Respuesta :
Answer:
8.6 m/s
Explanation:
The observer is stationary
The source is moving
Let v₀ be the speed of sound in the air
Let v be the speed of the ambulance
As the siren approaches
f₁ = 2000(v₀ / (v₀ - v))
As the siren departs
f₁' = 2000(v₀ / (v₀ + v))
f₁ - f₁' = 100
2000(v₀ / (v₀ - v)) - 2000(v₀ / (v₀ + v)) = 100
v₀ / (v₀ - v) - v₀ / (v₀ + v) = 100/2000
v₀(v₀ + v) / (v₀ - v)(v₀ + v) - v₀(v₀ - v) / (v₀ - v)(v₀ + v) = 0.05
(v₀² + vv₀) - (v₀² - vv₀) / (v₀² - v²) = 0.05
2vv₀ / (v₀² - v²) = 0.05
2vv₀ = 0.05 (v₀² - v²)
0.05v² + 2vv₀ - 0.05v₀² = 0
v² + 40vv₀ - v₀² = 0
quadratic formula positive answer
v = (-40v₀ + √((40v₀)² - 4(1)(v₀²))) / 2
v = (-40v₀ + √(1604v₀²)) / 2
v = (0.049968v₀) / 2
v = 0.02498439v₀
If we assume the speed of sound in air is 343 m/s
v = 8.56964... = 8.6 m/s
Approaching frequency heard is 2051 Hz
Departing frequency heard is 1951 Hz
