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Sagot :
Ignoring the malformed character, it looks like you're saying you have particle initially located at (1, 1, 3) that travels in a straight line to (5, 9, 4) with initial speed 6 and constant acceleration vector 4i + 8j + k.
Use the fundamental theorem of calculus to determine the velocity function for the particle:
[tex]\vec v(t) = \vec v(0) + \displaystyle \int_0^t \vec a(u) \, du[/tex]
The particle moves in the same direction as the vector
(5i + 9j + 4k) - (i + j + 3k) = 4i + 8j + k
which has magnitude
√(4² + 8² + 1²) = √81 = 9
Normalize the direction vector by dividing it by its magnitude:
(4i + 8j + k)/9 = 4/9 i + 8/9 j + 1/9 k
The particle has initial speed 6, so we must scale this unit vector by a factor of 1/6 to get the initial velocity vector:
6 (4/9 i + 8/9 j + 1/9 k) = 8/3 i + 16/3 j + 2/3 k
Solve for v(t) :
[tex]\vec v(t) = \dfrac83\vec\imath + \dfrac{16}3\vec\jmath + \dfrac23\vec k + \displaystyle \int_0^t \left(4\vec\imath + 8\vec\jmath+\vec k\right) \, du[/tex]
[tex]\vec v(t) = \dfrac83\vec\imath + \dfrac{16}3\vec\jmath + \dfrac23\vec k + \left(4t\,\vec\imath + 8t\,\vec\jmath+t\,\vec k\right)[/tex]
[tex]\vec v(t) = \left(\dfrac83+4t\right)\vec\imath + \left(\dfrac{16}3+8t\right)\vec\jmath + \left(\dfrac23+t\right)\vec k[/tex]
Use the fundamental theorem again to find the position vector r(t) :
[tex]\vec r(t) = \vec r(0) + \displaystyle \int_0^t \vec v(u) \, du[/tex]
[tex]\vec r(t) = \vec\imath+\vec\jmath+3\vec k + \displaystyle \int_0^t \left(\left(\dfrac83+4u\right)\vec\imath + \left(\dfrac{16}3+8u\right)\vec\jmath + \left(\dfrac23+u\right)\vec k\right) \, du[/tex]
[tex]\vec r(t) = \vec\imath+\vec\jmath+3\vec k + \left(\left(\dfrac83 t+2t^2\right)\vec\imath + \left(\dfrac{16}3t+4t^2\right)\vec\jmath + \left(\dfrac23t+\dfrac12t^2\right)\vec k\right)[/tex]
[tex]\vec r(t) = \left(1+\dfrac83 t+2t^2\right)\vec\imath + \left(1+\dfrac{16}3t+4t^2\right)\vec\jmath + \left(3+\dfrac23t+\dfrac12t^2\right)\vec k[/tex]
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