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Train and Speed of Sound in Air versus Track

  • Thread starter Schoomy
  • Start date
  • #1
42
0

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...
 

Answers and Replies

  • #2
161
0
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.
 
  • #3
42
0
I end up with this:

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

I'm not quite sure how to solve...
 
  • #4
161
0
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! :)
 
  • #5
42
0
Now I end up with something like this:

Ts + VoTs = 3.1Vo / Vs

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

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