Problem 297 - Elementary Algebra

Question

In the following exercises, solve. Lorena walks the path around the park in 30 minutes. If she jogs, it takes her 20 minutes. Her jogging speed is 1.5 miles per hour faster than her walking speed. Find Lorena's walking speed and jogging speed.

Step-by-Step Solution

Answer

Lorena's walking speed is 3 mph and her jogging speed is 4.5 mph.

Step 1: Identify variables

Let Lorena's walking speed be denoted as \( w \) in miles per hour. Her jogging speed will then be \( w + 1.5 \) mph, since it is 1.5 mph faster than her walking speed.

Step 2: Convert time to hours

Convert the time Lorena takes to walk and jog from minutes to hours. Since 30 minutes is half an hour, it can be written as 0.5 hours. Similarly, 20 minutes is one-third of an hour, or approximately 0.333 hours.

Step 3: Write equations for distance

Since distance equals speed times time, the distance Lorena covers while walking is \( w \times 0.5 \), and the distance she covers while jogging is \( (w + 1.5) \times 0.333 \). These distances are equal since it's the same path around the park. Set up the equation: \[ w \times 0.5 = (w + 1.5) \times 0.333 \]

Step 4: Solve for walking speed

Expand and simplify the equation: \[ 0.5w = 0.333w + 0.4995 \]Subtract \(0.333w\) from both sides: \[ 0.167w = 0.4995 \]Solve for \( w \) by dividing both sides by 0.167: \[ w = \frac{0.4995}{0.167} \approx 3 \text{ miles per hour} \]

Step 5: Calculate jogging speed

Since jogging speed is 1.5 mph faster than walking speed, add 1.5 to the walking speed: \[ w + 1.5 = 3 + 1.5 = 4.5 \text{ miles per hour} \]

Key Concepts

Distance-Speed-Time Relationship

To solve Lorena's walking and jogging speed, it's crucial to understand the relationship between distance, speed, and time. This relationship is simple yet powerful and can be expressed using the formula:
\textrm{Distance} = \textrm{Speed} \times \textrm{Time}

Given that Lorena covers the same distance whether she's walking or jogging, we can set up equations to express this. In our example, we converted Lorena's walking time of 30 minutes to hours, which is 0.5 hours. Similarly, her jogging time of 20 minutes converts to approximately 0.333 hours. Using the distance-speed-time relationship, we formed the equation:
\[w \times 0.5 = (w + 1.5) \times 0.333 \ \]
Here, we have expressed the distances in terms of speed \(w\) (walking speed in miles per hour) and the given times. Understanding this relationship is a foundational step in solving problems involving speed and time. Remember that consistent units (e.g., converting all time to hours) are critical for correctness.

Algebraic Manipulation

Next, let’s focus on algebraic manipulation, which is key to solving the given problem. Once we form the equation
\[w \times 0.5 = (w + 1.5) \times 0.333 \ \]
we need to isolate and solve for \(w\). Expanding and simplifying this equation involves distributing and combining like terms. First, expand both sides:
\[0.5w = 0.333w + 0.4995 \ \]
Then, we subtract 0.333w from both sides to get:
\[0.167w = 0.4995 \ \]
Finally, we solve for \(w\) by dividing both sides by 0.167:
\[w = \frac{0.4995}{0.167} = \approx 3 \textrm{ miles per hour} \ \]
Algebraic manipulation allows us to isolate \(w\) and find its value efficiently. Ensuring each step is clear and logical is crucial to avoid mistakes.

Understanding Rates

Understanding rates, such as Lorena's walking and jogging speeds, is vital in many real-world contexts. A rate is a measure of one quantity per unit of another, like miles per hour (mph). Here, we know that jogging speed is 1.5 mph faster than walking speed. By representing walking speed as \(w\), jogging speed becomes \(w + 1.5\) mph.

• When we solved for \(w\), we found it to be 3 mph.
• Adding 1.5 mph gives us the jogging speed of 4.5 mph.

This means Lorena walks at 3 miles per hour and jogs at 4.5 miles per hour. Grasping how rates work and how they differ helps in understanding various problems in math and science. Whether comparing speeds, costs, or any other rates, the concept remains the same:
A rate=Amount of Change/Time.