Question

After opening the parachute, the descent of a parachutist follows a linear model. At 2:08 P.M., the height of the parachutist is 7000 feet. At \(2: 10\) P.M., the height is 4600 feet. (a) Write a linear equation that gives the height of the parachutist in terms of the time \(t\). (Let \(t=0\) represent 2:08 P.M. and let \(t\) be measured in seconds.) (b) Use the equation in part (a) to find the time when the parachutist will reach the ground.

Short answer

The linear equation that gives the height of the parachutist in terms of time is \(y = -20t + 7000\). The parachutist will reach the ground after 350 seconds from 2:08 P.M.

Step-by-step solution

Identify the two points in the problem

The two points given in the problem are (0,7000) and (120,4600), where the x-coordinate represents time in seconds and y-coordinate represents the height in feet.

Calculate the slope of the linear equation

The slope 'm' for the linear equation is calculated by using the formula \(m=\frac{{y_{2}-y_{1}}}{{x_{2}-x_{1}}}\). By substituting the values of given points in this formula, \(m=\frac{{4600-7000}}{{120-0}} = -20\) feet/second.

Find the y-intercept of the linear equation

From the linear equation formula \(y = mx + c\), 'c' is the y-intercept. Substituting the values of one point (let's take (0,7000)) and the found slope into the equation, the y-intercept 'c' is found to be 7000.

Write the complete linear equation

After calculating values for the slope 'm' and the y-intercept 'c', the linear equation that represents the parachutist's height in relation to the time can be written as \(y = -20t + 7000\).

Find the time when the parachutist will reach the ground

We know that when the parachutist reaches the ground, the height 'y' will be 0. So the equation will be \(0 = -20t + 7000\). Solving this equation for 't', we get \(t = \frac{7000}{20} = 350\) seconds.

Key concepts

Slope Calculation

Understanding the slope calculation is crucial when dealing with linear equations. The slope, often denoted as 'm', indicates the rate at which one variable changes with respect to another. In essence, it represents the steepness or incline of a line on a graph.

To calculate the slope between two points, you can use the formula:
\[\begin{equation} m = \frac{y_2 - y_1}{x_2 - x_1}\end{equation}\]
Here, \( (x_1, y_1) \) and \( (x_2, y_2) \) represent the coordinates of two points on the line. For the parachutist example, we found the points on a time-height graph as (0, 7000) and (120, 4600). Substituting these values into our slope formula, the rate of descent was determined to be -20 feet per second, indicating a downward trend, as expected for a descending parachutist.

Y-Intercept

The y-intercept is the point where a linear equation crosses the y-axis on a graph. It's an essential part of a linear equation, as it provides a starting value when the independent variable, usually 'x' or in our example 't' for time, is zero. To find the y-intercept, one can either graph the equation and observe where it intersects the y-axis or use the coordinates of a known point and the slope to solve for the y-intercept in the equation of the line, \( y = mx + c \).

In our parachutist scenario, using the initial height at the zero time mark, we identified the y-intercept as 7000, which means that at time \( t = 0 \) seconds, the parachutist was at a height of 7000 feet.

Modeling with Linear Equations

Creating a Linear Model

Modeling with linear equations involves representing a real-world scenario with a mathematical formula that describes a linear relationship between variables. This process starts by identifying two variables that change in a consistent manner. Next, gather data points that show this relationship, and then determine the slope and y-intercept from these points to write the equation.

In the parachutist exercise, the time and height follow a linear model. By using the given times and corresponding heights, we formulated a linear equation that accurately represents the descent of the parachutist over time. These models are incredibly powerful tools for making predictions and understanding behavior within systems.

Linear Equations in Real-World Scenarios

Applying Linear Equations

Linear equations are not just for math textbooks; they are used to model and solve problems in various real-world scenarios. For example, they can predict expenses over time, the speed of objects, and the change in temperature, among other things. These equations are invaluable as they allow for making predictions about outcomes based on initial values and rates of change, aiding in decision-making and planning.

The parachutist problem is a prime example of a real-world application. By developing a linear equation from the observed data, we were able to predict how long it would take for the parachutist to reach the ground, a vital piece of information for ensuring a safe landing.