A driver in a car, approaching a vertical wall notices that the...

Given,

Frequency of sound after reflection from the wall is

{f}^{' '} = 480Hz

Frequency of sound emitted by the horn of car is

$f = 440Hz$

Speed of sound is $v = 345m{s}^{- 1}$

The diver in the car (as a source) approaching the vertical wall which is the observer in this case, The frequency of sound received by the stationary wall is given by

${f}_{r}^{'} = \left(\dfrac{v}{v - {v}_{c}}\right)f . . . ( 1 )$

As the frequency ${f}^{'}$ is received by wall will act as a source now, and driver in the car is an observer, the frequency of sound received by driver is ,

{f}_{r}^{' '} = \left(\dfrac{v + {v}_{c}}{v}\right){f}^{'} . . . ( 2 )

Equating equation (1) and (2) we get,

{f}_{r}^{' '} = \left(\dfrac{v + {v}_{c}}{v - {v}_{c}}\right)f . . . ( 3 )

Substitute the given value in the equation (3), we get
$⇒ 480 = \left[\dfrac{345 + {v}_{c}}{345 - {v}_{c}}\right]440$

$⇒ \dfrac{12}{11} = \dfrac{345 + {v}_{c}}{345 - {v}_{c}}$

Velocity of car is given by

$⇒ {v}_{c} = 54km{hr}^{- 1}$