Distance Calculation: Speed Drop & Time Increase

The problem asks for the distance traveled when a person's speed decreases, resulting in a longer travel time.

• Initial Speed, \(S_1 = 20\) kmph
• Reduced Speed, \(S_2 = 18\) kmph
• Additional Time Taken, \(\Delta T = 8\) minutes

First, convert the additional time from minutes to hours:

\(\Delta T = \frac{8}{60} \text{ hours} = \frac{2}{15} \text{ hours}\)

Solving for Distance

Let the distance traveled be \(D\) km.

The time taken at the initial speed is \(T_1 = \frac{D}{S_1} = \frac{D}{20}\) hours.

The time taken at the reduced speed is \(T_2 = \frac{D}{S_2} = \frac{D}{18}\) hours.

The difference in time is given as 8 minutes (\(\frac{2}{15}\) hours), so \(T_2 - T_1 = \Delta T\).

Set up the equation:

\(\frac{D}{18} - \frac{D}{20} = \frac{2}{15}\)

To solve for \(D\), find a common denominator for the terms on the left side (which is 180):

\(\frac{10D}{180} - \frac{9D}{180} = \frac{2}{15}\)

Simplify the left side:

\(\frac{D}{180} = \frac{2}{15}\)

Now, isolate \(D\) by multiplying both sides by 180:

\(D = 180 \times \frac{2}{15}\)

Calculate the final distance:

\(D = 12 \times 2\)

\(D = 24\) km

The distance traveled is 24 km.