A driver in a car, approaching a vertical wall notices that the frequency of his car horn has changed from $440 \mathrm{Hz}$ to $480 \mathrm{Hz}$, when it gets reflected from the wall. If the speed of sound in air is $345 \mathrm{m} {\mathrm{s}}^{-1}$, then the speed of the car is:

Given,

Frequency of sound after reflection from the wall is 

$f^{\text{'}\text{'}}=480 \mathrm{Hz}$

Frequency of sound emitted by the horn of car is 

$f=440 \mathrm{Hz}$

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

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

$f_r^{\text{'}}=\left(\frac{v}{v-v_c}\right)f ...\left(1\right)$

As the frequency $f^{\text{'}}$ 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 ,

$f_r^{\text{'}\text{'}}=\left(\frac{v+v_c}{v}\right)f^{\text{'}} ...\left(2\right)$

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

$f_r^{\text{'}\text{'}}=\left(\frac{v+v_c}{v-v_c}\right)f ...\left(3\right)$

Substitute the given value in the equation (3), we get
$\Rightarrow 480=\left[\frac{345+v_c}{345-v_c}\right]440$

$\Rightarrow \frac{12}{11}=\frac{345+v_c}{345-v_c}$

Velocity of car is given by

$\Rightarrow v_c=54 \mathrm{km} {\mathrm{hr}}^{-1}$