Doppler and electromagnetic waves

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The discussion focuses on calculating the speed of a speeder moving away from a police car using the Doppler effect. The police car emits an electromagnetic wave at a frequency of 6.0 x 10^9 Hz, which reflects off the speeder's car and returns with a frequency 316 Hz lower. The formula used is fo = fs * (1 ± vrel/c), where fo is the emitted frequency, fs is the observed frequency, vrel is the relative speed, and c is the speed of light. By applying the values into the formula, the calculated speed of the speeder with respect to the ground is approximately 498 m/s. This demonstrates the practical application of the Doppler effect in determining the speed of moving objects.
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A speeder is pulling directly away and increasing his distance from a police car that is moving at 23 m/s with respect to the ground. The radar gun in the police car emits an electromagnetic wave with a frequency of 6.0 109 Hz. The wave reflects from the speeder's car and returns to the police car, where its frequency is measured to be 316 Hz less than the emitted frequency. Find the speeder's speed with respect to the ground.

fo = fs * (1 +- vrel/c)

I am stressing out and can't get this at all
 
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.Answer: The speed of the speeder with respect to the ground is vrel = (fo - fs)c/fo = (6.0x109-316x106)c/6.0x109 = 4.98 x 10^2 m/s
 
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