Problem 46 - College Algebra with Corequisite Support

Question

Near the surface of the moon, the distance that an object falls is a function of time. It is given by \(d(t)=2.6667 t^{2},\) where \(t\) is in seconds and \(d(t)\) is in feet. If an object is dropped from a certain height, find the average velocity of the object from \(t=1\) to \(t=2\)

Step-by-Step Solution

Answer

The average velocity from \(t=1\) to \(t=2\) is 8.0001 feet per second.

Step 1: Understand the Given Function

We are given the distance function for an object falling on the moon: \(d(t) = 2.6667 t^2\), where \(t\) is time in seconds and \(d(t)\) is distance in feet. We need to find the average velocity over a time interval.

Step 2: Recall Average Velocity Formula

The average velocity \(v_{avg}\) over an interval \([t_1, t_2]\) can be calculated using the formula: \(v_{avg} = \frac{d(t_2) - d(t_1)}{t_2 - t_1}\).

Step 3: Determine Position at \(t=1\)

Substitute \(t = 1\) into the distance function: \(d(1) = 2.6667 \times 1^2 = 2.6667\) feet.

Step 4: Determine Position at \(t=2\)

Substitute \(t = 2\) into the distance function: \(d(2) = 2.6667 \times 2^2 = 2.6667 \times 4 = 10.6668\) feet.

Step 5: Calculate the Average Velocity

Using the formula for average velocity, \(v_{avg} = \frac{d(2) - d(1)}{2 - 1} = \frac{10.6668 - 2.6667}{1} = 8.0001\) feet per second.

Key Concepts

Distance Function

A distance function describes the distance an object travels over time. In this problem, the distance function is given by the equation \( d(t) = 2.6667t^2 \). This means that the distance \( d \), in feet, that an object falls is directly related to the square of the time \( t \), measured in seconds.
This quadratic function suggests that as time passes, the distance falls faster, creating a parabolic trajectory.

• The coefficient \( 2.6667 \) determines how fast the distance increases over time.
• A higher coefficient would mean the object is falling faster for the same amount of time.

Understanding this function helps us calculate important metrics such as how fast the object moves, specifically over any given time interval.

Time Interval

The concept of a time interval is crucial for calculating changes over time, such as velocity. In this context, the time interval is the span between two moments, \( t_1 \) and \( t_2 \). Here, we consider the time interval from \( t = 1 \) second to \( t = 2 \) seconds.
This specific interval of one second allows us to measure how the object's position changes over time.

• Time intervals can vary depending on what you want to analyze, allowing for flexibility.
• They help in determining how much distance is covered and, subsequently, the average velocity over that time.

Understanding time intervals lays the foundation for calculating different aspects of an object's motion.

Velocity Formula

Calculating the velocity of an object involves understanding its motion over time. The formula for average velocity \( v_{avg} \) is:
\[v_{avg} = \frac{d(t_2) - d(t_1)}{t_2 - t_1}\] This equation finds the change in distance divided by the change in time, giving us the average velocity.

The formula helps us understand the object's speed and direction.

Using points \( t_1 = 1 \) second and \( t_2 = 2 \) seconds, the distance changes from \( 2.6667 \) feet to \( 10.6668 \) feet.

Substituting in these values, we find the average velocity between these times to be \( 8.0001 \) feet per second.

This calculation is crucial for understanding how fast objects move over given periods.

Object Falling

The object in this scenario is falling on the moon, providing an interesting environment different from Earth. The moon has roughly one-sixth of Earth's gravity, thus affecting how objects fall.
This exercise explores how an object behaves when dropped under such conditions, providing insight into the physics of different celestial bodies.

• Understanding the falling motion involves knowing the gravitational forces at play.
• The distance function \( d(t) = 2.6667t^2 \) models how gravity impacts the falling object on the moon.
• As the object falls, it accelerates due to the moon's gravity, causing it to cover more distance over time.

Studying the concept of an object falling is key to grasping fundamental physics principles like velocity and acceleration.