A submarine in the sea, sends ultrasonic ping and a stopwatch is started simultaneously. The stopwatch stops on receiving the reflected wave from an obstacle and reads 1 minute 40 seconds. Calculate the distance of the obstacle from the submarine (Speed of sound in water 1500 ms-1)

Given: time on the stopwatch = 1 minute 40 seconds = 100 s

speed of sound in water v = 1500 m s⁻¹

The stopwatch runs from the moment the ping is sent until the reflection returns, so the 100 s covers the round trip.

2d = v × t 2d = 1500 × 100 2d = 150 000 d = 150 000 ÷ 2 d = 75 000 m

Distance of the obstacle from the submarine = 75 000 m (75 km)

Convert the time first: 1 minute 40 seconds is 100 s, not 1.40 s or 140 s. Mixed time units are there to catch you.

Then the usual echo relation, 2d = vt. The stopwatch was started when the ping left and stopped when it came back, so it timed both legs of the journey — halve the total distance to get the range. Skipping that gives 150 km.

Note 1500 m s⁻¹ for sound in water, over four times the speed in air. Sound travels fastest in solids, slower in liquids and slowest in gases.