Distance around a Triangle

1. Jan 28, 2010

adk

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

A particle moves from rest and has impressed a uniform velocity of 10 meters per second parallel to one side of an equilateral triangle, and a uniform acceleration of 10 meters/sec2 parallel to an adjacent side of the triangle. Find the distance of the particle from its original position at the end of 5 seconds.

2. Relevant equations

3. The attempt at a solution
Hi All - My teacher posted this problem but I'm not quite sure how to get going with it. Is this just a distance kinematic problem?

2. Jan 28, 2010

Lok

Just that, don't be fooled by particles might have been a pigeon in wind instead.

3. Jan 28, 2010

Staff: Mentor

The question is not fully defined as stated. There are two adjacent sides, so it depends on which one you are talking about. Is there a figure showing the direction of motion and which side they are talking about? Is a coordinate system shown?

I suppose you could just solve it for each of the adjacent sides and both of the initial directions of travel...

4. Jan 28, 2010

adk

no figure or coordinate system was given unfortunately. what do you mean by solve for adjacent sides?
thanks.

5. Jan 28, 2010

tiny-tim

That doesn't make sense …

how can it have a uniform velocity in one direction, and an acceleration in a non-perpendicular direction?

6. Jan 28, 2010

Lok

Just like an electron deviated from it;s path by a magnetic field. it has a velocity in one direction at first but the force will accelerate it in another. Just simpler in the problem.

7. Jan 28, 2010

Lok

The thing is that there are 2 versions. A 60 deg one and a 120 deg one. Which is it really?

8. Jan 28, 2010

kriegera

not sure what you mean here - all the information from the question is given. can you start me out on the first step?

9. Jan 28, 2010

inutard

You could start a vector diagram with equations vi*t and 0.5at^2, finding the resultant vector when t = 5.

10. Jan 28, 2010

adk

i solved this equation: (10)(5) + (1/2)(9.81)(5)^2 = 172.625

where does this fit in though - not really sure what i was solving for?

11. Jan 29, 2010

Lok

The adiacent side of the triangle could be any one so the vector from the acceleration might be at 60 deg or at 120 deg. The forward motion of 10m/s is not in the same angle direction vector, whatever does it for you, as the acceleration.

12. Jan 29, 2010

kriegera

i'm sorry, i'm not following this.

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