Answer:
40.13491 m/s
Explanation:
[tex]v_r[/tex] = My speed = 35 m/s
v = Speed of sound in air = 343 Hz
[tex]v_s[/tex] = Speed of the police car
When the car is approaching
[tex]f=f'\dfrac{v-v_r}{v-v_s}\\\Rightarrow 1340=f'\dfrac{343-35}{343-v_s}[/tex]
When the car is receding
[tex]f=f'\dfrac{v+v_r}{v+v_s}\\\Rightarrow 1300=f'\dfrac{343+35}{343+v_s}[/tex]
Dividing the equations
[tex]\dfrac{1340}{1300}=\dfrac{f'\dfrac{343-35}{343-v_s}}{f'\dfrac{343+35}{343+v_s}}\\\Rightarrow \dfrac{1340}{1300}=\dfrac{22\left(v_s+343\right)}{27\left(-v_s+343\right)}\\\Rightarrow -36180v_s+12409740-12409740=28600v_s+9809800-12409740\\\Rightarrow \frac{-64780v_s}{-64780}=\frac{-2599940}{-64780}\\\Rightarrow v_s=\frac{129997}{3239}\\\Rightarrow v_s=40.13491\ m/s[/tex]
The speed of the police car is 40.13491 m/s