Question

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.

Answer 1

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

  $f_1 = \dfrac{f_0(u + v)}{u}$..........(1)

u is the speed of sound = 340 m/s

v is the speed of the car

Frequency of the police car received

  $f_2= \dfrac{f_1(u)}{u-v}$

now, inserting the value of equation (1)

  $f_2= f_0\dfrac{u+v}{u-v}$

  $1275=1240\times \dfrac{340+v}{340-v}$

  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