Question

You are hiking a six-mile trail at a constant rate in Topanga Canyon. You begin at 10 A.M. At noon you are two miles from the end of the trail. Write a linear equation that gives the distance \(d\) (in miles) from the end of the trail in terms of time \(t .\) Let \(t\) represent the number of hours since 10 A.M.

Short answer

The linear equation that gives the distance \(d\) from the end of the trail in terms of time \(t\) is \(d = -2t + 6\).

Step-by-step solution

Identify the known values

The total length of the trail is 6 miles. At noon (2 hours after beginning the hike), the hiker is 2 miles from the end of the trail. Therefore, the hiker has covered \(6 - 2 = 4\) miles in 2 hours, implying a speed of \(4/2 = 2\) miles per hour.

Formulate linear relationship

The hiker's distance from the end of the trail, \(d\), decreases by 2 miles each hour, so this variable changes at a rate of -2 miles per hour. The hike begins at 10 A.M, which is designated as time \(t=0\) hours. At this time, the hiker is 6 miles from the end. So the initial value of \(d\), or the y-intercept, is 6.

Write the linear equation

Insert these values into the equation \(d = mt + c\) to get \(d = -2t + 6\). This linear equation represents the distance \(d\) (in miles) from the end of the trail for given time \(t\) (in hours) since 10 A.M.

Key concepts

Algebraic Expressions

Algebraic expressions are mathematical phrases that can include numbers, variables (letters), and operation symbols such as plus, minus, times, and divide. They represent specific amounts even when we don't know the values of the variables. For example, the hike described in the exercise can be represented by an algebraic expression that incorporates variables to represent time and distance.

To create an expression that relates the hiker's distance from the end of the trail to the time elapsed since 10 A.M., we use the variable 'd' for distance and 't' for time. In this case, we discovered that the hiker travels at a steady rate, meaning the distance changes at a constant pace relative to time. This relationship is captured by the linear equation (d = -2t + 6), which is an algebraic expression showing how ‘d’ depends on ‘t’.

Rate of Change

The rate of change in a linear relationship is the speed at which one variable changes in relation to another. In the context of our hiking problem, the rate of change is the rate at which the hiker's distance from the end of the trail decreases over time. It's essentially the hiker's speed, and because it's constant, it tells us that the trail is being covered at a uniform pace.

Calculating the rate of change is as simple as comparing the change in distance with the change in time, which in this scenario is (2) miles per hour. We express this rate as a negative when we write the algebraic equation (d = -2t + 6) because the distance to the end of the trail decreases as time increases. Thus, the rate of change is -2, reflecting the hiker's consistent pace towards the end of the trail.

Y-Intercept

In linear equations, the y-intercept is the point where the line crosses the y-axis on a graph. This occurs when the x-variable (which is often time, 't', in our case) is zero. The y-intercept represents the starting value of the dependent variable before any changes have occurred in the independent variable.

In the hiking example, the y-intercept represents the distance from the end of the trail when the hike begins at 10 A.M. or (t=0) hours. Here, it's 6 miles, as given in the problem. On the graph of the hiker's journey, this is the point at which the line intersects the y-axis. The equation representing the hiker’s progress, (d = -2t + 6), clearly shows the y-intercept as 6. This value is critical for understanding the initial conditions of any problem related to linear relationships.