You are driving along a highway at 25 m/s when you hear the siren of an emergency vehicle traveling at constant speed in the opposite direction on the other side of the highway.When the vehicle is approaching you,you hear the frequency of the siren as 2380 Hz,but when it is past you the frequency becomes 1680 Hz.What is the speed of the vehicle? The speed of sound in air is 343 m/s.
A)35 m/s
B)32 m/s
C)30 m/s
D)41 m/s
E)50 m/s

Question

Quizplus · Accepted Answer

The speed of the vehicle can be found using the Doppler effect formula for sound. When the source and observer are moving towards each other, the observed frequency $f'$ is given by $f' = f \frac{v + v_o}{v - v_s}$, where $f$ is the emitted frequency, $v$ is the speed of sound, $v_o$ is the speed of the observer, and $v_s$ is the speed of the source. When moving away, the formula becomes $f' = f \frac{v - v_o}{v + v_s}$.Given that the observed frequencies are 2380 Hz and 1680 Hz when the vehicle is approaching and moving away, respectively, and the speed of sound $v = 343$ m/s, and the observer's speed $v_o = 25$ m/s, we can set up two equations and solve for $v_s$ (the speed of the vehicle) and $f$ (the emitted frequency).For the approaching scenario:$$2380 = f \frac{343 + 25}{343 - v_s}$$For the moving away scenario:$$1680 = f \frac{343 - 25}{343 + v_s}$$By solving these two equations simultaneously, we can find the value of $v_s$. However, to simplify the calculation without explicitly solving for $f$, we can use the ratio of the two observed frequencies to eliminate $f$ and directly solve for $v_s$.$\frac{2380}{1680} = \frac{343 + 25}{343 - v_s} \cdot \frac{343 + v_s}{343 - 25}$Solving this equation for $v_s$, we find that $v_s = 35$ m/s.

You Are Driving Along a Highway at 25 M/s When