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Stuck Again Grade 11 physics

  1. Oct 25, 2009 #1
    1. The problem statement, all variables and given/known data
    A train is 120m long. Gaurav was standing 200m from the front of the train when it began to accelerate uniformly from rest. He noticed that the front of the train was moving at 8.00m/s when it passed him. How fast would the back of the train be going when it passed Gaurav?


    2. Relevant equations



    3. The attempt at a solution
    Distance from front=200m
    v1=0.00m/s
    v2=8.00m/s
    Step 1
    a=?
    V2^2= v1^2 + 2ad
    a= (((V2^2-v1^2 ))/2)/d
    a=(((8^2-0^2 ))/2)/200
    a=(64/2)/200
    a=32/200
    a=0.16m/s/s
    Step 2
    t=?
    t=v2/a
    t=8/0.16
    t=50s


    after that last part im completly lost...
     
  2. jcsd
  3. Oct 25, 2009 #2
    forgot to include equations :
    http://img196.imageshack.us/img196/7408/captureie.png [Broken]
     
    Last edited by a moderator: May 4, 2017
  4. Oct 25, 2009 #3

    Delphi51

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    Homework Helper

    Good job on that acceleration!
    I suggest you just use the same 2ad equation again, applying it to the whole 200+120 meters. Since you know the acceleration now, you can use the formula to find the final velocity.

    It would be interesting to apply the formula to the 120 m distance only and see if you get the same answer.
     
  5. Oct 25, 2009 #4
    LOL sorry about this but I'm kinda confused on what your asking me to do. What I don't get is to get the final speed I would need to know the amount of time it took to for the train to get 320m or I would need to know the speed of the train at 320m...Thanks in advance
     
  6. Oct 25, 2009 #5

    Delphi51

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    The speed of the train is v2 in
    v2² = v1² +2*a*d and you know all the variables on the right side.
     
  7. Oct 25, 2009 #6
    Ohh...Thanks...I'm gonna try it again and post the fixed solution
     
  8. Oct 25, 2009 #7
    ok i did it but I think something might be wrong...I don't believe that the speed has only increased by that much after 120m
    Distance from front=200m
    v1=0.00m/s
    v2=8.00m/s
    Step 1
    a=?
    v2^2= v1^2 + 2ad
    a= (((V2^2-v1^2 ))/2)/d
    a=(((8^2-0^2 ))/2)/200
    a=(64/2)/200
    a=32/200
    a=0.16m/s/s
    Step 2
    v2^2= v1^2 + 2ad
    v2^2= 0^2 + 2*0.16*320
    v2^2= 102.4
    v2=√102.4
    v2=10.11m/s/s
     
  9. Oct 25, 2009 #8

    Delphi51

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    Your physical intuition may well fail here. You may think the Δv would be about 1.5 times as much in 320 m as it was in 200 m, but when you think this way you are using v = a*d. In fact it is v = a*t and the time for the extra 120 m is much less than half the time for the 200 m because the train is moving faster.
     
  10. Oct 25, 2009 #9
    Oh, so does that mean the answer was right? Sorry for the stupid questions...I'm doing these half asleep
     
  11. Oct 25, 2009 #10

    Delphi51

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    Homework Helper

    yes, looks good!
     
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