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baddinlol

  • 2 years ago

An object is moving with constant speed in a straight line relative to O. the objects initial and final position vectors are respectively: u = i + 2j + 3k, v = 2i + j -4k, Find expression for q the position vector of the object when it has travelled a distance of x.

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  1. baddinlol
    • 2 years ago
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    Ok, so I am having a bit of trouble with this one.

  2. shana91
    • 2 years ago
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    is the diagram clear? |dw:1331119632568:dw|

  3. shana91
    • 2 years ago
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    do you knwo how to take vector UV?

  4. baddinlol
    • 2 years ago
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    Yes -u + v

  5. shana91
    • 2 years ago
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    |dw:1331119715424:dw|

  6. baddinlol
    • 2 years ago
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    Yep.

  7. baddinlol
    • 2 years ago
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    I was thinking maybe q = (v-u) + xu

  8. shana91
    • 2 years ago
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    ok so it gives, UV as i-j-7k. when it is at distance 0, the vector is U, when it is in its end, the vector is V so when it has travelled X it should be in between right? so the answer when it is after x distance is, |dw:1331119912201:dw|

  9. shana91
    • 2 years ago
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    when it is at x distance, it is the UNIT VECTOR*X i have found the unit vector by dividing by the scalar value of the vetor which is \[\sqrt{1^2+1^2+7^2}\]

  10. baddinlol
    • 2 years ago
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    Why are we using the unit vector though?

  11. shana91
    • 2 years ago
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    Unit vector means= what's the direction and the magnitude a particle will have when x=1(it has gone by one UNIT) so to get x units you have to multiply by X

  12. baddinlol
    • 2 years ago
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    Oh.. ok That makes it clear.. Thanks for your help.

  13. shana91
    • 2 years ago
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    welcome :)

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