Problem 301 - Elementary Algebra

Question

In the following exercises, solve. At 1:30 Marlon left his house to go to the beach, a distance of 7.6 miles. He rode his skateboard until \(2: 15,\) then walked the rest of the way. He arrived at the beach at 3:00. Marlon's speed on his skateboard is 2.5 times his walking speed. Find his speed when skateboarding and when walking.

Step-by-Step Solution

Answer

Walking speed: 2.9 mph, Skateboarding speed: 7.25 mph.

Step 1: Define Variables

Let Marlon's walking speed be denoted as \( w \) (in miles per hour). Therefore, his skateboarding speed will be \( 2.5w \) as he rides 2.5 times faster on his skateboard.

Step 2: Calculate Time Spent

Marlon skateboarded from 1:30 to 2:15, which is 45 minutes or 0.75 hours. He walked from 2:15 to 3:00, which is another 45 minutes or 0.75 hours.

Step 3: Set Up Distance Equation

The total distance to the beach is 7.6 miles, with the skateboarding distance being \( 2.5w \times 0.75 \) and the walking distance being \( w \times 0.75 \). Thus, the equation is: 2.5w \times 0.75 + w \times 0.75 = 7.6

Step 4: Simplify and Solve for w

Combine the terms in the equation: 1.875w + 0.75w = 7.6 Combine to get: 2.625w = 7.6 Solve for \( w \): \( w = \frac{7.6}{2.625} \approx 2.9 \) miles per hour

Step 5: Calculate Skateboarding Speed

Using the value of \( w \), calculate the skateboarding speed: Skateboarding speed \( = 2.5w = 2.5 \times 2.9 \approx 7.25 \) miles per hour

Key Concepts

Speed Calculation

Understanding speed is essential in solving distance-speed-time problems. Speed measures how fast an object moves and is usually expressed in miles per hour (mph) or kilometers per hour (kph). In this exercise, Marlon’s skateboarding speed is 2.5 times his walking speed. This means if we know his walking speed, we can easily find his skateboarding speed by multiplying the walking speed by 2.5.
For example, if Marlon walks at 2.9 mph, he skateboards at \[2.5 \times 2.9 = 7.25\] mph. This relationship allows us to transition between speeds based on different modes of travel, such as skateboarding or walking. With the formula for speed \[ \text{Speed} = \frac{\text{Distance}}{\text{Time}}\], we can calculate how fast Marlon is traveling for any segment of his journey.

Algebraic Equations

To solve distance-speed-time problems, setting up the right algebraic equations is crucial. These equations often involve defining variables to represent unknowns. Here, we defined Marlon's walking speed as \[w\], making his skateboarding speed \[2.5w\].
Using the times given in the problem (0.75 hours for skateboarding and 0.75 hours for walking), we can set up an equation relating these speeds to the total distance:
\[2.5w \times 0.75 + w \times 0.75 = 7.6\].
Combining terms, we get
\[1.875w + 0.75w = 7.6\].
This simplifies to
\[2.625w = 7.6\],
allowing us to solve for \[w\] by dividing both sides by 2.625. Algebra makes it possible to systematically find unknown quantities using known values and relationships.

Distance Problem Solving

Solving distance problems involves understanding how distance, speed, and time interact. In Marlon's case, the total distance to the beach is 7.6 miles. By dividing the journey into segments—skateboarding and walking—we can calculate the time and distance each activity covers.
Marlon skateboarding from 1:30 to 2:15 covers 0.75 hours, while walking from 2:15 to 3:00 covers another 0.75 hours. Knowing these times, and with the distances tied to their speeds, we can form equations:

• Distance skated: \[2.5w \times 0.75\]


• Distance walked: \[w \times 0.75\]

The sum of these distances equals the total distance: \2.5w \times 0.75 + w \times 0.75 = 7.6\.
By solving this equation step-by-step, we found Marlon’s walking speed (2.9 mph) and skateboarding speed (7.25 mph). Analyzing the distance, speed, and time in this structured method helps break down complex problems into manageable parts.