How Far Does a Particle Travel in an Equilateral Triangle with Mixed Motion?

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The discussion revolves around a physics problem involving a particle moving with a uniform velocity and acceleration in an equilateral triangle. Participants express confusion about the problem's setup, particularly regarding the lack of clarity on which sides of the triangle are being referenced and the absence of a coordinate system. There is debate over the implications of having a uniform velocity in one direction while experiencing acceleration in a non-perpendicular direction. Suggestions for solving the problem include using vector diagrams and kinematic equations, but participants remain uncertain about the correct approach and the angles involved. The conversation highlights the need for clearer problem definitions and visual aids to facilitate understanding.
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Homework Statement



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.

Homework Equations





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

Homework Statement



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.

Homework Equations





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?

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

That doesn't make sense …

how can it have a uniform velocity in one direction, and an acceleration in a non-perpendicular direction? :confused:
 
tiny-tim said:
That doesn't make sense …

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

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

not sure what you mean here - all the information from the question is given. can you start me out on the first step?
 
You could start a vector diagram with equations vi*t and 0.5at^2, finding the resultant vector when t = 5.
 
  • #10
inutard said:
You could start a vector diagram with equations vi*t and 0.5at^2, finding the resultant vector when t = 5.

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

The adiacent side of the triangle could be anyone 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
i'm sorry, I'm not following this.
 
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