Two submarines located one above the other both emit sonar pulses, which return from an underwater cliff in 1.02 seconds and 1.24 seconds respectively. If the speed of sound in salt water is 1531 m/s, find the distance between the two submarines (to the nearest integer in meters).

First, we need to determine the distance of each submarine from the underwater cliff.

For the first submarine:
- Total travel time for the sonar pulse: 1.02 seconds
- Speed of sound in salt water: 1531 m/s

The sonar pulse travels to the cliff and back, so the actual one-way distance to the cliff is half the total travel time multiplied by the speed of sound.

$d_1=\frac{1.02\text{ s}\times 1531\text{ m/s}}{2}$

$$\begin{gathered} d_1=\frac{1561.62\text{ m}}{2} \\ {d}_1=780.81\text{ m} \end{gathered}$$

For the second submarine:
- Total travel time for the sonar pulse: 1.24 seconds
- Speed of sound in salt water: 1531 m/s

Similarly, we calculate the one-way distance for the second submarine.

$$\begin{gathered} d_2=\frac{1.24\text{ s}\times 1531\text{ m/s}}{2} \\ {d}_2=\frac{1898.44\text{ m}}{2} \\ {d}_2=949.22\text{ m} \end{gathered}$$

The distance between the two submarines is the difference in their respective distances from the cliff.

$$\begin{gathered} \text{Distance between submarines}=d_2-d_1 \\ \text{Distance between submarines}=949.22\text{ m}-780.81\text{ m} \\ \text{Distance between submarines}=168.41\text{ m} \end{gathered}$$

Therefore, the distance between the two submarine is approx 168 meters.