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Prove r(t) moves in a line, if a and v are parallel

  1. Sep 29, 2016 #1
    1. The problem statement, all variables and given/known data
    A point moves on a curve [itex]\vec { r } [/itex] with constant acceleration [itex]\vec { A } [/itex], initial velocity [itex]\vec { { V }_{ 0 } } [/itex], and initial position [itex]{ \vec { { P }_{ 0 } } }[/itex]

    b. if [itex]\vec { A } [/itex] and [itex]\vec { { V }_{ 0 } } [/itex] are parallel, prove [itex]\vec { r } [/itex] moves in a line

    c. Assuming [itex]\vec { A } [/itex] and [itex]\vec { { V }_{ 0 } } [/itex] are not parallel, prove [itex]\vec { r } [/itex] lies in a plane.

    2. Relevant equations


    3. The attempt at a solution
    part a asked for the position function, so here it is:
    [itex]\vec { r(t) } =\frac { 1 }{ 2 } \vec { A } { t }^{ 2 }+\vec { { V }_{ 0 } } t+{ \vec { { P }_{ 0 } } }[/itex]

    my attempt at part b:
    [itex]\vec { A } [/itex] must be parallel to [itex]\vec { { V }_{ 0 } } [/itex], so [itex]\vec { A } =a\vec { { V }_{ 0 } } [/itex], where [itex]a[/itex] is some constant.
    so [itex]\vec { r(t) } =\frac { 1 }{ 2 } \vec { a{ V }_{ 0 } } { t }^{ 2 }+\vec { { V }_{ 0 } } t+{ \vec { { P }_{ 0 } } }[/itex]
    ???
     
  2. jcsd
  3. Sep 29, 2016 #2

    olivermsun

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    So how do you show that something moves in a straight line? (Hint: you already showed similar for ##\vec{A}## and ##\vec{V_0}##).
     
  4. Sep 29, 2016 #3
    well I was thinking if it's acceleration and velocity are in the same direction, then it's already
    moving in a straight line, but that's not a proof
     
  5. Sep 29, 2016 #4

    olivermsun

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    What about position?
     
  6. Sep 29, 2016 #5
    I'm thinking a cross product between r and r prime? It should equal 0.

    edit: maybe this doesn't work, b/c I'm left with P x V, and I dunno if they are parallel
     
  7. Sep 29, 2016 #6

    olivermsun

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    Think more simply than that. We already know intuitively that A and V parallel means the object will move in a straight line. We just need to show this formally.

    What does it mean to move in a straight line?

    You formulated "##\vec{A}## parallel to ##\vec{V_0}##" as "##\vec{A}## = a##\vec{V_0}##," which shows that ##\vec{A}## and ##\vec{V_0}## lie on the same line passing through the origin.

    Can you show that ##\vec{r(t)}## also satisfies the equation for a straight line, passing through some initial point (but not necessarily the origin)?
     
  8. Sep 29, 2016 #7
    something like this?
    [itex]let\vec { { P }_{ 0 } } =\left< 0,0,0 \right> \\ \vec { { r(t) } } =\frac { a{ t }^{ 2 } }{ 2 } \vec { { V }_{ 0 } } +t\vec { { V }_{ 0 } } \\ \vec { { r(t) } } =\left( \frac { a{ t }^{ 2 } }{ 2 } +t \right) \vec { { V }_{ 0 } } =k(t)\vec { { V }_{ 0 } } [/itex]Some varying scalar k times V, which is straight line motion? Because it fits the standard vector equation for a line.
     
  9. Sep 29, 2016 #8

    olivermsun

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    Seems reasonable, right?
     
  10. Sep 29, 2016 #9
    Yep, at first I thought it couldn't be a "line" because there was a t squared term. Thanks.
     
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