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Basic Physics ?

  1. Sep 9, 2005 #1
    Basic Physics...?!?!

    Hey Everyone, I just started the course and the first problem I come to, so very confused me!! I don't need an answer just a basic response on exactly HOW I can approach this problem because I am just totally blown away by it :surprised .

    Two trains, each having a speed of 30 km/hr are headed at each other on the same straight track. A bird that can fly 60 km/h flies off the front of one train when they are 60 km apart and heads directly for the other train. On reaching the other train, the bird flies directly back to the first train and so forth. What is the total distance the bird travels before the trains collide?

    Please help the lost and confused!! :cry:
     
  2. jcsd
  3. Sep 9, 2005 #2

    quasar987

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    You could do it using the equations of position at constant acceleration (here a=0). Draw a diagram of the position of every object (train 1, train 2 and Bird) at the initial time. Then use the equations to find the time at which the bird meets a train for the first time. Calculate the position of each object at that time. Draw the diagram again, then find the time when the bird meets the other train for the first time, etc, etc, etc.
     
  4. Sep 9, 2005 #3
    Still don't get it because I mean both the train and the bird are moving so how would I make 2 separate equations correlate? I'm assuming since you said a=0 and that I need to find the time that I am using v^2=vi^2+2a(x-xi) equation. Yet, I don't have the distance unless it is the 60km but that changes constantly until the bird meets the train.

    Sorry we have only done straight forward equations and now they're pulling ones that we haven't done or have no examples in the book for.
     
  5. Sep 9, 2005 #4

    quasar987

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    The equation of position at constant acceleration is

    [tex]x(t) = x_0+v_0t+\frac{1}{2}at^2[/tex]

    Write the 3 equations. One for train 1, one for train 2 and one for the bird. Personally, I chose my coordinate system such that at t=0, train 1 is at x_0=0 and voyaging at v_0 = +30 km/h. On the other hand, train 2 is at x_0 = 60 km and is voyaging at v_0 = -30 km/h. As for the bird, it is also located at x_0 = 60 km but it is voyaging at -60 km/h. Given these informations, can you write down the 3 equations of position

    [tex]x_1(t) = ...[/tex]

    [tex]x_2(t) = ...[/tex]

    [tex]x_B(t) = ...[/tex]

    ???
     
  6. Sep 9, 2005 #5

    David

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    There are two ways to solve this problem. One is to actually work out the full path of the bird between the trains. You end up calculating the limit of an infinite sum. It's do-able but messy.

    The other is to think creatively as to how to solve the problem using the information given in a different way than just working out the path of the bird. And then you'll realise that it is actually pretty simple. Here is a hint: the bird is always flying at the same speed (even though it changes directions back and forth).
     
  7. Sep 9, 2005 #6

    quasar987

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    David's a genius, listen to him, not me :tongue: .
     
  8. Sep 9, 2005 #7
    with t=0, the entire equation would be 0? I tried that but the time is what throws me over because I don't know what time they will meet (both the train and bird).
     
  9. Sep 9, 2005 #8
    I solved up to when the trains are 5 miles apart...I drew the problem out and used ratios but I know there has to be a more direct way. LoL, I'm sorry I'm a complete blank slate when it comes to this (calc teacher sucked and my last math before that was 3 years ago).
     
  10. Sep 9, 2005 #9

    David

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    Here is another hint: When will the two trains meet each other?
     
  11. Sep 9, 2005 #10
    at 30 km, dead center.
     
  12. Sep 9, 2005 #11

    quasar987

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    He meant in terms of time. Once you've found that, ask yourself, how far has the bird flew during that interval? :wink:
     
  13. Sep 9, 2005 #12
    it took the bird 30 minutes to get to the collision point BUT the bird flies to the second train going -30km/hr and meets the second train at the 40km mark in 40 minutes...that's what I got and then flies back and meets the first train at the 27.5km mark in 15 minutes.

    and an hour for the trains to meet....

    ...so I have 5 minutes left before the trains crash and they're 5km apart so that would mean that the bird would fly the last 5km to the other train and get there at the same time as the collision thus making the answer....

    55km?
     
    Last edited: Sep 9, 2005
  14. Sep 9, 2005 #13

    quasar987

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    The problem is actually much simpler than that. Re-read the teachings of Master Dave:

    Suffice to understand that it doesn't matter that the bird goes from train to train. The crucial point is that it flies at constant speed. Hence the total distance it travelled from the initial moment to the moment when the two trains collide is simply the distance it would have traveled in going in a straight line during that time. So find that time*, multiply by the speed of the bird, and there you go.
     
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