Question

A speeder is pulling directly away and increasing his distance from a police car that is moving at 25.1 m/s with respect to the ground. The radar gun in the police car emits an electromagnetic wave with a frequency of 6.98×109 Hz. The wave reflects from the speeder's car and returns to the police car, where its frequency is measured to be 312 Hz less than the emitted frequency. Calculate the speeder's speed with respect to the ground.

Answer 1

given

a police car that is moving at 25 m/s = V_p   

with a frequency of 6.98 * 10^9 Hz = f_s

measured to be 322 Hz less than the emitted frequency = f_s - f_o

V_r = relative velocity

V_r = ( f_s - f_o / 2 f_s ) X C

V_r = ( 312 / ( 2 * 6.98 * 10^9) ) * 3 * 10^8

V_r = 6.71

the speeder's speed with respect to the ground is = V_s

but we have V_r = V_s - V_p

then V_s = V_r + V_p

V_s = 6.71 + 25.1

V_s = 31.81 m/sec

the speeder's speed with respect to the ground is = V_s = 31.81 m/sec