A research submarine moving underwater at a constant speed of 12.0\text{ m/s} directly tow… — JAMB UTME

A research submarine moving underwater at a constant speed of $12.0\text{ m/s}$ directly toward a vertical underwater cliff face emits an ultrasonic acoustic pulse. The echo reflected from the cliff face is detected by the submarine's receiver $2.50\text{ s}$ after emission. If the speed of sound in seawater is $1500\text{ m/s}$ , what is the distance between the submarine and the cliff face at the exact moment the echo is detected?

Answer: 1860 m

Answer

The distance between the submarine and the cliff face at the exact moment the echo is detected is

1860\text{ m}

During the

2.50\text{ s}

transit time of the acoustic signal, the sound covers a total path of

3750\text{ m}

(

1500\text{ m/s} \times 2.50\text{ s}

) while the submarine moves

30\text{ m}

closer to the cliff face (

12.0\text{ m/s} \times 2.50\text{ s}

). The total path of the sound consists of the outward journey to the cliff (

d + 30\text{ m}

) and the return journey to the submarine (

d

). Setting

(d + 30) + d = 3750

gives

2d + 30 = 3750

, leading to

d = 1860\text{ m}

Step-by-Step Solution

Calculate total sound travel distance and submarine displacement during the 2.50 s window.

Sound distance

d_{\text{sound}} = 1500\text{ m/s} \times 2.50\text{ s} = 3750\text{ m}

; Submarine displacement

d_{\text{sub}} = 12.0\text{ m/s} \times 2.50\text{ s} = 30.0\text{ m}

Both the acoustic wave and the submarine move continuously throughout the total elapsed transit time.

Establish the geometric equation for the sound path relative to the final distance d.

d_{\text{sound}} = 2d + d_{\text{sub}}

, where

d

is the remaining distance to the cliff face at detection time.

The sound pulse travels forward across the initial separation

(d + d_{\text{sub}})

and reflects back across the remaining separation

d

Solve the linear equation for the final separation distance d.

3750 = 2d + 30 \implies 2d = 3720 \implies d = 1860\text{ m}

Subtracting the submarine's forward displacement from the total sound path gives twice the distance to the obstacle at the instant of signal reception.

Key Concept

Echo distance calculations with moving receiver and source

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