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Grade 12 Science · Earth & Space Science

The ocean: seafloor, currents and ocean–atmosphere interaction

Use this for “A sonar echo returns in 6 s, and sound travels 1,500 m/s in seawater. How deep is the ocean there?”

Diagram

02,0004,0006,0008,00010,000Depth (m)shelfsloperiseabyssal plainmid-ocean ridgetrench4,500 mdepths stretched, not to scale across
The seafloor is 4,500 m under the ship.
This is an example. Type your own values to start a new problem.
d (m)
=
v (m/s)
×
t (s)
÷
2

Formulas

d
=
v
×
t
2
4,500
=
1,500
×
6
2

Assumptions

•
The ping goes down and back, so halve the path.
•
Sound travels about 1,500 m/s in seawater.
•
The shelf is under 200 m deep and trenches reach almost 11,000 m.

Step-by-step

WE KNOW
t = 6 s, v = 1,500 m/s
FIND
depth (d)
Step 1 · Find depth (d)
Use d = v × t ÷ 2
Speed times time is the path down and back; the depth is half of it.
d
=
1,500
×
6
2
d
=
9,000
2
d
=
4,500
m
Check
Put the numbers back in. Both sides should match.
4,500
=
1,500
×
6
2
✓
Topic check
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