Answer:
r(t) = (-t^2 -16/√33t +4)i +(-5/2t^2 -40/√33t +6)j +(-t^2 -16/√33t +3)k
Step-by-step explanation:
The equation of motion for constant acceleration is ...
r(t) = (1/2)at^2 +v0·t +r0
where a is acceleration, v0 is the initial velocity, and s0 is the initial position.
We are told the initial speed, but we need to find the corresponding velocity vector. The direction of motion is found by subtracting the starting point from the ending point:
(2, 1, 1) -(4, 6, 3) = (-2, -5, -2)
The unit vector in that direction will be the direction vector divided by its magnitude:
(-2i -5j -2k)/√((-2)^2 +(-5)^2 +(-2)^2) = (-2i -5j -2k)/√33
So, the initial velocity is ...
v0 = (8/√33)(-2i -5j -2k) = (-16/√33)i -(40/√33)j -(16/√33)k
Substituting the given values, you have ...
r(t) = (1/2)(-2i -5j -2k)t^2 +((-16/√33)i -(40/√33)j -(16/√33)k)t +(4i +6j +3k)
r(t) = (-t^2 -16/√33t +4)i +(-5/2t^2 -40/√33t +6)j +(-t^2 -16/√33t +3)k
_____
The end point is reached in about t = 0.59216940615 seconds.
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<em>Comment on exact answer</em>
We have left the coefficients with the √33 in the denominator. If you need to have the denominator be rational, you can replace 1/√33 with (√33)/33. For the purpose here, that makes the difficult expression even harder to read, so we did not make that substitution.
