A student stands between two tall, parallel vertical walls and claps her hands once. She h… — JAMB UTME

A student stands between two tall, parallel vertical walls and claps her hands once. She hears the first echo after $1.0\text{ s}$ and the second echo after $1.5\text{ s}$ . Taking the speed of sound in air to be $340\text{ m s}^{-1}$ , what is the distance between the two walls?

$425\text{ m}$ Answer
B
$850\text{ m}$
C
$510\text{ m}$
D
$255\text{ m}$

Answer

The distance between the two walls is

425\text{ m}

Sound travels from the student to each wall and reflects back. The distance to the first wall is

\frac{340 \times 1.0}{2} = 170\text{ m}

and to the second wall is

\frac{340 \times 1.5}{2} = 255\text{ m}

. Adding both distances gives the total separation between the walls as

425\text{ m}

Step-by-Step Solution

Calculate the distance from the student to the closer wall (

d_1

)

d_1 = \frac{v \times t_1}{2} = \frac{340 \text{ m s}^{-1} \times 1.0 \text{ s}}{2} = 170\text{ m}

Sound travels to the wall and reflects back, so the time given corresponds to twice the distance.

Calculate the distance from the student to the further wall (

d_2

)

d_2 = \frac{v \times t_2}{2} = \frac{340 \text{ m s}^{-1} \times 1.5 \text{ s}}{2} = 255\text{ m}

The sound for the second echo travels to the second wall and back.

Determine the total distance between the two parallel walls

D = d_1 + d_2 = 170\text{ m} + 255\text{ m} = 425\text{ m}

Since the student is positioned between the two walls, the separation of the walls is the sum of both individual distances.

Key Concept

Echo distance relation

2d = v t

for sound reflection from barriers.

Estimated Time:1m 30s

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Step-by-Step Solution

Key Concept

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