Determining Distance and Final Velocity for an Object with 3D Vectors

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The discussion focuses on calculating the final velocity and distance traveled by an object subjected to acceleration in a 3D vector context. The final velocity is determined using the equation Vf = Vi + at, resulting in Vf = 10i + 10j + 100k. The distance traveled is calculated using the Pythagorean theorem, with some participants suggesting the use of the equation s = ut + 0.5at^2 for position vector determination. There is confusion regarding the distance from the z-axis, with participants debating the correct application of formulas to find this distance. Overall, the calculations and methods discussed illustrate the complexity of 3D motion analysis.
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An object has an initial velocity given by V = vi + vj, where v=10m/s. If an acceleration of A = ak, where a = 10 m/s^2 is applied for 10 seconds determine the final velocity for the object.

How far did the object travel in 10s?

How far from the z axis is the object?


Well to start. I used the equation:
Vf = Vi + at
Vf = (10i + 10j) + (10k)(10s)
Vf = 10i + 10j + 100k


Then to determine distance, I used:
D^2 = (change in x)^2 + (change in y)^2 + (change in z)^2
so.

D = sqrt.(10,200) m?
I am not sure if I calculated this right.


Then for the last part of the question, I am not really sure how to determine this. I know that the object moved 100m in the x direction, and 100m in the y direction, and I think it moved 500m in the z direction given the Vf= 100m/s in z direction and Vi = 0m/s in z direction.

So I am assuming we have to use pythagorean theorem again? I am not sure. Please help, and comment on any of the previous answers please.
 
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for the second part, i think you should use
s = ut + half*at^2 to find its position vector and then use the distance formula to find its distance from the origin.

for the third part, that will be the z-component of the above position vector that you find.
 
so if I use that equation:

i get:

s(t) = 100i + 100j + 500K

D = sqrt. (270,000) m?Then for distance from z axis is simply sqrt. (500) m?
 
i did a mistake here.

for the second part, i guess so...

for the third part, it is square root of (x-component^2+y-component^2) = sqrt(10000+10000). if you draw the final coordinate of the particle, i think you will see why.
 
i am not really understanding how you determined the 3rd part...The way you did it, it simply looks like the displacement between the x and y components...not the distance from the z axis.
 
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