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Find an expression of acceleration

  1. Sep 20, 2011 #1
    1. The problem statement, all variables and given/known data
    A particle has mass m = 0.1 kg , has a velocity which is given as: v = 3.2i+4.3j+5.4k m/s
    It's acted by force f = 7.1i+8.2j+9.3k

    The questions which arise are:
    i) Find the speed of the particle.
    ii) Write down a unit vector in the direction of the velocity.
    iii)Find the angle between the projection of the velocity in the x,y plane and the x - axis.
    iv) Find an expression of acceleration.
    v) Find the velocity at a time of 2s.
    2. Relevant equations
    Now the equations that come to my mind are:
    F = ma and f = change of momentum / time


    3. The attempt at a solution

    To get the speed: wouldn't that be the same as finding out an r.m.s value as such particle speed becomes: ~7.6m/s ?
    For part (ii)is this how we find out a unit vector ? (ii)[itex]\vec u / ||\vec u||[/itex]
    Now I'm a little lost about part (iii). You see I know how to find an angle between two vectors but here, I suppose since it's speaking of x,y axis within the same vector so I will use the following equation: A .B = |A| x |B| cos theta Can someone please clarify this for me?

    Part iv doesn't seem that tricky... expression for acceleration has to be: a = f/m ...
    v) Finding out velocity could be find out by using f = momentum /time...


    Some background information: I have just started my course in Astrophysics, and this question comes out of a tutorial sheet where they don't expect us to know much and just see how much knowledge we have retained.(Proper lessons haven't even started, however I want to finish this early)


    P.S: I'd like to apologize in advance of not using proper notation or over the lack of latex usage.


    -ibysaiyan
     
  2. jcsd
  3. Sep 20, 2011 #2
    Re: Vectors

    Anyone ?
     
  4. Sep 20, 2011 #3

    HallsofIvy

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    Re: Vectors

    Why in the world would you want to trick your teacher into thinking you know things you don't? You say that "this question comes out of a tutorial sheet where they don't expect us to know much and just see how much knowledge we have retained." If you cannot do this problem yourself, then your teacher needs to know that so he/she can teach it!
     
  5. Sep 20, 2011 #4

    Hootenanny

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    Re: Vectors

    Patience is a virtue. Your question has only been up for a few hours.
    You have a vector in [itex]\mathbb{R}^3[/itex] and you want to find the angle between its projection onto the xy-plane and the x-axis. In other words, you first need to project the vector onto the xy-plane, giving you a vector in [itex]\mathbb{R}^2[/itex] and then find the angle between this projection and the x-axis. Does that make sense?
    Since the particle is being acted on by a constant force, I would use SUVAT equations.
     
  6. Sep 20, 2011 #5
    Re: Vectors

    Thanks for your reply.I understand where you'e coming from but I am trying to make myself useful here by not wasting around time, I have just found an article describing vectors in good detail.Of what I have found so far... most of my answers on the above post are wrong.
    In this case unit vector will be (3.2 4.3 5.4) for part (ii)
    So do I need to find out the magnitude of Ox and Oy and then use the above formula to the respective angle ? Am I right ? [for part (iii) ]

    But how can I use a formula which is for an angle between two vectors on this....... I am confused can someone shed some light on this.. i am not asking for direct answers, a tip maybe ? Thanks
     
  7. Sep 20, 2011 #6

    Hootenanny

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    Re: Vectors

    What is the projection of the velocity vector onto the xy-plane?
     
  8. Sep 20, 2011 #7
    Re: Vectors

    I am not sure rather I have no idea. hm... I have the question posted on the OP.
    My current working for all the bits is as following

    (i)
    speed of particle = square root of [(3.2)^2 + (4.3)^2 + (5.4)^2 ] which gives me an approximate speed of 7.6 m/s

    (ii) For unit vector: 3.2 i + 4.3 j + 5.4 k / [7.60 ]

    (iii)
    theta = tan^-1 ( 4.3/ 3.2) = 53 degree since only axis x and y are mentioned for this part.

    (iv) a = f/m

    (v) find the velocity at a time of 2s.
    Using f = rate of change of momentum / time , we get velocity : force x time / mass
    to find force we need to get the modulus of vector force: which is : square root of [ (7.1)^2 + (8.2)^2 + (9.3)^2 ] = ~14.3 N

    Velocity at t = 2 ,
    14.3 x 2 / 0.1 = 286 m/s.

