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(A) Given:Time taken for the echo to return,$t = 4 \, s$.Speed of ultrasound in seawater,$v = 1550 \, m \, s^{-1}$.The ultrasound travels from the shi

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$3100$

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(A) Given:Time taken for the echo to return,$t = 4 \, s$.Speed of ultrasound in seawater,$v = 1550 \, m \, s^{-1}$.The ultrasound travels from the ship to the seabed and back to the ship. Therefore,the total distance covered is $2d$,where $d$ is the distance of the seabed from the ship.Using the formula: $	ext{Distance} = 	ext{Speed} 	imes 	ext{Time}$.$2d = v 	imes t$$2d = 1550 \, m \, s^{-1} 	imes 4 \, s$$2d = 6200 \, m$$d = 6200 / 2 = 3100 \, m$.Thus,the distance of the seabed from the ship is $3100 \, m$ or $3.1 \, km$.

$A$ ship sends out ultrasound waves that are detected after $4 \, s$ upon reflection from the seabed. If the speed of ultrasound through seawater is $1550 \, m \, s^{-1}$,find the distance of the seabed from the ship. (in $, m$)

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