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Homework Help: 3D Motion

  1. May 17, 2010 #1
    1. The problem statement, all variables and given/known data

    A boat heading south initially with a speed of 10 m/s has a constant acceleration of a= (10,5,10) m/s^2

    What is the velocity, speed, position and distance travelled after 30 minutes?

    t=30 min = 1800

    2. Relevant equations

    v = u + at

    3. The attempt at a solution

    a^2= 10^2 + 5^2 + 10^2

    v= u + at
    = 10 + 15x1800
    =27010 m/s south

    b) speed is just the scalar quantity so 27010 m/s (NOT SOUTH)
  2. jcsd
  3. May 17, 2010 #2


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    I am not sure what kind of "boat" this is that has non-zero acceleration in all three directions. Assuming that it can fly, then you need to write the kinematic equations in the three directions independently, find each component after thirty minutes separately, then combine as needed using the Pythagorean theorem.
  4. May 17, 2010 #3
    Hahaha, sorry I ment a hot air balloon. Can you expand a bit more please, I still don't quite understand, sorry. I really need help because I'm in grade 12 and I would like to get into university, so I need a good mark in physics.
  5. May 18, 2010 #4


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    Say "South" is x, "East" is y and "Up" is z. What expression will give you the x-component of velocity at any time t? What about the x-component of position at any time t?

    If you can figure these out, then you should be able to apply the same reasoning to the other two directions. So start from there.
  6. May 18, 2010 #5
    Also is this an acutal question set by someone? As a constant acceleation of 10m/s for 30 mins puts you beyond the escape velocity of earth.

    You've got the right idea with the kinematic equations, but think about how you can describe the movement in 3 directions.

    It's good practise to write out everything you know for each step, not just what you are given in the question. So for example each direction you have a:

    Starting velocity
    Staring Position.

    You may have to define your own things, that aren't strictly necessary to solve the question but will help you to think through it step by step. So for example, it may be a good idea to define the start at coordinates (0,0,0).
  7. May 18, 2010 #6
    So Vx=Ux + axt
    Vy=Uy + ayt
    Vz=Uz + azt

    Where V=final velocity, U=initial velocity, a=acceleration and t=time

    That seems about right, no? and then you use the other 4 kinematic equations in the same manner?

    And chris, I don't know what that is lol. I think they just wanted you to apply the kinematic equations then anything else.
  8. May 18, 2010 #7
    Using those equations (and the displacemnt ones), do you know how to calcualte the final position?

    I suspect that you have made the mental 'short cut' past what i've said, as it's fairly intutative, but I just find it easier to put every step down in full until I know im sure what the steps are. It's just really hard to describe without giving the answer away.
  9. May 18, 2010 #8
    No, lol. I'm sorry, physics isn't my thing. I'm more of a chem guy.
  10. May 18, 2010 #9
    Ok well can work through it from the very start.

    Remember the path travelled can be done by vectors, so we can split the problem up into each axis.

    For example if I walk 10 paces north and 10 paces west I end up in a certain place north west of my starting position. I can go north first or west first it doesnt matter.

    So the first thing to do is define the coordiantes where you start from.
    You then work each direction out sepeately, as it doesnt matter which of the directions you travel first.
  11. May 18, 2010 #10
    Yeah I know that, that's very basic. It's the acceleration vector that's screwing me up because I don't know how to apply it to find velocity, speed e.t.c
  12. May 18, 2010 #11
    Do you know how to calcualte the velocity, and distance in 1 dimension using the kinematic equations?
  13. May 18, 2010 #12
    If your given initial velocity, acceleration and time u can use v=u + at and then using that answer, you use s=(v+u/2)t. Correct?
  14. May 18, 2010 #13
  15. May 18, 2010 #14
    so is this right?
  16. May 18, 2010 #15
    Well yes, but you are skipping steps which is the reason you can't find the distance etc. Also the numbers are simply silly, that could be another reason it's not making sense to you.

    You are trying to do it all as 1 vector, when you need to find the three components of the final travelled distance. That's why I talked about going through all the steps.

    doing what you are doing gives you the right answers, but if you dont know why it's the right answer then you need to go through each step.
  17. May 18, 2010 #16
    Can you use this reasoning?


    where r=initial position, u=speed, a=acceleration

    Am I right here?
  18. May 18, 2010 #17
    Yeah, you've got the answer so we can talk about it now.

    I assume you added the squares af the accelerations to find the resultant acceleration because of pythagoras? What you did there, what what I was going to do at the end. You found the resultant acceleration, I was going to find the resultsnats at the end.

    If you did then, you got the right answer for the right reason.
  19. May 18, 2010 #18
    So 207010 is the right answer? and then to find distance you use s=(v+u/2)t which is s=(207010+10/2)1800 is your answer? But then I'm stuck on how to find position now.

    Also, I have another question.

    In my notes I suspect these equations having importance:

    x=Xo + uxt + axt^2/2
    y=Yo + uyt + ayt^2/2
    z=Zo + uzt + azt^2/2

    Does this relate to the equation s=ut + at^2/2? If so, what is the significance of Xo and x?
  20. May 18, 2010 #19
    Which is why you should have worked out the distance travelled in each component. you would have have 3 distances giving you the 3 coordiantes needed to define position.

    Yes they are component form of the kinematic equation.

    The X and X0 are a standard notation.

    They subscript 0 means (position/velocity/acceleration) at time 0. The starting conditions in each axis x,y,z.

    So where as you can use V and U for inital and final velocity you can use V0 and V.
  21. May 18, 2010 #20
    So thats whats screwing me up. How would you determine the distance travelled in each component?

    And for some reason they substituted s for initial and final velocity in the equation s=ut + at^2/2?
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