Physics 2019 JAMB Past Questions - Question 9: A ship travelling toward 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 the sound in air under the prevailing conditions is 250ms-¹, how much closer is the ship to the cliff?

Correct Answer

A

Explanation

Difference in time = 1 sececho time = ½ = 0.5secsDistance of cliff from ship = 0.5 x 250
To determine how much closer the ship is to the cliff, we can use the speed of sound and the time it takes for the echo to reach the ship.

Let's denote the distance between the ship and the cliff as "d". When the ship emits a sound, it travels to the cliff and then reflects back to the ship. The total distance traveled by the sound is 2d.

We know that the speed of sound is 250 m/s, and the time it takes for the echo to reach the ship is 3.5 seconds and 2.5 seconds, respectively.

Using the formula distance = speed × time, we can calculate the total distance traveled by the sound in both cases:

For the first echo:
Distance = Speed × Time
2d = 250 m/s × 3.5 s

For the second echo:
Distance = Speed × Time
2d = 250 m/s × 2.5 s

Now, let's solve these equations to find the value of "d":

For the first echo:
2d = 250 m/s × 3.5 s
2d = 875 m
d = 875 m / 2
d = 437.5 m

For the second echo:
2d = 250 m/s × 2.5 s
2d = 625 m
d = 625 m / 2
d = 312.5 m

Therefore, the ship is 437.5 m - 312.5 m = 125 m closer to the cliff.