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

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1. Sep 29, 2016

### Sho Kano

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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 $\vec { r(t) } =\frac { 1 }{ 2 } \vec { a{ V }_{ 0 } } { t }^{ 2 }+\vec { { V }_{ 0 } } t+{ \vec { { P }_{ 0 } } }$
???

2. Sep 29, 2016

### olivermsun

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

3. Sep 29, 2016

### Sho Kano

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

4. Sep 29, 2016

### olivermsun

5. Sep 29, 2016

### Sho Kano

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

6. Sep 29, 2016

### 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)?

7. Sep 29, 2016

### Sho Kano

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.

8. Sep 29, 2016

### olivermsun

Seems reasonable, right?

9. Sep 29, 2016

### Sho Kano

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