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Vectors problem.

  1. Mar 8, 2009 #1
    1. The problem statement, all variables and given/known data

    If a car has an initial velocity of 8.0m/s, 25degrees and turns in 4.0 s so it ends up traveling at 11.0m/s, 140degrees find:
    a)the acceleration
    b)the displacement for the 4.0 s

    2. Relevant equations
    Im not sure on what equations are relevant for vector problems, I understand everything in the unit besides this because I wasent there for it so if you really need equations I can give you all of he equations for the unit and the right one will be somewhere in it.

    just say it if you need them.
     
  2. jcsd
  3. Mar 8, 2009 #2

    LowlyPion

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    Homework Helper

    Acceleration is change in Velocity over time. You can resolve the two velocities into their x,y components and then determine their change with respect to time.

    If the presumption is that the acceleration is uniform over the time of 4s, then you should be able to use a kinematic equation to resolve the displacement.
     
  4. Mar 8, 2009 #3
    ok I will try that and put it down on here, and then you can correct me if I make a mistake

    what do you mean by acceleration is uniform over the time?
     
  5. Mar 8, 2009 #4
    I think its x=-15.676 and y=3.69 but what do I do once I get these?
     
  6. Mar 9, 2009 #5
    use these results to find x,y components of the acceleration.
     
  7. Mar 9, 2009 #6
    wait how do I find the x y components of the acceleration do I find out the x and y for 8.0m/s, 25degrees and the x and y for 11.0m/s, 140degrees
     
  8. Mar 9, 2009 #7

    Astronuc

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    These are the differences in speeds (magnitude of velocity) in the two principal directions.

    So [tex]\vec{v_2} - \vec{v_1} = \Delta{v}[/tex], and as LowlyPion suggested one may assume uniform acceleration, so look at Δv/Δt.

    The one has initial velocities or speeds vx and vy, and one must solve for the two components of displacement in the x and y directions, and add them vectorily.

    Take the reference (0,0) at time 0, when the acceleration is introduced.

    See - http://hyperphysics.phy-astr.gsu.edu/hbase/acca.html#c1

    http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html

    http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html
     
  9. Mar 9, 2009 #8
    ax = (x component of final v - x component of initial v)/t
    ay = ...

    a=...
     
  10. Mar 9, 2009 #9
    this is what we were doing in class for these types of question though:
    we would set up this first
    Velocity 1-8.0m/s, 25degrees
    Velocity 2-11m/s,140degrees
    displacement-
    acceleration-
    time-4.0s


    and the I think the right equation would be V2=V1+a*t or

    11m/s,140degrees=8.0m/s, 25degrees+a*4.0s

    and then thats where I would be lost and I dont remember learning anything like what you guys are saying so Im a little lost.
     
  11. Mar 9, 2009 #10
    ok so do I just change the v1 and v2 to components? and then what would do I do put it overtop of time.
     
  12. Mar 9, 2009 #11

    Astronuc

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    If one refers to http://hyperphysics.phy-astr.gsu.edu/hbase/acca.html#c1

    then one will find the definition of acceleration in terms of the change (difference) in velocity over time.

    [tex]\vec{a} = \frac{\Delta{v}}{\Delta{t}} = \frac{\vec{v_2} - \vec{v_1}}{{t_2} - {v_1}}[/tex], which assumes that a is constant or uniform.

    but since velocity is a vector, one has two independent components in Cartesian coordinates, so one must consider vx2, vx1, vy2 and vy1 in the above equation for acceleration.

    Please refer to http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html#vec2 and
    http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html#vec3 on combining vector components.

    With respect to displacement, what is the equation for displacement x from a starting point with an initial velocity vx0 and a uniform acceleration ax in terms of time?
     
  13. Mar 9, 2009 #12

    LowlyPion

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    Homework Helper

    I always like to simply resolve the vectors into i,j , so for your situation you can write:

    Vi = 8*cos25 i + ... j
    Vf = ... i + ... j

    Then you can add the i's and j's (in this case with vector difference subtract the i from the j). That gives you the ΔV. Since the time is a scalar then just divide each i and j by 4s.

    Armed with your acceleration then I would apply (since they don't say otherwise) the x = 1/2*a*t2 kinematic relationship between displacement, acceleration and time. (Again t2 being a scalar.)
     
  14. Mar 9, 2009 #13
    ok I went to write it but hes letting us come back at lunch to finish now I got the acceleration right on the test because the teacher helped me and Im pretty sure I got the displacement question on the same question right to but for c)it said find the velocity for the time given so if it was this question that were doing it would be find the velocity at 4.0s? Im unsure of how to do this.


    Im more worried about this question below because there will be more marks involved


    There was also one more question that I couldnt get it this one would be just kinematics and it was just like this question in my review so I will give you it

    If a half back accelerates from rest with constant acceleration and travels 13.75m in the sixth second find the acceleration.

    and heres the answer for it to help you its 2.5m/s
     
    Last edited: Mar 9, 2009
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