Question

A ship sends out ultrasound that returns from the seabed and is detected after 3.42s. If the speed of ultrasound through seawater is 1531m/s. What is the distance of the seabed from the ship?​

Answer 1

Given:-

t = 3.42s

v = 1531m/s

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Distance travelled by the ultrasound

= 2× depth of the sea

= 2d

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where d is the depth of the sea

2d = speed of sound × time

= 1531m/s × 3.42s

= 5236m

d = $\frac{5236}{2}$

= 2618m

Answer 2

Given:

Time = 3.42 seconds

Velocity = 1531 m/s.

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To find:

Distance = ?

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Solution:

An ultrasound from ship hits the seabed and comes back.

So the distance gets doubled

We know that,

$velocity = \frac{distance}{time}$

$distance = velocity \times time$

Since distance is doubled,

$2 \times distance = velocity \times time$

$\therefore \: distance = \frac{velocity \times time}{2}$

$= \frac{1531 \times 3.42}{2}$

$= 2618.01$

Distance from the ship to seabed

$= \boxed{ \large{2618.01 \: m}}$

Hope it helps you