How Fast is a Particle Moving at 15 Meters in Two Dimensions?

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A particle starts at the origin with an initial velocity of 9.0 m/s in the positive y direction and experiences a constant acceleration of (2.0i - 4.0j) m/s². To determine the speed when the particle's x-coordinate reaches 15 m, kinematic equations for constant acceleration can be applied separately for the x and y motions. The x-component acceleration is 2 m/s², and the y-component acceleration is -4 m/s². By calculating the time and final velocities in both dimensions, the overall speed can be derived, leading to a final speed of approximately 8.91 m/s. The discussion emphasizes the importance of using the correct kinematic equations to solve for unknown variables.
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Homework Statement


At t=0, a particle leaves the origin with a velocity of 9.0 m/s in the positive y direction and moves in the xy plane with a constant acceleration of (2.0i - 4.0j)m/s^2. At the instant the x coordinate of the partice is 15 m, what is the speed of the particle?


Homework Equations


Basic Equations


The Attempt at a Solution


I'm not sure how to begin this problem, any help would be awesome. Keep in mind this is my first physics class so keep it as simple as possible. Thank you.
 
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Welcome to PF im4rheal,

I'm sure you know the kinematics equations for constant acceleration. You can treat the x and y motions separately. So you just have to apply these equations for the motions in each direction. For the x direction:

- you know 'a'
- you know 'd' (or Δx)
- you know 'v0' (initial velocity)

You don't know t, or v, at the position in question, but all you need is two kinematics equations and you can solve for these two unknowns.
 


Amongst your 'basic equations', do you have one that tells you a relationship between a constant acceleration, an initial speed, a distance travelled, and a final speed? If so, can you write this out with known values in the x-direction?
 


Ok, can I use the Vxf (velocity in the x direction, final) = Vxi + ax*t to find t?
 


That's one of the equations you need to use, but you don't know what vxf is, so you need another equation as well.

How about one incorporating the distance traveled Δx, which you DO know?
 


Is it v^2xf = v^2xi + 2ax(xf-xi) ?
 


im4rheal said:
Is it v^2xf = v^2xi + 2ax(xf-xi) ?
Confusing notation, but that's basically it. (If you can't be bothered to use superscript and subscript, at least use '*' for multiply and maybe mixed case to represent subscripting, like Vx.)
You'll also need to figure out the time so as to get the velocity in the y direction.
 


im4rheal said:
Is it v^2xf = v^2xi + 2ax(xf-xi) ?

Well, yeah. I mean, you know all but one of the things in that equation, and it is the thing you are trying to solve for. So, of course this should work.
 


Sorry I don't know how to do superscript and subscript. I should be able to find time from the first equation for solving for t right?
 
  • #10


im4rheal said:
Sorry I don't know how to do superscript and subscript. I should be able to find time from the first equation for solving for t right?

Yup! You can get t from the equation in post #4, once you have vxf.

Try the sup and sub tags:

[noparse] vxf2 [/noparse]

to get this result:

vxf2

OR just use the buttons marked X2 and X2 that appear above the reply box, when you are in advanced reply mode (not quick reply)
 
  • #11


Ok thanks
 
  • #12


cepheid said:
Well, yeah. I mean, you know all but one of the things in that equation, and it is the thing you are trying to solve for. So, of course this should work.

For the value of ax is it 2 m/s2?
 
  • #13


im4rheal said:
For the value of ax is it 2 m/s2?

Yeah, this is given in the problem. The x-component of the vector is one with unit vector "i", and the y-component has unit vector "j".
 
  • #14


I got 8.91 m/s. Correct?
 
  • #15
I get a different answer. Can you post your whole solution please? Otherwise we can't help you fix any errors that might be there...

What do you get for the individual components of the velocity (vx and vy)?
 
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