Acceleration is (-2.0i + 4.0j), velocity is 12 m/s

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Homework Help Overview

The problem involves a particle moving in the xy plane with an initial velocity of 12 m/s in the positive x direction and a constant acceleration of (-2.0i + 4.0j) m/s². The goal is to determine the x coordinate of the particle when its y coordinate reaches 18 m.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations to relate position, velocity, and acceleration. Some explore the relationships between acceleration, velocity, and time, while others express uncertainty about their calculations and seek clarification on the correct application of these equations.

Discussion Status

The discussion is ongoing, with participants attempting to manipulate the equations and share their thought processes. Some have offered insights into integrating acceleration to find velocity as a function of time, while others are questioning their own reasoning and calculations.

Contextual Notes

Participants are working within the constraints of the problem as stated, with specific attention to the initial conditions and the requirement to find the x coordinate at a given y coordinate. There is a noted uncertainty regarding the correct application of calculus in this context.

new^2^physics
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Homework Statement



At T=0 a particle leaves the origin with a velocity of 12 m/s in the positive x direction and moves in the xy plane with a constant acceleration of (-2.0i + 4.0j)m/s^2. At the instant the y coordinate of the particle is 18 m, what is the x coordinate of the particle.

Homework Equations



rf=ri + vit + 1/2at2
=(12m/s)(t) + 1/2 (-2.0i + 4.0j)t2

So stuck. Any direction or help would be greatly appreciated!
 
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so you know initial velocity to be 12.0i..
now you also know a = dv/dt
and v = dr/dt (r being position)..

Which mathematical (or calculus) tool can you use to get your answer now? and how?
 
Okay, so I am going to play with these equations and try to get closer...

a = dv/dt
(-2.0i + 4.0j)= 12i / dt

dt = 12i / -6.0i + 4.0j
dt = -6.0i - 4.0 j

v = dr/dt
12 i = xi - 18j / -6.0i - 4.0j
12i(-6.0i - 4.0 j) = xi - 18j

oh dear, now I am stuck again and I am not even sure I have done this part correctly.
 
new^2^physics said:
Okay, so I am going to play with these equations and try to get closer...

a = dv/dt
(-2.0i + 4.0j)= 12i / dt


that isn't correct...

a=dv/dt
(-2.0i+4.0j)dt=dv
integrating

-(2t)i + (4t)j = v-12

so you get v as a function of time... now proceed in a similar manner...
 
new^2^physics said:
At T=0 a particle leaves the origin with a velocity of 12 m/s in the positive x direction...

So, what are V_{ix}[/tex] and V_{iy}[/tex]?
 

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