How Does a Particle's Position Change with Acceleration in the XY Plane?

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SUMMARY

The discussion centers on a physics problem involving a particle that departs from the origin with an initial velocity of 7.2 m/s in the positive y-direction and experiences a constant acceleration of (3.0i - 2.0j) m/s². To determine the x-coordinate when the particle crosses the x-axis (y=0), users confirm that initial positions are (xi, yi) = (0, 0). The relevant equations for solving the problem include kinematic equations for motion in two dimensions.

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  • Understanding of kinematic equations in two dimensions
  • Knowledge of vector components in physics
  • Ability to perform substitutions in equations
  • Familiarity with the concept of acceleration and its effects on motion
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This discussion is beneficial for physics students, educators, and anyone interested in understanding motion dynamics in two dimensions, particularly those dealing with kinematic problems involving acceleration and initial conditions.

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Homework Statement



A particle leaves the origin with a velocity of 7.2 m/s in the positive y direction and moves in the xy plane with a constant acceleration of (3.0i - 2.0j) m/s2. At the instant the particle moves back across the x-axis (y=0), what is the value of its x coordinate?

Homework Equations



vxf=vxi+axt
xf=xi+.5(vxf+vxi)t
xf=xi+vxit+.5axt2
v2xf=v2xi+2ax(xf-xi)


The Attempt at a Solution



Are my x and y initial (0,0) because it says origin? Do I have to used to substitution as well?
 
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Starting at the origin does mean that xi=yi=0, yes.

I'm not entirely sure what your second question means, but you should be able to formulate an expression for both the x and y position of the particle at any given time and then use substitution to find the answer.
 

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