(t)\) yields the velocity function \(v(t)\): \(3t^2 - 12t\).
• A second differentiation of \(v(t)\) gives the acceleration \(a(t)\): \(6t - 12\), demonstrating how differentiation helps in transitioning from one motion metric to another.

This concept explores beyond simple calculations, providing the ability to model systems dynamically and predicting how they change over time.