Solving Motion Equations with Integration

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SUMMARY

The discussion focuses on solving motion equations using integration, specifically the equation for position as a function of time, pos(t) = -0.65t² i + 0.13t² j + 14ti + 13tj. A key point is the importance of correctly interpreting the components of the equation, particularly that the j-component should not be set to zero. The user successfully resolved their confusion by applying the quadratic formula to the x-component, leading to a valid solution.

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  • Understanding of vector components in motion equations
  • Familiarity with integration techniques in physics
  • Knowledge of quadratic equations and their solutions
  • Basic concepts of motion in two dimensions
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  • Study the application of integration in kinematics
  • Learn about vector decomposition in physics
  • Explore the quadratic formula and its applications in motion problems
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Students and educators in physics, particularly those focusing on kinematics and motion analysis, as well as anyone looking to deepen their understanding of integration in motion equations.

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Homework Statement
A particle leaves the origin with its initial velocity given by v⃗ 0=14i+13jm/s, undergoing constant acceleration a⃗ =−1.3i+0.26j m/s2. Find when and where the particle crosses the y axis.
Relevant Equations
Δx=V_0 t+1/2 at^2
and other kinematics equations
I'm not sure where to start, when I tired using integration of the initial equation to get pos(t)=-.65t^2 i + .13t^2 j + 14ti +13tj but after separating each component, i and j, and setting j equal to zero I got 0 or -100 seconds which doesn't seem like a reasonable answer.
 
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First, you do not set j = 0. I assume you mean setting the j-component to zero.

Second, y=0 is the x-axis, not the y-axis. The y-axis corresponds to x=0.
 
OK, I tried it out and it works! I was confused with the wording of the question, but setting the x equation, x=-.65t^2 i + 14t i, equal to zero and using the quadratic worked. Thanks for your help!
 
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