Problem 67 - Algebra 1

Question

You are supervising the construction of a roller coaster for young children. For the first 20 feet of horizontal distance, the track must rise off the ground at a constant rate. After your crew has constructed 5 feet of horizontal distance, the track is 1 foot off the ground. After 20 feet of horizontal distance is constructed, you are at the highest point of your roller coaster. How high off the ground is the track?

Step-by-Step Solution

Answer

The track would be 4 feet off the ground.

Step 1: Identify the constant rate of increase

We know that after 5 feet of horizontal distance, the track is 1 foot above the ground. This means that with every 5 feet, the height increases by 1 foot. This gives us a ratio of 1 foot height per 5 feet of horizontal distance, or \( \frac{1}{5} = 0.2 \) feet of height per foot.

Step 2: Apply the rate to the 20 feet distance

For 20 feet, thus, the height of the track would be the constant rate of increase (0.2 feet per foot) times the total horizontal distance (20 feet). Thus, height would be \( 0.2 \times 20 = 4 \) feet.

Step 3: Confirm the result

After 20 feet of horizontal distance, the track would be 4 feet above the ground, given the constant rate of increase.

Key Concepts

Algebraic Concepts

Algebraic concepts are the foundation stones of higher mathematics, underpinning the processes used to solve a range of mathematical problems. Take the problem we're considering: calculating the height of a roller coaster track based on a constant rate of change. This scenario involves foundational algebraic concepts such as ratios and proportional relationships. In fact, by understanding that for every 5 feet of horizontal distance, the track rises 1 foot, we're observing a direct proportionality between the horizontal distance and the height of the track.

This relationship is expressed algebraically as a ratio, \( \frac{1}{5} \), which can then be applied to a new value – in this case, 20 feet – to find the corresponding height of the coaster. Algebraic manipulation of these expressions, including multiplying the rate by the new distance, is essential to finding the solution and showcases the power of algebra in problem-solving.

Linear Functions

Linear functions are arguably one of the simplest and most important concepts in algebra. They describe a constant rate of change and are represented by a straight line when graphed on a coordinate plane. The roller coaster problem illustrates a real-world application of a linear function. The track's rise in height is a linear function of the horizontal distance traveled. We observe that for every unit the horizontal distance increases, the height increases at a constant rate.

The general form of a linear function is \( y = mx + b \), where \( m \) represents the slope, or the rate of change, and \( b \) represents the y-intercept. In our case, the slope is the constant rate of increase of the track's height, 0.2, and the y-intercept would be zero since the track starts at ground level. Consequently, the equation describing the rise of the roller coaster's track is \( y = 0.2x \).

Rate of Increase

Understanding the 'rate of increase' is crucial when it comes to predicting future values based on current data. In the context of our roller coaster, the 'rate of increase' refers to the speed at which the track gains height as it extends horizontally. It is a measure of the relationship between two varying quantities. Here, for every additional foot of horizontal track laid, the vertical height increases by 0.2 feet. This is a constant rate of increase, meaning it doesn't change regardless of the length of track already built.

It's important to distinguish that a 'rate of increase' does not always have to be steady or linear. However, in the context of linear functions and in many practical situations like the construction of this roller coaster, the rate remains constant. Applying this steady rate of increase allows us to extrapolate and predict that at 20 feet of horizontal distance, the track will be 4 feet off the ground - providing a straightforward method to solve for unknown values in linear relationships.