Solving a Walking Speed Problem: A Comprehensive Guide

In this article, we will solve a walking speed problem involving multiple segments of a journey. This problem will involve setting up equations based on distance and time relationships and solving them using algebraic methods, specifically the quadratic equation. Additionally, we will walk through the reasoning and calculations step-by-step.

Problem Statement

A woman walks 20 km at a certain speed and 10 km at a speed 3 km/h faster. If the total time for the journey is 12 hours, what is her original speed?

Symbols and Equations

Let the woman's original speed be x km/h. Then her speed for the second part of the journey is x 3 km/h.

We can set up the equations for the time taken for each part of the journey using the formula:

Time Distance ÷ Speed

Time for the first part (20 km): Time_1 20 ÷ x Time for the second part (10 km): Time_2 10 ÷ (x 3)

According to the problem, the total time for both parts of the journey is 12 hours. Therefore, we have:

Time_1 Time_2 12

Substituting the expressions for Time_1 and Time_2, we get:

20 ÷ x 10 ÷ (x 3) 12

Solving the Equation

To solve the equation, we first find a common denominator, which is x(x 3):

(20(x 3) 1) ÷ x(x 3) 12

Multiplying both sides by x(x 3), we get:

20(x 3) 1 12x(x 3)

Expanding and simplifying the equation:

2 60 1 12x^2 36x

Combining like terms:

3 60 12x^2 36x

Rearranging the equation:

12x^2 36x - 3 - 60 0

Simplifying further:

12x^2 6x - 60 0

Dividing the entire equation by 6:

2x^2 x - 10 0

Solving the Quadratic Equation

Now, we use the quadratic formula x (-b ± sqrt(b^2 - 4ac)) ÷ 2a, where a 2, b 1, and c -10:

x (-1 ± sqrt(1^2 - 4 ? 2 ? (-10))) ÷ (2 ? 2)

Calculating the discriminant:

1^2 - 4 ? 2 ? (-10) 1 80 81

Applying the quadratic formula:

x (-1 ± 9) ÷ 4

This gives us two potential solutions:

x (8 ÷ 4) 2 and x (-10 ÷ 4) -2.5

Since speed cannot be negative, we take x 2 km/h.

Conclusion

The woman's original speed is 2 km/h.

Verification

Verification steps:

Time for the first part (20 km): 20 ÷ 2 10 hours Time for the second part (10 km): 10 ÷ (2 3) 2 hours Total time: 10 2 12 hours, which matches the given condition.

Additional Problem

Let the speed be x km/h, then:

20/x 10/(x 3) 12

Multiplying both sides by x(x 3), we get:

20(x 3) 1 12x(x 3)

Expanding and simplifying the equation:

2 60 1 12x^2 36x

Combining like terms:

3 60 12x^2 36x

Rearranging the equation:

12x^2 36x - 3 - 60 0

Simplifying further:

12x^2 - 6x - 60 0

Dividing the entire equation by 6:

2x^2 - x - 10 0

Solving the Quadratic Equation

Now, we use the quadratic formula x (-b ± sqrt(b^2 - 4ac)) ÷ 2a, where a 2, b -1, and c -10:

x (-(-1) ± sqrt((-1)^2 - 4 ? 2 ? (-10))) ÷ (2 ? 2)

Calculating the discriminant:

1 80 81

Applying the quadratic formula:

x (1 ± 9) ÷ 4

This gives us two potential solutions:

x (10 ÷ 4) 2.5 and x (-8 ÷ 4) -2

Since speed cannot be negative, we take x 2.5 km/h. However, the problem states that the original speed is 4 km/h, which implies a different setup and verification.

Alternative Solution

Let s be the original speed and s_2 be the faster speed. Then:

16/s 12/s_2 6

Multiplying both sides by s_2s, we get:

16s_2 12s 6s_2s

Rearranging the equation:

6s_2 12s 6s_2s

Multiplying and simplifying:

6s_2 - 8s - 16 0

Dividing the entire equation by 3:

2s_2 - 8/3s - 16/3 0

Solving for s and s_2, we get:

s 4 km/h and s_2 6 km/h

The original speed is 4 km/h.

Distance-Time Relationship

Total time: 6 hours

Total distance: 28 km

Average speed of the entire trip: 4.66 km/h

Distance: 16 km Answer: Original speed was 4 km/h, Time: 4 hours Distance: 12 km Speed: 6 km/h Time: 2 hours