A ship travelling towards a cliff receives the echo of its whistle after 3.5…
A ship travelling towards a cliff receives the echo of its whistle after 3.5 seconds. A short while later, it receives the echo after 2.5 seconds. If the speed of sound in air under the prevailing conditions is 250 m/s, how much closer is the ship to the cliff
To properly understand this question, please observe this sketch:
The aim is to find Z, this is because Z is how much closer the ship is when compared to the initial position.
We would solve for Z by firstly solving for X and then for Y,
Z = X – Y
To solve for X, please note that the total distance travelled by the sound is 2X.
Speed = 2X/time
250 = 2X/3.5
250 × 3.5 = 2X
X = (250 × 3.5)/2 = 875/2 = 437.5m
To solve for Y, the total distance travelled by the sound = 2Y
Speed = 2Y/time
250 = 2Y/2.5
250 × 2.5 = 2Y
Y = (250 × 2.5)/2 = 625/2 = 312.5
Hence, Z = X – Y = 437.5 – 312.5 = 125m
Now for the right answer to the above question:
- A is not the right answer.
- B is incorrect.
- C is incorrect.
- D is correct. the ship is 125m closer.
You may please note this/these:
- For echo, the right formula for speed will be: speed of sound = 2X/T.
- Where X = distance between the reflecting surface and the observer.
- T = time of echo.
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/ culled from 2020 JAMB-UTME physics past question 27 /