Bullets and train $d=v\cdot t$ problem

Question

Vinay fires two bullets from the same place at an interval of $12$ minutes but Raju sitting in a train approaching the place hears the second report $11$ minutes $30$ seconds after the first. What is the approximate speed of the train (if sound travels at the speed of $330$ meters per second)?

I'm not able to make an equation for this question, I know $\text{distance} = \text{speed} \cdot \text{time}$ but how to frame this question into mathematical equation.

Answer 1

We have that

$$d=12 v_S \quad d=11.5(v_S+v_T)$$

therefore

$$12v_S=11.5(v_S+v_T) \iff v_T= \frac{0.5}{11.5}v_S\approx 14.3 \, (m/s)$$

Answer 2

Say, initial distance between train and the place is $d$ and the speed of the train is $s$. $t$ is time for sound to reach the person in the train the first time.

Speed of sound in meter / minute = $5.5$

So using relative speed, $d = (s + 5.5) t$

For the second gunshot, the sound and the train travel together in opposite direction for $0.5$ minute less than the first time. Also the train travels for $12$ minutes without the sound.

So again, $d = (s + 5.5) \times (t -0.5) + 12s$

Replace $d$ from first equation into 2nd, you get $23s = 5.5$

Or, $s = \frac{11}{46}$ meter per minute or $\approx 14.35$ meter/second,