2D Kinematics; Flying Saucer observed with two velocities

[oddjobmj]

Homework Statement

A flying saucer maneuvering with constant acceleration is observed with the positions and velocities shown below:

http://imgur.com/gq1g0

What is the saucer's acceleration?

*note: I created the image in paint and I did forget to label the vectors. The one starting at the origin is '1' and the second vector is '2'.

Homework Equations

EDIT:
a(vector)=lim (as delta t approaches 0) delta v(vector)/delta t(vector)

That equals:
dv(vector)/dt

(dv_x/dt)i + (v_y/dt)j = a(vector)

The Attempt at a Solution

[STRIKE]The portion of the chapter that most closely relates to this, as far as I can see, is the bit on Galilean transformation of velocity. 'If we know an object's velocity measured in one reference frame, we can transform it into the velocity that would be measured by an experimenter in a different reference frame.'

That method simply converts between two reference points which is how I interpret the two different vectors. Same object w/ different reference points yields two different vectors for velocity.

The issue here is that I don't know how to find acceleration of which I understand to be a change in velocity over a change in time. Having not been provided any information on time elapsed I cannot figure out how to find acceleration. I have the answer in the back of the book and see that it's not some trick where it is not accelerating or the acceleration is 0.[/STRIKE]

None of that applies >_< The section this points to suggests you can combine two vectors into one and then somehow come up with instantaneous velocity but it doesn't provide any examples of how to do it; just images. I'm checking google to figure it out but am having a hard time coming up with a reasonable search.

Thank you!
~Odd

 

[Delphi51]

It looks like you need to know the time that elapses between the first velocity measurement and the second. Then you could use
a = Δv/Δt in each dimension to find the components of acceleration.

We don't know time, but we do also have distance info. If you assume constant acceleration, then in each dimension you have an a = Δv/Δt equation with unknowns t and a, plus a d=Vi*t +½at² equation with unknowns t and a. Hey, that gives you 2 equations with two unknowns so you can find the acceleration!
Remember to do all that for both x and y directions.

 

[oddjobmj]

I edited my original post.

I appreciate your help but I don't see how that would help me. The equation d=Vi*t +½at² is familiar but never used in this chapter/section. Shouldn't d=0? If so, it doesn't work out to anything useful. If it equals the only other attainable distance, 1000sqrt(5) (from the overall chance in position between point 1 and 2) it still doesn't help me out.

I must be missing something here because this problem is, I'm sure, very simple.

 

[Delphi51]

On the graph, you can see that vector 1 is at x=0, while vector 2 is at x=2000. So the horizontal distance from time 1 to time 2 is 2000 m.
I worked out the horizontal acceleration and it comes to about 10 m/s².

That velocity formula is the very basic accelerated motion formula from a first high school physics course. It applies to any situation where there is constant acceleration. I think that is a reasonable assumption here because there likely are an infinite number of solutions with varying acceleration.

If you wish to do the x and y parts together, you could use the same formulas with your i and j unit vectors.