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

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 = (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= (f_1(u))/(u-v)`

now, inserting the value of equation (1)

`f_2= f_0(u+v)/(u-v)`

`1275=1240* (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