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Damn locomotive

  1. Sep 9, 2011 #1
    A locomotive is accelerating at 1.7 m/s2. It passes through a 20.0-m-wide crossing in a time of 2.8 s. After the locomotive leaves the crossing, how much time is required until its speed reaches 33 m/s?

    I really don't know how to approach this equation, I feel like it will involve more than one equation. I'd really love a step by step explanation :)

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 9, 2011 #2

    Doc Al

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    Hint: How fast is it going after it leaves the crossing (assuming it started from rest)?
     
  4. Sep 9, 2011 #3
    Would it be going 4.76 m/s? v=at?
     
  5. Sep 9, 2011 #4

    Doc Al

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    Exactly! (But now that I think of it, you really didn't need this step.)

    Now use the same formula to figure out how long it takes to get to the final speed from the starting point.
     
  6. Sep 9, 2011 #5
    would i do v= vo + at and use vo as 4.76 and v as 33m/s and use the a that i'm given and solve for t?
     
  7. Sep 9, 2011 #6

    Doc Al

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    Sure. That works!
     
  8. Sep 9, 2011 #7

    Doc Al

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    Oops! I think I messed you up a bit.

    My assumption that the locomotive started through the crossing at zero speed was wrong. Of course, you're supposed to use the information given to figure out the initial speed. (Not just assume it, like I did. D'oh!)

    Do this: Given the distance, acceleration, and time, figure out how fast the locomotive was going when it entered the crossing. You'll need another kinematic formula for this. Then you can finish up just as you were doing.

    Sorry about that!
     
  9. Sep 9, 2011 #8
    it comes out to 16.6 seconds and apparently that's not the right answer :/
     
  10. Sep 9, 2011 #9

    Doc Al

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    See my post above. I misled you! :redface:
     
  11. Sep 9, 2011 #10
    Okay the only other equation I found that uses distance, acceleration and time is x= vot + .5at^2 and when i solve for vo it still comes out to 4.76. am i missing something?
     
  12. Sep 9, 2011 #11

    Doc Al

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    Yes, v0 is 4.76 m/s (not 0). Now, using that value, figure out the total time it took for the locomotive to go from 4.76 m/s to 33 m/s. Use that value to figure the time between when the locomotive left the crossing to when it reached its final speed--that's the answer they want.
     
  13. Sep 10, 2011 #12
    Alright I got it!! thanks for all your help :)
     
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