Calculating Time of Force Application in Non-Uniform Motion

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To calculate the time of force application for a 230N force accelerating a 4.0kg mass from 5.0m/s North to 3.0m/s East, the initial and final velocities must be treated as vector components. The change in velocity can be represented as Δv = (3, -5), allowing for the use of the equation dvec{v}/dt = F/m to derive two separate equations for each velocity component. The force can be broken down into its x and y components, with the relationship F² = Fx² + Fy² helping to solve for the unknowns. By applying the equations F_xΔt = mΔvx and F_yΔt = mΔvy, the time of force application can be determined. This approach effectively addresses the challenge of non-uniform motion in different directions.
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A 230N force causes a 4.0kg to accelerate from 5.0m/s North to 3.0m/s east. for how much time does hte force act?

t = m(vf-vi) / F

i got this but the trouble is vf and vi are in different directions...and since t is not a vector i can't break it down to components. what should i do?
 
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Maybe you can do it by this way.
let \vec{v_i} =(0,5) \vec{v_f}=(3,0)
so \Delta \vec{v} =(3,-5)
 
You can integrate

\frac{d\vec{v}}{dt} = \frac{\vec{F}}{m}

Your solution will be two equations (one for each component of velocity). You know the initial and final velocity vectors. Your unknowns are the two components of \vec F and t but you also know

F^2 = F_x^2 + F_y^2

where F = 230 N. You should be able to handle the details.
 
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using
F_x\Delta t\ =\ m\Delta v_x\ =\ a
F_y\Delta t\ =\ m\Delta v_y\ =\ b
and then
a^2\ +\ b^2\ =\ \Delta t^2 \mid F\mid^2
 
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