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2 votes
A train approaching a hill at a speed of 18 km/h sounds a whistle when it is 1600 m away from a hill. Wind with speed 18 km/h is blowing in the direction of motion of train. The train sound whistle again when first echo is heard by the driver. The distance from the hill at which the second echo from the hill is heard by the driver is approximately [Speed of sound is 315 m/s]​

2 Answers

3 votes

Answer: speed of sound in air is c and as the air is also moving the velocity of sound is c+v

a

where v

a

is the speed of air

The train sounds the whistle when it is at a distance of x from the hill. Sound moving with velocity v with respect to ground, takes time t to reach the hill

t=

v

x

=

c+v

a

x

after the reflection from the sound waves travels towards the train hence now the velocity of sound is c−v

a

traveledlet the distance concerning by reflected wave be x

therefore time the by it t

=

c−v

a

x

the total time between echoing the whistle and its reflection t+t

meanwhilethe is taken, the train travels the distance x−x

with constant speed v

s

x−x

=(t+t

)v

s

x−x

=

c+v

a

v

s

x+

c−v

a

v

s

x

x

c+v

A

c+v

a

−v

s

=x

c−v

a

c−v

a

+v

s

x

=x

(c+v

a

)×(c−v

a

+v

s

)

(c−v

a

)×(c+v

a

−v

s

)

Step-by-step explanation:

User Frenchi In LA
by
7.7k points
5 votes
Let's first find the time taken by the sound to travel from the train to the hill, and then from the hill back to the train.

The speed of the train is 18 km/h, which is equivalent to 5 m/s (since 1 km/h = 1000 m/3600 s = 5/18 m/s).

The distance between the train and the hill when the whistle is sounded is 1600 m. Let's call this distance d1.

Using the formula for time, t = d / v, where d is distance and v is speed, we get:

t1 = d1 / v
= 1600 / 5
= 320 s

During this time, the train moves a distance of d2 = v * t1 = 5 * 320 = 1600 m closer to the hill.

The speed of sound in air is 315 m/s.

The wind is blowing in the direction of motion of the train, so the effective speed of sound in the direction of the train is:

v_eff = v_sound + v_wind
= 315 + 18
= 333 m/s

Let's call the distance from the train to the hill when the first echo is heard as d3.

The sound wave travels from the train to the hill, and then reflects off the hill and travels back to the train. During this time, the train has moved a distance of d2 = 1600 m closer to the hill.

So the total distance traveled by the sound wave is:

d_total = d1 + d2 + d3 + d2

At the moment the second echo is heard by the driver, the sound wave has traveled twice the distance d_total. So we can write:

2 * d_total = v_eff * t2

where t2 is the time taken for the sound wave to travel from the train to the hill and back to the train again.

Using the formula for time, t = d / v, we can write:

t2 = (d1 + d3 + 2 * d2) / v_eff

Substituting this expression for t2 in the equation above, we get:

2 * d_total = (d1 + d3 + 2 * d2) * v_eff / 2

Simplifying this equation, we get:

d_total = (d1 + d3 + 2 * d2) * v_eff / (2 * 2)

Substituting the values we have calculated, we get:

3200 = (1600 + d3 + 2 * 1600) * 333 / 4

Simplifying this equation, we get:

d3 = 2472.6 m

Therefore, the distance from the hill at which the second echo from the hill is heard by the driver is approximately 2472.6 m
User SSG
by
7.2k points