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A stationary police car emits a sound of frequency 1240 HzHz that bounces off of a car on the highway and returns with a frequency of 1275 HzHz. The police car is right next to the highway, so the moving car is traveling directly toward or away from it.

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Answer

given,

frequency from Police car= 1240 Hz

frequency of sound after return  = 1275 Hz

Calculating the speed of the car = ?

Using Doppler's effect formula

Frequency received by the other car

  [tex]f_1 = \dfrac{f_0(u + v)}{u}[/tex]..........(1)

u is the speed of sound = 340 m/s

v is the speed of the car

Frequency of the police car received

  [tex]f_2= \dfrac{f_1(u)}{u-v}[/tex]

now, inserting the value of equation (1)

  [tex]f_2= f_0\dfrac{u+v}{u-v}[/tex]

  [tex]1275=1240\times \dfrac{340+v}{340-v}[/tex]

  1.02822(340 - v) = 340 + v

   2.02822 v = 340 x 0.028822

   2.02822 v = 9.799

   v = 4.83 m/s

hence, the speed of the car is equal to v = 4.83 m/s

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