Answers:
a)The balloon is 68 m away of the radar station
b) The direction of the balloon is towards the radar station
Explanation:
We can solve this problem with the Doppler shift equation:
[tex]f'=\frac{V+V_{o}}{V-V_{s}} f[/tex] (1)
Where:
[tex]f=250,000 Hz[/tex] is the actual frequency of the sound wave
[tex]f'=240,000 Hz[/tex] is the "observed" frequency
[tex]V=340 m/s[/tex] is the velocity of sound
[tex]V_{o}=0 m/s[/tex] is the velocity of the observer, which is stationary
[tex]V_{s}[/tex] is the velocity of the source, which is the balloon
Isolating [tex]V_{s}[/tex]:
[tex]V_{s}=\frac{V(f'-f)}{f'}[/tex] (2)
[tex]V_{s}=\frac{340 m/s(240,000 Hz-250,000 Hz)}{240,000 Hz}[/tex] (3)
[tex]V_{s}=-14.16 m/s[/tex] (4) This is the velocity of the balloon, note the negative sign indicates the direction of motion of the balloon: It is moving towards the radar station.
Now that we have the velocity of the balloon (hence its speed, the positive value) and the time ([tex]t=4.8 s[/tex]) given as data, we can find the distance:
[tex]d=V_{s}t[/tex] (5)
[tex]d=(14.16 m/s)(4.8 s)[/tex] (6)
Finally:
[tex]d=68 m[/tex] (8) This is the distance of the balloon from the radar station
