Problem 29 - Algebra A Combined Function

Question

Kevin Briley began a 186 -mile bicycle trip to build up stamina for a triathlon competition. Unfortunately, his bicycle chain broke, so he finished the trip walking. The whole trip took 6 hours. If Kevin walks at a rate of 4 miles per hour and rides at 40 miles per hour, find the amount of time he spent on the bicycle.

Step-by-Step Solution

Answer

Kevin spent 4.5 hours on the bicycle.

Step 1: Understand the Problem

Kevin completes a total trip of 186 miles in 6 hours. He rides a bicycle part of the way and walks the rest. We need to find out how much time he spent bicycling.

Step 2: Set Up Equations

Let the time spent bicycling be \( t_b \) hours and the time spent walking be \( t_w \) hours. We know that \( t_b + t_w = 6 \) because the total time is 6 hours.

Step 3: Formulate Distance Equations

The distance bicycled is \( 40t_b \) since he travels at 40 miles per hour. The distance walked is \( 4t_w \) since he walks at 4 miles per hour. The total distance is 186 miles, so we have \( 40t_b + 4t_w = 186 \).

Step 4: Substitute Time Equation

Substitute \( t_w = 6 - t_b \) into the distance equation: \( 40t_b + 4(6 - t_b) = 186 \).

Step 5: Solve for \( t_b \)

Simplify the equation: \( 40t_b + 24 - 4t_b = 186 \). Combine like terms to get \( 36t_b + 24 = 186 \). Subtract 24 from both sides to get \( 36t_b = 162 \). Divide by 36 to find \( t_b = \frac{162}{36} = 4.5 \).

Step 6: Conclusion

Kevin spent 4.5 hours riding the bicycle.

Key Concepts

Distance Formula

To understand Kevin’s trip, we first need to grasp the concept of the distance formula. The distance formula relates distance traveled, speed (or rate), and time. This fundamental relationship can be expressed as:\[ \text{Distance} = \text{Rate} \times \text{Time} \]In simple terms, this means the distance Kevin covers can be calculated by multiplying how fast he travels (his rate) with the time he takes. During his trip, Kevin rides a bicycle part of the way and walks the rest. - When riding, his rate is 40 miles per hour.- When walking, his rate is 4 miles per hour.Using these rates, we establish separate equations for the distance bicycled and walked. It's important because it allows us to solve problems by converting word problems into mathematical expressions. With the distance formula, we can explore how Kevin’s entire distance, 186 miles, is split between riding and walking.

Rate and Time Problems

Rate and time problems often require identifying the given rates and the relationship between time variables. In Kevin's scenario:- **Rate for bicycling:** 40 miles per hour- **Rate for walking:** 4 miles per hourKevin's total travel time is given as 6 hours. Therefore, the time spent biking plus the time spent walking must equal 6 hours.This is represented as:\[ t_b + t_w = 6 \]In this context:- \( t_b \) is the time spent bicycling- \( t_w \) is the time spent walkingFrom this equation, we learn that every moment not spent biking is spent walking. This equation is central to forming the next steps of solving the problem. By substituting into distance equations, we use this relationship to figure out exactly how long Kevin biked and walked to complete his journey.

Equation Solving

To find out how much time Kevin spent on his bicycle, we need to solve for the unknown time variable using simultaneous equations. This problem can initially seem daunting, but breaking it into steps makes solving more manageable.1. **List Equations:** - Total time: \( t_b + t_w = 6 \) - Total distance: \( 40t_b + 4t_w = 186 \)2. **Substitute the Time Equation:** Replace \( t_w \) with \( 6 - t_b \) in the distance equation: \[ 40t_b + 4(6 - t_b) = 186 \]3. **Simplify and Solve:** Simplify the equation to combine like terms: \[ 36t_b + 24 = 186 \] Subtract 24: \[ 36t_b = 162 \] Solve for \( t_b \): \( t_b = \frac{162}{36} = 4.5 \)By following these steps, we determine that Kevin spent 4.5 hours on his bicycle. The process of setting up and manipulating these equations is a crucial algebraic skill that simplifies solving word problems.