Problem 86 - Intermediate Algebra

Question

Solve each problem by writing a variation model. Free Fall. An object in free fall travels a distance \(s\) that is directly proportional to the square of the time \(t\). If an object falls \(1,024\) feet in 8 seconds, how far will it fall in 10 seconds?

Step-by-Step Solution

Answer

The object will fall 1,600 feet in 10 seconds.

Step 1: Understand the Problem

We need to find the distance fallen, which is directly proportional to the square of time (\( s \propto t^2 \)). Given the distance an object falls in a known time, we must derive the proportionality constant and use it to find the distance at a different time.

Step 2: Formulate the Variation Model

The problem states that distance \( s \) is directly proportional to the square of time \( t \), so we can express the relationship as \( s = k t^2 \) where \( k \) is the proportionality constant.

Step 3: Find the Proportionality Constant

Given that the object falls \( 1,024 \) feet in \( 8 \) seconds, we can substitute these values into the equation \( s = k t^2 \). Thus, we have \( 1,024 = k \times 8^2 \). Solving for \( k \), we get \( k = \frac{1,024}{64} = 16 \).

Step 4: Apply the Proportionality Constant to the New Situation

Now, we need to find how far the object falls in \( 10 \) seconds. Substitute \( t = 10 \) and \( k = 16 \) into the formula \( s = k t^2 \). This gives us \( s = 16 \times 10^2 = 16 \times 100 = 1,600 \).

Step 5: Verify the Calculation

Rechecking our calculations ensures accuracy. We have derived that for \( t = 10 \) seconds and \( k = 16 \), the distance \( s = 1,600 \) feet. All values check out correctly with our initial model.

Key Concepts

Free Fall

Free fall is a fascinating concept in physics where an object moves under the influence of gravity alone, without any resistance from air or other forces. When an object is in free fall, it accelerates at a constant rate, usually approximated as 9.8 m/s² (or about 32 ft/s²) on Earth. This acceleration causes the object to cover more distance over time as it continues falling.

In the context of this exercise, we're exploring the distance an object travels in free fall over time. The item in question experiences this gravitational pull, leading to a direct increase in the distance it covers as time passes. For example, in our problem, we know that an object falls 1,024 feet in 8 seconds. We're interested in understanding how these variables connect through a proportion that includes time squared.

Direct Proportionality

Direct proportionality is a simple and powerful mathematical concept used to describe the relationship between two variables. In a directly proportional relationship, as one variable increases, the other increases at a constant rate, and vice versa.

In this exercise, the distance fallen, denoted as "s," is directly proportional to the square of the time "t". This is a special type of direct proportionality called quadratic proportionality, expressed as:

• \[ s = k t^2 \]

Here, "s" is the distance, "k" is our proportionality constant, and "t" is the time squared.

Quadratic proportionality implies that if you double the time, the distance increases by four times (since 2² = 4). Understanding this principle aids in predicting how the distance changes as time progresses without requiring complex calculations every time.

Proportionality Constant

The proportionality constant "k" is a critical element in variation models, acting as the link that ties two quantities together. You can think of it as the "glue" that makes direct proportionality work.

In the equation \( s = k t^2 \), the value of "k" determines how "t^2" translates into "s" with accuracy. To find "k," you can substitute the known values into the equation and solve. In our exercise, we find "k" by setting the distance the object falls (1,024 feet) equal to "k" times the time squared (8 seconds squared):

• \[ k = \frac{1,024}{8^2} = \frac{1,024}{64} = 16 \]

With "k" calculated, it becomes easy to apply this constant to new situations, such as determining how far the object will fall in a different time frame. This simple step makes solving variation problems a breeze, as it allows for quick adjustments and predictions based on consistent relationships.