# Train and Speed of Sound in Air versus Track

• Schoomy

## Homework Statement

Stan and Ollie are standing next to a train track. Stan puts his ear to the steel track to hear the train coming. He hears the sound of the train whistle through the track 3.1 s before Ollie hears it through the air. How far away is the train? (Use 5,790 m/s as the speed of sound in steel and 343 m/s as the speed of sound in air.)

## Homework Equations

distance = velocity*time
v-steel= 5,790m/s
v-air= 343m/s

## The Attempt at a Solution

distance-stan = 3.1seconds*5790m/s = 17,949m
distance-olie = 3.1seconds*343 = 1063.5m

Not sure what to do from here. I don't quite understand the concept relationship they're trying to establish...

I tried subtracting Olie's distance from Stan's, but that didn't get me the right answer...

The assumption you need to make is that Stan and Ollie are standing next to each other. ie the distance is the same.

Then you can set up the equation: v(stan) x t(stan) = v(ollie) x t(ollie)

Remember that the question states that Ollie hears the sound 3.1 seconds after Stan, so t(ollie) = t(stan) + 3.1

Substitute this in, expand brackets, solve for t(stan).

Then use d = vt for this time and speed of sound in steel.

Hope this helps.

I end up with this:

T(stan) = (V(olie) x (T(stan)+3.1s)) / V(stan)

I'm not quite sure how to solve...

Leave t(stan)*v(stan) = v(ollie)*(t(stan) + 3.1)

Now expand the brackets.

Then get the t(stan) factors on the same side of the equals sign.

Then factorise

Keep trying! :)

Now I end up with something like this:

Ts + VoTs = 3.1Vo / Vs

I don't know how to get the Ts's alone...