Let the time taken by 'A' to reach point Q be $t$ seconds. Since 'A' starts his journey at 1:00 p.m. with a speed of 1 m/s, the distance $PQ$ traveled by 'A' will be:

$\text{Distance} = \text{Speed} \times \text{Time} = 1 \times t = t \, \text{meters}.$

'B' starts 10 minutes (or 600 seconds) later, i.e., at 1:10 p.m. So, by the time 'B' starts, 'A' has already traveled for 600 seconds. When 'B' meets 'A' at point Q, both would have covered the same distance, $PQ$.

The time taken by 'B' to reach point Q will be $t - 600$ seconds, as he starts 600 seconds after 'A'. Since 'B' travels at a speed of 2 m/s, the distance covered by 'B' is:

$\text{Distance} = \text{Speed} \times \text{Time} = 2 \times (t - 600).$

Since both 'A' and 'B' meet at point Q, the distances covered by both must be equal:

$t = 2 \times (t - 600).$

Now, solve this equation for $t$:

$t = 2t - 1200,$ $1200 = t.$

So, $t = 1200$ seconds.

Thus, the distance $PQ$ is:

$\text{Distance} = t = 1200 \, \text{meters}.$

Therefore, the distance $PQ$ is 1200 meters.

Would you like further details or explanations?

Here are some related questions:

1. How would the solution change if 'B' started earlier?
2. What if 'A' and 'B' had different speeds?
3. Can this problem be solved using graphical methods?
4. How does the time delay affect when 'B' meets 'A'?
5. What is the significance of relative speed in such problems?

Tip: In problems involving relative motion, consider the time difference and speeds carefully to set up equations.