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One Dimensional Kinematics Jet Problem

  1. Dec 14, 2007 #1
    1. The problem statement, all variables and given/known data

    A Boeing 747 "Jumbo Jet" has a length of 59.7 m. The runway on which the plane lands intersects another runway. The width of the intersection is 29.9 m. The plane decelerates through the intersection at a rate of 6.33 m/s2 and clears it with a final speed of 42.3 m/s. How much time is needed for the plane to clear the intersection?



    2. Relevant equations

    V^2 = V0^2 + 2a (X - X0)

    X = 1/2 (V0 + V) t



    3. The attempt at a solution

    Knowns:
    X0 = 0m
    X = 89.6m
    V0 = Unknown
    V = 42.3 m/s
    a = -6.33 m/s^2
    t = Unknown

    I began by trying to solve for the initial velocity of the jet using the first of the above equations. I am getting 95.7 m/s but I am not sure if I am simply doing the algebra incorrectly or something else. It would help to get step by step help on the algebra as well and not skip any steps.

    I then used 95.7 m/s in the second equation to solve for time. For this I am getting
    1.3s but the homework webpage says that isn't correct.

    I have tried repeatedly to work this through and I just can't seem to get it.

    Thank you
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 14, 2007 #2

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    Do a trick here. Time reverse the situation. Imagine the jet is moving backward with the same acceleration (not deceleration) of +6.33 m/s^2. It's initial velo is what is given as the final velo of +42.3 m/s. Then you'll get the answer in one shot.

    X = Vi*t + (1/2)at^2.

    You can do this because the time for going from Vi to Vf with deceleration a is the same for going from Vf to Vi with accn a.
     
  4. Dec 14, 2007 #3

    You're doing the algebra incorrectly. Can you show your steps, and we'll see where you're going wrong?
     
    Last edited: Dec 14, 2007
  5. Dec 14, 2007 #4
    I tried plugging in the numbers to that formula but I can't seem to get the right answer.

    I did it as follows.

    89.6m = 42.3 m/s*t + (1/2) 6.33 m/s^2*t^2

    First off, was that what you meant?

    I tried to solve it using the quadratic forumula but it doesn't seem to be working out.
    0= -89.6m + 42.3m/s*t +(1/2) 6.33m/s^2*t^2

    t= ((-42.3 + √(42.3^2-4(-89.6)(3.17))) / (2(-89.6)))

    What might I be doing wrong?
     
  6. Dec 14, 2007 #5

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    I meant exactly that.

    I got 5.89 s. Find the positive root of the quadratic eqn in 't'. A matter of one minute!
     
  7. Dec 14, 2007 #6
    Hmmm...

    I tried checking that answer on the WileyPlus webpage we use and it says 5.89s is not a correct answer.
     
  8. Dec 14, 2007 #7

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    Maybe I did some arithmetical error. You don't know how to solve quad eqn? Why do you have to go to some web page?

    I did it again. I got 0.93 s.
     
  9. Dec 14, 2007 #8

    I got those answers too. The webpage is how my teacher assigns us homework which is why I am using that. It doesnt like 0.93 s either. I guess I give up. Thanks for your help though.
     
  10. Dec 14, 2007 #9

    rl.bhat

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    V^2 = V0^2 + 2a (X - X0)
    42.3^2 = Vo^2 - 2*6.33*89.6
    Vo = 54.1m/s
    V = Vo + at or 42.3 = 54.1 - 6.33t
    t = (54.1-42.3)/6.33 = 1.86s
     
    Last edited: Dec 14, 2007
  11. Dec 14, 2007 #10
    The equation you used first were the easy way to start. if you follow your steps again you should find that the initial velocity is 54.07m/s.


    for the quadratic equation you need to used the initial speed not the final.

    X-Xo = Vo*t + (1/2)*a*t^2.

    also you are applying the quadratic formula incorrectly. you are divining by (2*constant) were you should be doing (2*a), where a is the coefficient of the squared term of the variable which is t^2 in our case
     
  12. Dec 14, 2007 #11

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    All right, I divided by an extra 2 -- the 2a in the denominator of the qudratic roots. So, 2*0.93 = 1.86.
     
  13. Dec 14, 2007 #12

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    Width + length of plane.
     
  14. Dec 14, 2007 #13

    robphy

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    I saw it after posting... but didn't delete it quick enough. :rolleyes:
     
  15. Dec 14, 2007 #14

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    See if the web page likes the correct answer now. And never give up.
     
  16. Dec 14, 2007 #15
    Thanks for the help. That answer is correct.
     
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