Let's figure out the distance the person traveled using the information about their speeds and the difference in time taken.

Distance, Speed, and Time

The relationship between distance, speed, and time is a fundamental concept in motion problems. It is given by the formula:

\( \text{Distance} = \text{Speed} \times \text{Time} \)

From this formula, we can also express time as:

\( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \)

Setting Up the Problem Variables

Let the unknown distance be \(d\) kilometers.

We are given two different speeds at which the person walks:

• Speed 1 (\(S_1\)) = 2 km/hour
• Speed 2 (\(S_2\)) = 3 km/hour

Let \(T_1\) be the time taken to cover the distance \(d\) at Speed 1, and \(T_2\) be the time taken to cover the same distance \(d\) at Speed 2.

Calculating Time Taken at Each Speed

Using the formula \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \), we can write the time taken for each speed:

• Time taken at 2 km/hour: \( T_1 = \frac{d}{2} \) hours
• Time taken at 3 km/hour: \( T_2 = \frac{d}{3} \) hours

Using the Time Difference Information

The problem states that the person takes one hour more to cover the distance walking at 2 km/hour compared to walking at 3 km/hour. This means the difference between the two times is 1 hour.

So, we can write the equation:

\( T_1 - T_2 = 1 \)

Substitute the expressions for \(T_1\) and \(T_2\) we found earlier:

\( \frac{d}{2} - \frac{d}{3} = 1 \)

Solving for the Distance \(d\)

Now, we need to solve this equation to find the value of \(d\).

To subtract the fractions on the left side, we need a common denominator. The least common multiple of 2 and 3 is 6.

Multiply the first term by \( \frac{3}{3} \) and the second term by \( \frac{2}{2} \):

\( \frac{d}{2} \times \frac{3}{3} - \frac{d}{3} \times \frac{2}{2} = 1 \)

\( \frac{3d}{6} - \frac{2d}{6} = 1 \)

Combine the fractions on the left side:

\( \frac{3d - 2d}{6} = 1 \)

\( \frac{d}{6} = 1 \)

To isolate \(d\), multiply both sides of the equation by 6:

\( d = 1 \times 6 \)

\( d = 6 \)

Final Distance

The distance covered is 6 kilometers.