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A ship travelling towards a cliff receives the echo of its whistle after 3.5…

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PHYSICS

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

A. 10m

B. 350m

C. 175m

D. 125m

QUICK ANSWER…

D

DETAILS…  

To properly understand this question, please observe this sketch:

A ship travelling towards a cliff receives the echo of its whistle after 3.5 seconds

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.

Hence,

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

Hence,

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:

  1. A is not the right answer.
  2. B is incorrect.
  3. C is incorrect.
  4. D is correct. the ship is 125m closer.

KEY-POINTS…

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 /

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