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Vector Question on motorist

  1. Oct 4, 2006 #1
    I am just aweful at this vector stuff, i can NEVER get it right.

    A motorist drives south at 20.0 m/s for 3.00 min, then turns west and travels at 25.0 m/s for 2.20 min, and finally travels northwest at 30.0 m/s for 1.00 min. For this 6.20 min trip, find the following values.

    (a) total vector displacement

    (b) average speed i got this part already, easy

    (c) average velocity (include the magnitude)

    total displacement i got 6291 meters (rounded). i did this by figuring out each x and y coordinate for the vectors, then adding them up, and finding the hypotenuse of the resulting vector. apparentlys its not right. As for the velocity part, i was unsure of how to go about starting it...
  2. jcsd
  3. Oct 4, 2006 #2
    also for this question:

    A particle initially located at the origin has an acceleration of = 3.00 m/s2 and an initial velocity of 0 = 4.00 m/s.

    (a) Find the vector position at any time t (where t is measured in seconds).

    i put that the vector was (3.00t)i + (4.00t^2)j, apparently the j part is wrong and im not sure why...
  4. Oct 4, 2006 #3


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    In general, the vector of position for a particle moving in the xy plane is given by [tex]\vec{r}(t)=x(t)\vec{i}+y(t)\vec{j}[/tex], where [tex]x(t)=x_{0}+v_{x0}t+\frac{1}{2}a_{x}t^2[/tex] and [tex]y(t)=y_{0}+v_{y0}t+\frac{1}{2}a_{y}t^2[/tex]. Now, you gave values of the acceleration and initial velocity, but you did not give any directions, which are needed to find the position vector.

    Regarding the first post, question (a), I think the total displacement means the displacement with respect to the origin (i.e. the point from which the travel started), so it equals the resultant vector.
    Last edited: Oct 4, 2006
  5. Oct 4, 2006 #4
    i copied/pasted the question exactly from my webassign, i guess it may be an error in the book? theres been a few errors before, but they really dont give you any directions associated w/ that problem.
  6. Oct 4, 2006 #5


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    Assuming that there is no error, you should be able to conclude from the context what the directions are. I would assume it is motion among a straight line, i.e. the direction is understood to have the direction of [tex]\vec{i}[/tex].
  7. Oct 4, 2006 #6
    if that's the case, what am i doing wrong with the vector?
  8. Oct 4, 2006 #7


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    If that's the case, you only have [tex]\vec{r}(t)=x(t)\vec{i}=(x_{0}+v_{0}t+\frac{1}{2}at^2)\vec{i}[/tex]. Plug in the initial speed, the acceleration, and the initial position (equals zero if the particle starts from the origin).
  9. Oct 4, 2006 #8
    theres definetely an error in the software or something (i do my homework via webAssign)...it said that, for the i component of the vector, just 3.00 was right.

    ill print that one out and ask my teacher tommorow, in the meantime, any suggestions on the first question i posted?
  10. Oct 4, 2006 #9
    If you can figure out the lengths travelled, you know the directions and can sovle much of it graphically. The point being, for vector problems, always draw them to scale. Use graph paper, a ruler and a protractor.
    Last edited: Oct 4, 2006
  11. Oct 5, 2006 #10
    got it, thanks for the advice :)
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