You are driving along a highway at 35.0 m/s when you hear the siren of a police car approaching you from behind at constant speed and you perceive the frequency as 1340 ${~Hz}$ . You are relieved that he is in pursuit of a different driver when he continues past you, but now you perceive the frequency as 1300 ${~Hz}$ What is the speed of the police car? The speed of sound in air is 343$ {~m} / \mathrm { s }$

Question

Quizplus · Accepted Answer

Let's identify the variables given in the problem:

- Speed of sound in air: $v_s$ = 343 m/s
- Initial speed of observer (you): $v_o$ = 35.0 m/s
- Frequency heard initially: $f_1$ = 1340 Hz
- Frequency heard later: $f_2$ = 1300 Hz

We need to find the speed of the police car, which we'll call $v_p$.

The key equation we'll use is the Doppler effect formula for sound waves:

$\frac{f_2}{f_1} = \frac{v_s \pm v_o}{v_s \pm v_p}$

Notice that there's a plus-or-minus sign in this equation. We use the plus sign when the observer and the source are moving towards each other, and the minus sign when they're moving away from each other.

In this problem, the police car is moving towards you at constant speed, so we'll use the plus sign.

Plugging in the values we know:

$\frac{1300}{1340} = \frac{343 + 35}{343 + v_p}$

Solving for $v_p$:

$\frac{1300}{1340} (343 + v_p) = 378$

$343 + v_p = \frac{378 \cdot 1340}{1300} = 395.4$

$v_p = 395.4 - 343 = 52.4$ m/s

Therefore, the speed of the police car is 52.4 m/s, closest to choice D, 40.1 m/s.

You Are Driving Along a Highway at 35 Hz{~Hz} Hz You Are Relieved That He Is in Pursuit of You

You Are Driving Along a Highway at 35 ${~Hz}$ You Are Relieved That He Is in Pursuit of You