Cutnell & Johnson Physics 7th Edition, volume 2. Chapter 24,question 32. A speed

Question

Cutnell & Johnson Physics 7th Edition, volume 2. Chapter 24,question 32.

A speeder is pulling directly away and increasing his distance froma police car that is moving at 28m/s with respect to the ground.The radar gun in the police car emits an electromagnetic wave witha frequency of 7.0 x 10^9 Hz. The wave reflects from the speeder'scar and returns to the police car, where its frequency is measuredto be 230Hz less than the emitted frequency. Find the speeder'sspeed with respect to the ground.

I used the equation fo=fs(1-Vrel/c)
fo=7.0 x 10^9 Hz - 230 Hz
fs=7.0 x10^9 Hz
c=3x10^8 m/s


I thought Vrel would be difference between two cars' speedbut am not getting the right answer.

Explanation / Answer

   for this we use the doppler effect from which thedoppler frequency for an electromagnetic    radiation is given by    fo = fs[1 ±(vrel / c)]    where vrel is the relativespeed between the source and the observer of the radiation    according to our pbm as the distance betweenthe police car and the speeders car is increasing    so the doppler frequency will changeto    fo - fs = -fs (vrel / c)    from the given problem after the wavereflects from the speeders car it returns to thepolice car    then its frequency will befo'    so the second doppler shift is    fo' - fo=  - fo (vrel /c)      adding the above tow equations we get    (fo - fs) +(fo' - fo) = - fo(vrel / c) - fs (vrel / c)    fo' - fo =- [fo (vrel / c) + fs(vrel / c)]               =- 2 fs (vrel / c)    but fo ˜ fs    so we get    fs - fo' ˜2 fs (vrel / c)    the relative speed will be    vrel = [(fs -fo') / 2 fs] c          = [230Hz / 2 (7.0 x 109 Hz)] (3 x 108 m /s)             =....... m / s    the speeders speed with respect to theground will be    vspeeder = vrel +vpolice                 =  vrel + 28 m / s                = ........ m / s

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