Calculating Speed After Meeting Point

This problem involves two individuals, A and B, travelling towards each other from two different points, P and Q. They meet somewhere between P and Q and then continue their journeys to the other's starting point. We are given the time each person takes to reach their destination *after* they have met and the speed of one person. We need to find the speed of the other person.

This specific type of problem has a well-known formula relating the speeds of the two individuals to the times they take to cover the remaining distance after meeting.

Formula for Speed and Time After Meeting

If two people, A and B, start at the same time from points P and Q and travel towards each other, meeting at a point M, and if A takes \(t_A\) hours to reach Q from M, and B takes \(t_B\) hours to reach P from M, then the ratio of their speeds (\(v_A\) and \(v_B\)) is given by:

$$ \frac{v_A}{v_B} = \sqrt{\frac{t_B}{t_A}} $$

This formula is derived from the fact that they both travel for the same amount of time until they meet, and the distance covered by each before meeting is proportional to their speed. After meeting, A covers the distance that B covered before meeting, and B covers the distance that A covered before meeting.

Applying the Given Information

We are given the following information:

• Time taken by A after meeting to reach Q (\(t_A\)) = \(6\frac{1}{8}\) hours.
• Time taken by B after meeting to reach P (\(t_B\)) = 8 hours.
• Speed of B (\(v_B\)) = 16.8 km/h.

First, let's convert the time taken by A to an improper fraction:

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

Step-by-Step Calculation

Now, we can plug the values into the formula:

$$ \frac{v_A}{v_B} = \sqrt{\frac{t_B}{t_A}} $$

Substitute the given values:

$$ \frac{v_A}{16.8} = \sqrt{\frac{8}{\frac{49}{8}}} $$

Simplify the expression under the square root:

$$ \frac{8}{\frac{49}{8}} = 8 \times \frac{8}{49} = \frac{64}{49} $$

So, the equation becomes:

$$ \frac{v_A}{16.8} = \sqrt{\frac{64}{49}} $$

Calculate the square root:

$$ \sqrt{\frac{64}{49}} = \frac{\sqrt{64}}{\sqrt{49}} = \frac{8}{7} $$

Now the equation is:

$$ \frac{v_A}{16.8} = \frac{8}{7} $$

Solve for \(v_A\):

$$ v_A = 16.8 \times \frac{8}{7} $$

Calculate the value:

$$ v_A = \frac{16.8 \times 8}{7} $$

We can perform the division first:

$$ \frac{16.8}{7} = 2.4 $$

Then multiply by 8:

$$ v_A = 2.4 \times 8 $$

$$ v_A = 19.2 $$

The speed of A is 19.2 km/hr.

Let's summarize the values:

Verification

Let's quickly check the ratio of speeds:

$$ \frac{v_A}{v_B} = \frac{19.2}{16.8} = \frac{192}{168} $$

Divide both by 24:

$$ \frac{192 \div 24}{168 \div 24} = \frac{8}{7} $$

Now let's check the ratio of square root of times in reverse order:

$$ \sqrt{\frac{t_B}{t_A}} = \sqrt{\frac{8}{\frac{49}{8}}} = \sqrt{\frac{64}{49}} = \frac{8}{7} $$

The ratios match, confirming our calculation is correct.

Revision Table: Speed and Time Problems

Additional Information: Understanding the Speed After Meeting Formula

Let's understand the derivation of the formula \(\frac{v_A}{v_B} = \sqrt{\frac{t_B}{t_A}}\).

Assume A starts from P and B starts from Q at the same time and they meet at point M after time \(T\).

Distance PM = \(v_A \times T\)

Distance QM = \(v_B \times T\)

After meeting, A travels from M to Q, taking time \(t_A\). The distance MQ is covered by A at speed \(v_A\).

Distance MQ = \(v_A \times t_A\)

After meeting, B travels from M to P, taking time \(t_B\). The distance MP is covered by B at speed \(v_B\).

Distance MP = \(v_B \times t_B\)

Notice that the distance PM is the same as the distance covered by B after meeting (from M to P). So,

\(v_A \times T = v_B \times t_B\) -- (Equation 1)

Also, the distance QM is the same as the distance covered by A after meeting (from M to Q). So,

\(v_B \times T = v_A \times t_A\) -- (Equation 2)

Now, divide Equation 1 by Equation 2:

$$ \frac{v_A \times T}{v_B \times T} = \frac{v_B \times t_B}{v_A \times t_A} $$

The \(T\) cancels out on the left side:

$$ \frac{v_A}{v_B} = \frac{v_B \times t_B}{v_A \times t_A} $$

Rearrange the terms to group \(v_A\) and \(v_B\) on one side and \(t_A\) and \(t_B\) on the other:

$$ v_A \times v_A \times t_A = v_B \times v_B \times t_B $$

$$ v_A^2 \times t_A = v_B^2 \times t_B $$

$$ \frac{v_A^2}{v_B^2} = \frac{t_B}{t_A} $$

$$ \left(\frac{v_A}{v_B}\right)^2 = \frac{t_B}{t_A} $$

Taking the square root of both sides (since speeds and times are positive):

$$ \frac{v_A}{v_B} = \sqrt{\frac{t_B}{t_A}} $$

This confirms the formula used in the solution.

The final answer for the speed of A is 19.2 km/hr.