Understanding the Journey Problem

This question involves two individuals, A and B, traveling towards each other from points X and Y, respectively. They meet at some point and then continue their journeys to the opposite starting points. We are given the time each takes to complete the remaining part of their journey after they have met and the speed of one of them. We need to find the speed of the other.

Key Concept: Time After Meeting

When two people or objects start simultaneously from two points and travel towards each other, and after meeting, they take time \(t_A'\) and \(t_B'\) respectively to reach their destinations, the ratio of their speeds (\(v_A\) and \(v_B\)) is related to the square root of the inverse ratio of the times taken after meeting.

The formula that relates their speeds and the time taken after meeting is:

\[ \frac{v_A}{v_B} = \sqrt{\frac{t_B'}{t_A'}} \]

Where:

• \(v_A\) is the speed of person A
• \(v_B\) is the speed of person B
• \(t_A'\) is the time taken by A to complete the remaining journey after meeting
• \(t_B'\) is the time taken by B to complete the remaining journey after meeting

Applying the Formula to Find the Speed of B

We are given the following information:

• Speed of A, \(v_A = 32\) km/h
• Time taken by A after meeting, \(t_A' = 6 \frac{1}{8}\) hours
• Time taken by B after meeting, \(t_B' = 8\) hours

First, let's convert the mixed fraction time \(t_A'\) into an improper fraction:

\(t_A' = 6 \frac{1}{8} = \frac{(6 \times 8) + 1}{8} = \frac{48 + 1}{8} = \frac{49}{8}\) hours.

Now, substitute the given values into the formula:

\[ \frac{32}{v_B} = \sqrt{\frac{8}{\frac{49}{8}}} \]

Simplify the expression inside the square root:

\[ \frac{32}{v_B} = \sqrt{8 \times \frac{8}{49}} \]

\[ \frac{32}{v_B} = \sqrt{\frac{64}{49}} \]

Calculate the square root:

\[ \frac{32}{v_B} = \frac{\sqrt{64}}{\sqrt{49}} \]

\[ \frac{32}{v_B} = \frac{8}{7} \]

Now, solve for \(v_B\). We can cross-multiply:

\(32 \times 7 = 8 \times v_B\)

\(224 = 8v_B\)

Divide both sides by 8:

\(v_B = \frac{224}{8}\)

\(v_B = 28\) km/h.

Therefore, the speed of B is 28 km/h.

Conclusion on B's Speed

Using the formula relating speeds and times taken after meeting, we found that the speed of B is 28 km/h.

Revision Table: Journey Problem Key Points

Additional Information: Derivation of the Formula

Let the point where A and B meet be M.

Distance covered by A before meeting = XM

Distance covered by B before meeting = YM

Let the time taken for A and B to meet at M be \(t\).

Then, \(XM = v_A \times t\) and \(YM = v_B \times t\).

After meeting at M, A travels from M to Y, taking time \(t_A'\). The distance MY = \(v_A \times t_A'\). But MY is the distance B covered before meeting (YM). So, \(v_A \times t_A' = YM\).

Similarly, after meeting at M, B travels from M to X, taking time \(t_B'\). The distance MX = \(v_B \times t_B'\). But MX is the distance A covered before meeting (XM). So, \(v_B \times t_B' = XM\).

We have:

• \(XM = v_A \times t = v_B \times t_B'\)
• \(YM = v_B \times t = v_A \times t_A'\)

From the first equation: \(t = \frac{v_B \times t_B'}{v_A}\)

Substitute this value of \(t\) into the second equation:

\(v_B \times \left(\frac{v_B \times t_B'}{v_A}\right) = v_A \times t_A'\)

\[ \frac{v_B^2 \times t_B'}{v_A} = v_A \times t_A' \]

Rearrange the terms to get the ratio of speeds:

\(v_B^2 \times t_B' = v_A^2 \times t_A'\)

\[ \frac{v_A^2}{v_B^2} = \frac{t_B'}{t_A'} \]

Taking the square root of both sides:

\[ \sqrt{\frac{v_A^2}{v_B^2}} = \sqrt{\frac{t_B'}{t_A'}} \]

\[ \frac{v_A}{v_B} = \sqrt{\frac{t_B'}{t_A'}} \]

This confirms the formula used to solve the problem. This formula is very useful for solving problems involving relative motion and time after meeting.