A police car is sounding a siren with a frequency of $1280$ Hz while traveling right towards you. At a certain time, you measure the frequency of the siren to be $1400$ Hz, and increasing at a rate of $1$ Hz/sec. What is the police car's speed in km/h and acceleration in km/h/sec at the time of the observation.
Hint: Recall that the Doppler effect measures the change in frequency of a wave for an observer moving relative to its source. In our case, the source of the siren is moving at velocity $v$ while the observer is standing still. The Doppler effect formula is $$f=\frac{v_s}{v_s+v}f_0$$ where $v_s$ is the speed of sound ($340$ m/s) in dry air at $20^\circ C$ and $f_0$ is the original frequency of the siren
This is a related rates problem involving the Doppler Effect! My teacher has told me that all I need to use to solve the problem is the provided information. I'm stuck and have no idea what to do, all I know is that at some point I'll be taking the derivative of the equation is given, but I'm not sure how to do that either. Any help would be very much appreciated!
Best Answer
We can just use the equation to find the velocity $v$. $$1400=\frac{343}{343+v}\cdot 1280$$ so $v=-29.4$ m/s. But the question asks for the speed (absolute value of velocity) in km/h. $$29.4 \ \text{m}/\text{s}=105.84 \ \text{km}/\text{h}$$ To find the acceleration (the derivative of velocity with respect to time) we take the derivative with respect to $t$ of the equation $$f=\frac{v_s}{v_s+v}f_0=v_sf_0(v_s+v)^{-1}$$$$\frac{\mathrm{d}f}{\mathrm{d}t}=-v_sf_0(v_s+v)^{-2}\frac{\mathrm{d}v}{\mathrm{d}t}$$ Plugging in known values we get $$1=-343\cdot1280(343+105.84)^{-2}\frac{\mathrm{d}v}{\mathrm{d}t}$$$$\frac{\mathrm{d}v}{\mathrm{d}t}=-.459 \ \text{km}/\text{h}/\text{sec}$$ The speed of the car is $105.84$ km/h and its acceleration is $-.459$ km/h/s.