Question

A police officer standing at the side of the road uses a radar emitting frequency of 24.15 GHz. A car is going away from the officer at a speed of 50mph. What will be the difference in frequency of the beam reflected by the car received back by the radar?

a. 4.0 kHz (lower frequency)
b. 1.8 kHz (higher frequency)
c. 4.0 kHz (higher frequency)
d. 1.8 kHz (lower frequency)

Answer 1

The correct answer is C

Step-by-step explanation:

From the question we are told that

The frequency of the radar is
`f = 24.15 \ GHz = 24.15 *10^(9) \ Hz`

The speed of the car is
`v = 50 mph = (50)/(2.237) = 22.35 \ m/s`

Generally the difference frequency reflected by the car and the frequency which the radar receives back is mathematically represented as

`\Delta f = (f * 2 * v )/( c )`

Here c is the speed of light with value
`c = 3.0 *10^(8) \ m/s`

=>
`\Delta f= (2 * 24.15 *10^(9) * 22.35)/( 3.0*10^(8))`

=>
`\Delta f = 4000 \ Hz`

=>
`\Delta f = 4.0 \ kHz`

Given that the value is positive then it a higher frequency