Question

A police car on the side of the road (at rest) uses a radar gun to catch speeders as they approach it. The frequency the radar gun emits is 8 x 109 Hz and the speed limit is 65 mi/hr. What is the difference in frequency of the emitted and returned radar wave

Answer 1

`\Delta f = 1.49 * 10^9 Hz`

Step-by-step explanation:

Apparent frequency that is received to the speeder is given as

`f_1 = f_0(v + v_o)/(v)`

`f_1 = (8 * 10^9)(340 + v_o)/(340)`

here we know that

`v_o = 65 mph = 29 m/s`

now we have

`f_1 = (8 * 10^9)(340 + 29)/(340)`

`f_1 = 8.68 * 10^9 Hz`

now the frequency that is received back from the speeder is given as

`f_2 = f_1(v)/(v- v_o)`

`f_2 = (8.68 * 10^9}(340)/(340 - 29)`

`f_2 = 9.49 * 10^9 Hz`

So difference is the frequency is given as

`\Delta f = 9.49 * 10^9 - 8 * 10^9`

`\Delta f = 1.49 * 10^9 Hz`