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

[Sho Kano]

Homework Statement

A point moves on a curve \vec { r } with constant acceleration \vec { A }, initial velocity \vec { { V }_{ 0 } }, and initial position { \vec { { P }_{ 0 } } }

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

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

Homework Equations

The Attempt at a Solution

part a asked for the position function, so here it is:
\vec { r(t) } =\frac { 1 }{ 2 } \vec { A } { t }^{ 2 }+\vec { { V }_{ 0 } } t+{ \vec { { P }_{ 0 } } }

my attempt at part b:
\vec { A } must be parallel to \vec { { V }_{ 0 } }, so \vec { A } =a\vec { { V }_{ 0 } }, where a is some constant.
so \vec { r(t) } =\frac { 1 }{ 2 } \vec { a{ V }_{ 0 } } { t }^{ 2 }+\vec { { V }_{ 0 } } t+{ \vec { { P }_{ 0 } } }
?

 

[olivermsun]

So how do you show that something moves in a straight line? (Hint: you already showed similar for $\vec{A}$ and $\vec{V_0}$).

 

[Sho Kano]

So how do you show that something moves in a straight line? (Hint: you already showed similar for $\vec{A}$ and $\vec{V_0}$).

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

 

[olivermsun]

What about position?

 

[Sho Kano]

What about position?

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 don't know if they are parallel

 

[olivermsun]

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)?

 

[Sho Kano]

Can you show that →r(t)r(t)→\vec{r(t)} also satisfies the equation for a straight line, passing through some initial point (but not necessarily the origin)?

something like this?
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 } }Some varying scalar k times V, which is straight line motion? Because it fits the standard vector equation for a line.

 

[olivermsun]

Seems reasonable, right?

 

[Sho Kano]

Seems reasonable, right?

Yep, at first I thought it couldn't be a "line" because there was a t squared term. Thanks.