A train approaching a detector emits a siren of frequency fs. The detector registers a frequency of 3148 Hz when the train is moving towards it, and 2073 Hz when it is moving away. If the speed of sou

The Doppler effect is the apparent change in the frequency (or pitch) of a sound wave detected by a listener, due to the relative motion between the source of the sound and the listener.

According to the Doppler effect, when a sound source moves towards a listener, the observed frequency is higher than the source frequency. From the graph, 3148 Hz is the higher observed frequency. Therefore, the detector records 3148 Hz when the train moves TOWARDS the detector.

Let $f_s$ be the frequency of the siren emitted by the train, $v$ be the speed of sound in air ($340 \text{ m/s}$), and $v_s$ be the speed of the train.
When the train moves towards the detector, the observed frequency $f_{L, \text{towards}} = 3148 \text{ Hz}$.

When the train moves away from the detector, the observed frequency $f_{L, \text{away}} = 2073 \text{ Hz}$.

Divide equation (1) by equation (2):

The speed of the train is approximately $\boxed{\text{70.01 m/s}}$.

The detector started recording when the train was $350 \text{ m}$ away. Time $t_1$ is when the train passes the detector, meaning it has covered the $350 \text{ m}$ distance.
Using the formula for constant speed: distance = speed $\times$ time.

Time $t_1$ is approximately $\boxed{\text{5.00 s}}$.