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Calculate the distance the aircraft has travelled ?

  1. Mar 24, 2012 #1
    Please HELP me solve this :

    An aircraft at rest experiences a uniform acceleration of 6 ms-2 for 4 seconds , It maintains a constant velocity for 12 seconds and is then brought to rest in 7 seconds by a uniform retardation .

    Calculate the distance the aircraft has travelled ?
     
  2. jcsd
  3. Mar 24, 2012 #2
    How have you approached the problem uptil now?
    We can help you only after you show us your work.
     
  4. Mar 24, 2012 #3
    I don't know much actually , I just wanted to get a quick detailed answer and then copy it , I will be thankful :"(
     
  5. Mar 24, 2012 #4

    DaveC426913

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    Well, at least you're honest about your intentions (if not about your education). :smile:

    But these are the rules you agreed to when you registered. You must follow them if you intend to stick around:
    Full text:
    https://www.physicsforums.com/showthread.php?t=414380
     
  6. Mar 24, 2012 #5
    I have love for all .
    I will try to solve , I know I wont but ... just sayin .
    ,
    I'm from saudi arabia ,
    I never studied physics seriously in my life !
    But now it's about getting a good job so I'm trying to learn
    " Colin " is my teacher name
    I thought if I entered by an english name , I may have better chances to get answers ,
    Anyhow ,
    I'm terribly clueless :"\
     
  7. Mar 24, 2012 #6
    I think the problem involves three stages:

    When the aircraft is accelerating, when the velocity is constant and then when it is braking. Using the Suvat equations, I got an answer, but the rules forbid me from disclosing. Waiting to see if I am correct.

    "Colin19", if you want to learn, a little bit of seriousness is essential :smile:. I am a learner too, try to use the Suvat equations and tell us the answer.
     
  8. Mar 24, 2012 #7
    I guess I'm in the wrong place then , I don't have a clue about " Suvat equations "
    I'm truly serious , all I wanted is to get the answer then to copy it ,
    I'm not looking to be a scientist or actually learn anything " even though it's great "
    What I'm looking for is to fill this exercise paper I have ,
    So if anyone can help please please please do so
    I'm trying to learn\find just to pass this period where we have to study physics .
     
  9. Mar 24, 2012 #8

    rcgldr

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    Homework Helper

    Suvat equations are a formal defnition of the equations of motion. You just need to separate this problmem into 3 steps. For the first step, I'll offer some basic information.

    If an object accelerates with a constant acceleration, then the change in velocity equals acceleration x time, and the average velocity is 1/2( (original velocity) + (final velocity) ).

    In this case, the acceleration is 6 m / s2, and the initial velocity is 0 m/s. What is the final velocity be after 4 seconds? What is the average velocity?

    Once you know the average velocity, then how far did the aircraft travel in 4 seconds?
     
  10. Mar 24, 2012 #9
    Thanks ,
    I will try this ,
    So it's 6 X 4 = 24 m/s2 , thats the final velocity after 4 seconds , Right ?
    va = ( v + v0 ) / 2 , So the avearge velocity is 12 m/s2
    So in 4 seconds the Aircraft travelled 24 m ?
    Is that correct ?
     
  11. Mar 24, 2012 #10

    rcgldr

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    Yes, but the unit is m/s (velocity) (you multiplied a m/s2 x t s) = v m/s), So 24 m/s final velocity, 12m/s average velocity.

    If average velocity is 12 m/s and time is 4 seconds, how far did the aircraft tavel?
     
  12. Mar 24, 2012 #11
    Sorry i'm a bit slow on understanding ,
    So the answer would be 12X4 = 48 meters
    The aircraft travelled 48 meters in 4 seconds
    I greatly appreciate this help ,
     
  13. Mar 24, 2012 #12

    DaveC426913

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    OK, now part II. The craft maintains a constant velocity.
    At the start of part II, it's travelling how fast? (based on the end part I), and how long is it travelling at this speed, resulting in what distant travelled?

    BTW, this is an international board. We are happy to answer your questions, regardless of your name or nationality. You should feel free to be who you are.
     
  14. Mar 24, 2012 #13
    Thanks Mr.Dave , should I re join by my real name " Yazeed " Or is it possible to change it from admins , I'm sorry I made such assumption ,
    And I'm sorry for the lazy attitude I had .

    Distance = velocity X time
    Craft is travelling on the constant speed of 24 m/s " Final velocity "
    Time is 12 seconds , So 12X24 = 288 m !
    So adding to the first step , 288+48=336 m
     
  15. Mar 24, 2012 #14

    rcgldr

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    For the third part, you are told it takes 7 seconds to slow down to a stop (0 m/s), and you know the initial velocity is 24 m/s. What is the rate of deceleration (retardation)?

    A lot of members on this board use nicknames, so Colin19 is fine. You can ask an admin to change this to your real name if you still want to change it.
     
    Last edited: Mar 24, 2012
  16. Mar 24, 2012 #15
    Ok ,
    acceleration = Final velocity/time
    24/7 = 3.42 m\s2 ?
     
    Last edited: Mar 24, 2012
  17. Mar 24, 2012 #16
    So How can I calculate Distance ?
     
  18. Mar 24, 2012 #17

    rcgldr

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    The same way you did for the intial acceleration, average velocity times the 7 seconds it took to stop. Actually you didn't need to calculate the deceleration in this case, since you only need to determine average velocity, which you've already done before (it's the same for deceleration as acceleration).
     
  19. Mar 25, 2012 #18
    Oh Ok , I thoght it would be the same but I hesitated and assumed stuff ,
    So with the same concept applied
    The avearage velocity X Time = Distanse
    12 X 7 = 84 meters .
    Adding all distance findings
    48+288+84= 420 m
    So I can say that the distance the Aircraft travelled is 420 m
     
  20. Mar 25, 2012 #19
    Yes, you got it Yazeed :-)
     
  21. Mar 25, 2012 #20
    You won't believe how happy I'm , I thank you all greatly ,
    This is truly beneficial ,
    I came here looking for a shortcut , but I actually end up learning
    I'm grateful to all who helped me !
    I'm excited to go for more than a pass now , I got couple more questions in this practice papers to crack which I'll be posting later on .
     
    Last edited: Mar 25, 2012
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