    Please criticize if any of my answer needs to be.
     
  9. Sep 20, 2011 #8

    Hootenanny

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    Re: Vectors

    All good.
    Not sure what you're doing here - velocity is a vector, not a scalar.

    I suspect that there is some information missing from the question. Could you copy it verbatim please?
     
  10. Sep 20, 2011 #9
    Re: Vectors

    Sure.

    A particle of mass m = 0.1 kg has a velocity vector, v, given by: 3.2 i + 4.3 j + 5.4 k m/s
    it's acted by a force
    f = 7.1 i + 8.2 j + 9.3 k N

    i) Find the speed of the particle
    ii) Write down a unit vector in the direction of velocity
    iii) Find the angle between the projection of the velocity in the x,y plan and the x-axis.
    iv) Find an expression of acceleration.
    v) Find the velocity at a time 2 seconds later.


    EDIT: Sorry HN I just noticed post no. 4 where you have mentioned suvat equation...
    Here's what I have got for the last two parts;
    iv) v = u+at => v-u/ t = a.
    v) a = f/ m , 14.3 / 0.1 = 143 m/s^2
    to find velocity we use the equation stated in part (iv) :
    v = u+at = 7.6 + (2 x 143)= ~ 294 m/s.

    How about now ?
     
    Last edited: Sep 20, 2011
  11. Sep 20, 2011 #10

    Hootenanny

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    Re: Vectors

    Excellent. So, for part (v), since the force is constant and you have the acceleration and initial velocity, I would use SUVAT equations.
     
  12. Sep 20, 2011 #11
    Re: Vectors

    I have posted the solution on the same post. Can you have a look please.Thank you very much for your contribution!
     
  13. Sep 20, 2011 #12

    Hootenanny

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    Re: Vectors

    Again, you have to be careful here. Velocity is a vector quantity, not scalar. You need to apply that SUVAT equation to each component individually. That will then give you the final velocity vector.
     
  14. Sep 20, 2011 #13
    Re: Vectors

    Hm.. I think I am getting this now...
    So I should use suvat equations to find out the initial velocity of the particle and in the same manner for acceleration , which will subsequently get me the final velocity at the time interval of 2 seconds?
     
  15. Sep 21, 2011 #14

    Hootenanny

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    Re: Vectors

    No, no. You have been given the initial velocity (v in the question) and you have already worked out the acceleration (a=f/m - remember this is a vector too). Now all you need to do is apply that SUVAT equation to x,y and z components individually to find the final velocity. Do you follow?
     
  16. Sep 21, 2011 #15
    Re: Vectors

    Ah I think I understand the flaw in my reasoning... was it wrong of me to take speed of 7.6m/s into account for my final velocity, since it's a scalar (speed) quantity,right ? but when you speak of components xyz , I presume them to be ijk vector. If this is right then how do I get initial 'v' value for each. Will finding out the magnitude of individual components give me those numbers ?
     
  17. Sep 21, 2011 #16

    Hootenanny

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    Re: Vectors

    I think its best if I show you an example. So, we have the initial velocity and acceleration vectors [itex]\boldsymbol{u} = [3.2, 4.3, 5.4]^\text{T}[/itex] and [itex]\boldsymbol{a} = [71, 82, 93]^\text{T}[/itex]. Now, lets look at the first (x-component), we have: [itex]u_x = 3.2[/itex] and [itex]a_x = 71[/itex]. So, now applying the SUVAT equation

    [tex]v = ut + \frac{1}{2}at^2[/tex]

    to just the x-component, we find

    [tex]v_x = 3.2\times2 + \frac{1}{2}\times 71 \times 2^2 = 148.4 \text{m/s}[/tex]

    So, the first component of the final velocity is 148.4 m/s. Now, you need to do the same for the y and z components.

    Do you follow?
     
  18. Sep 21, 2011 #17
    Re: Vectors

    Yes, I do now. Thank you very much!
    Expect more postage from me. Again I can't thank you enough...
    -ibysaiyan
     
  19. Sep 22, 2011 #18

    Hootenanny

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    Re: Vectors

    It was a pleasure. Glad to help out.
     
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