Answer:
48.99 m/s
Explanation:
So, to solve this problem, we just need to remember the Doppler effect equation, for an observer moving at speed [tex]v_o[/tex] relative to the medium of propagation, with a source moving at speed [tex]v_s[/tex] relative to the medium of propagation, emitting the sound with a frequency [tex]f_0[/tex], and for a wave speed in the medium of c, the measured frequency of the wave by the observer, f , will be:
[tex]f = (\frac{c \ \pm \ v_o}{ c\ \mp \ v_s} ) f_0[/tex],
where the signs depends of the relative movement between observer and source.
If the source its approaching the observer, you got an plus in the upside side of the fraction and a minus below, and if the source its moving away from the observer, you got an minus upside and a plus below.
Now, for our problem, we got:
Speed of the observer relative to the medium= 0 m/s
wave speed in air:
c = 343 m/s
So, our equation will be:
[tex]f = (\frac{ 343 \frac{m}{s} }{ 343 \frac{m}{s} \ \mp \ v_s} ) f_0[/tex].
Now, we got two parts of the problem.
In the first part, the car its approaching you, and you measure a frequency of 80.0 hz. So, the equation will be:
[tex] 80.0 \ hz = (\frac{ 343 \ \frac{m}{s} }{ 343 \ \frac{m}{s} \ - \ v_s} ) f_0[/tex].
In the second part, the car its getting away, and you measure a frequency of 60.0 hz. So, the equation will be:
[tex] 60.0 \ hz = (\frac{ 343 \ \frac{m}{s} }{ 343 \ \frac{m}{s} \+ \ v_s} ) f_0[/tex].
From this two equations, its easy to get:
[tex] 343 \ \frac{m}{s} \ - \ v_s = (\frac{ 343 \ \frac{m}{s} }{ 80.0 \ hz } ) f_0[/tex],
and
[tex] 343 \ \frac{m}{s} \ + \ v_s = (\frac{ 343 \ \frac{m}{s} }{ 60.0 \ hz } ) f_0[/tex].
We can add this equations to obtain:
[tex]343 \ \frac{m}{s} \ + \ v_s + 343 \ \frac{m}{s} \ - \ v_s = (\frac{ 343 \ \frac{m}{s} }{ 60.0 \ hz } ) f_0 + (\frac{ 343 \ \frac{m}{s} }{ 80.0 \ hz } ) f_0[/tex]
[tex]2 * 343 \ \frac{m}{s} = (\frac{ 343 \ \frac{m}{s} }{ 60.0 \ hz } + \frac{ 343 \ \frac{m}{s} }{ 80.0 \ hz } ) f_0[/tex]
[tex]686 \ \frac{m}{s} = (\frac{ 343 \ \frac{m}{s} }{ 60.0 \ hz } + \frac{ 343 \ \frac{m}{s} }{ 80.0 \ hz } ) f_0[/tex]
[tex]68.57 \ Hz = f_0[/tex]
Now we can just replace in one of the equations above
[tex] 343 \ \frac{m}{s} \ + \ v_s = (\frac{ 343 \ \frac{m}{s} }{ 60.0 \ hz } ) f_0[/tex].
[tex] 343 \ \frac{m}{s} \ + \ v_s = (\frac{ 343 \ \frac{m}{s} }{ 60.0 \ hz } ) * 68.57 \ Hz [/tex].
[tex] 343 \ \frac{m}{s} \ + \ v_s = 391.99 \ \frac{m}{s} [/tex].
[tex] v_s = 391.99 \ \frac{m}{s} - 343 \ \frac{m}{s} [/tex].
[tex] v_s = 48.99 \ \frac{m}{s} [/tex].
