A stationary police car emits a sound of frequency 1200 Hz that bounces off of a car on the highway and returns to the police car with a frequency of 1250 Hz. The police car is right next to the highway, so the moving car is traveling either directly towards or directly away from the parked officer. Determine the car’s velocity, and whether the car was traveling towards or away from the police car.

The answer said it was 7.02 m/s and not 14.3 m/s

Well, if the police car is emitting a sound at 1200 Hz and the sound bounces back with a frequency of 1250 Hz, it means that the frequency of the sound wave has increased.

Now, we know that if a sound source and an observer are moving towards each other, the sound wave will appear to have a higher frequency (due to compression of the wave) – like a pop singer hitting a high note when squeezed into tight jeans.

Conversely, if the source and observer are moving away from each other, the sound wave will appear to have a lower frequency (due to stretching of the wave) – like a pop singer going flat when wearing loose-fitting clothes.

So, since the frequency of the sound wave has increased, we can conclude that the car is moving towards the police car.

To determine the car's velocity, we can use the Doppler effect equation:

Velocity = (Change in frequency / Original frequency) * Speed of sound

Given that the change in frequency is 1250 Hz - 1200 Hz = 50 Hz, and the original frequency is 1200 Hz, we can plug these values into the equation and solve for velocity. However, to complete this calculation, we also need to know the speed of sound.

Now, instead of boring you with the calculations, let's spice things up a bit! Why did the car cross the road? Because it could finally confirm that it was moving towards the police car!

But hey, let's not forget that we need to solve this mystery. Assuming the speed of sound is about 343 meters per second at room temperature, we can calculate the car's velocity using the Doppler equation. Plugging in the values, we get:

Velocity = (50 Hz / 1200 Hz) * 343 m/s

Now, you can whip out your calculator to get the exact velocity. Just remember to be careful not to drop it on your foot!

To determine the car's velocity and whether it is traveling towards or away from the police car, we can use the Doppler effect equation.

The Doppler effect is the change in frequency or wavelength of a wave as perceived by an observer moving relative to the source of the wave. In this case, the police car is the observer and the moving car is the source.

The equation for the Doppler effect in terms of frequency is as follows:

f' = (v + vr) / (v + vs) * f

Where:
- f' is the observed frequency
- f is the emitted frequency
- v is the speed of sound in air
- vr is the velocity of the receiver (police car)
- vs is the velocity of the source (moving car)

In this scenario, we know that the emitted frequency (f) is 1200 Hz, and the observed frequency (f') is 1250 Hz. The speed of sound in air (v) is approximately 343 m/s.

Let's assume that the velocity of the moving car is vs. If the moving car is traveling towards the police car, then the observed frequency should be higher (f' > f). On the other hand, if the car is traveling away from the police car, then the observed frequency should be lower (f' < f).

Let's solve for vr and vs using the equation above:

1250 Hz = (343 m/s + vr) / (343 m/s + vs) * 1200 Hz

To simplify the equation, let's divide both sides by 1200 Hz:

1250 Hz / 1200 Hz = (343 m/s + vr) / (343 m/s + vs)

1.0417 ≈ (343 m/s + vr) / (343 m/s + vs)

Now, let's consider two scenarios:
1. If the car is traveling towards the police car, then the observed frequency (f') would be higher than the emitted frequency (f), so f' > f. In this case, we can assume that f' = 1250 Hz and f = 1200 Hz.

Plugging in the values, we have:

1.0417 ≈ (343 m/s + vr) / (343 m/s - vs)

2. If the car is traveling away from the police car, then the observed frequency (f') would be lower than the emitted frequency (f), so f' < f. In this case, we can assume that f' = 1200 Hz and f = 1250 Hz.

Plugging in the values, we have:

1.0417 ≈ (343 m/s + vr) / (343 m/s + vs)

Now we have two equations to solve for vr and vs. Subtracting 1 from both sides, we get:

0.0417 ≈ vr / (343 m/s - vs) (for scenario 1)
0.0417 ≈ vr / (343 m/s + vs) (for scenario 2)

Now we can solve for vr and vs in these two separate scenarios:

For scenario 1:
vr ≈ 0.0417 * (343 m/s - vs)

For scenario 2:
vr ≈ 0.0417 * (343 m/s + vs)

Comparing the two equations, we can see that in scenario 1, vr will be positive if the car is moving towards the police car, and vs will be negative. In scenario 2, vr will be negative if the car is moving away from the police car, and vs will be positive.

By comparing these equations, we can determine the car's velocity and whether it is traveling towards or away from the police car.

F = ((V + Vr) / (V + Vs)) * Fo = 1250.

((343 + Vr) / (343 + 0)) * 1200 = 1250,
Divide both sides by 1200:
(343 + Vr) / 343 = 1.0417,
Multiply both sides by 343:
343 + Vr = 357.29,
Vr = 357.29 - 343 = 14.3m/s Toward the
police car.

Vr = Velocity of the receiver(car).
Vs = Velocity of the source(police car.