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A stationary police car emits a sound of frequency of 1240 HZ that bounces off a car on the highway and returns with a frequency of 1270 HZ. The police car is right next to the highway, so the car is either moving right towards the police car, or away from it.
A) How fast was the car moving? Was it moving towards, or away from the police car?
B) What frequency would the police car have received if it had been moving toward the other car at 20 m/s?

Respuesta :

Answer:

Explanation:

Since the sound bouncing back from car has greater frequency , the car must be moving towards the police car . If v be the velocity of car

f = [tex]f_0\times\frac{V+v}{V-v}[/tex]

f is apparent frequency , f₀ is original frequency , V is velocity of sound and v is velocity of car

1270 = [tex]1240\times\frac{340+v}{340-v}[/tex]

1.02419 = [tex]\frac{340+v}{340-v}[/tex]

348.22-1.02419v =340 +v

2.02419v  = 8.22

v = 4.06 m /s

B)

In this case , we shall take relative velocity in place of velocity of car .So

v = 20+4.06

= 24.06 m /s

f = [tex]1240\times\frac{340+v}{340-v}[/tex]

= [tex]1240\times\frac{340+24}{340-24}[/tex]

1240 x [tex]\frac{364}{316}[/tex]

= 1428 Hz .