Respuesta :
Answer:
Step-by-step explanation:
First, we integrate the acceleration fiction to obtain the velocity function.
The acceleration function, 3i - j - k, in vector form is [tex]\left<3, -1, -1\right>[/tex]
So, [tex]\int \left<3, -1, -1\right>dt \\ \\= \left<3t, -t, -t \right> + V_0 \\ \\[/tex]
Where [tex]V_0 = (V_1, V_2, V_3)[/tex] is a constant vector. To obtain [tex]V_0[/tex], knowing that the particle has speed 4 at (5, 2, 7) at t = 0.
[tex]| V_0 |[/tex] = 4
The particle travels in a straight line to (8, 1, 6), and (8, 1, 6) - (5, 2, 7) = [tex]\left<3, -1, -1\right>[/tex]
There is a constant, s, such that [tex]V_1 = 3s, V_2 = -s, V_3 = -s[/tex]
So,
[tex]|V_0|= 4 = \sqrt{(3s)^2 + (-s)^2 + s^2} \\ \\\sqrt{11s^2} = 4 \\ \\s = \frac{4}{\sqrt{11}}[/tex]
Therefore, [tex]V_0 = \left<\frac{12}{\sqrt{11}}, - \frac{4}{\sqrt{11}}, - \frac{4}{\sqrt{11}}\right>[/tex]
and
[tex]\left<3t, -t, -t \right> + \left<\frac{12}{\sqrt{11}}, - \frac{4}{\sqrt{11}}, - \frac{4}{\sqrt{11}}\right> \\ \\= \left<3t +\frac{12}{\sqrt{11}}, -t - \frac{4}{\sqrt{11}}, -t - \frac{4}{\sqrt{11}}\right>[/tex]
[tex]v(t) = \left(3t +\frac{12}{\sqrt{11}}\right)i - \left(t + \frac{4}{\sqrt{11}}\right) j - \left(t + \frac{4}{\sqrt{11}}\right)k\\[/tex]
Next, we find the position function by integrating the velocity function.
[tex]\int{\left<3t +\frac{12}{\sqrt{11}}, -t - \frac{4}{\sqrt{11}}, -t - \frac{4}{\sqrt{11}}\right>}dt \\ \\= \left<\frac{3}{2}t^2 +\frac{12}{\sqrt{11}}t, -\frac{t^2}{2} - \frac{4}{\sqrt{11}}t, -\frac{t^2}{2} - \frac{4}{\sqrt{11}}t\right> + r_0[/tex]
Where [tex]r_0 = \left<r_1, r_2, r_3\right>[/tex] is a constant vector. To obtain [tex]r_0[/tex], knowing that the particle is at (5, 2, 7) at t = 0.
[tex]r(t) = \left<\frac{3}{2}t^2 +\frac{12}{\sqrt{11}}t, -\frac{t^2}{2} - \frac{4}{\sqrt{11}}t, -\frac{t^2}{2} - \frac{4}{\sqrt{11}}t\right> + \left<5, 2, 7\right> \\ \\= \left<\frac{3}{2}t^2 +\frac{12}{\sqrt{11}}t + 5, -\frac{t^2}{2} - \frac{4}{\sqrt{11}}t + 2, -\frac{t^2}{2} - \frac{4}{\sqrt{11}}t + 7 \right>r(t) = \left(\frac{3}{2}t^2 +\frac{12}{\sqrt{11}}t + 5\right)i - \left(\frac{t^2}{2} + \frac{4}{\sqrt{11}}t - 2\right)j - \left(\frac{t^2}{2} + \frac{4}{\sqrt{11}}t - 7 \right)k[/tex]
[tex]r(t) = \left(\frac{3}{2}t^2 +\frac{12}{\sqrt{11}}t + 5\right)i - \left(\frac{t^2}{2} + \frac{4}{\sqrt{11}}t - 2\right)j - \left(\frac{t^2}{2} + \frac{4}{\sqrt{11}}t - 7 \right)k[/tex]
The equation for the vector position of the particle at time [tex]t[/tex] is [tex]\vec r = (5,2,7) + t\cdot (3.617,-1.206, -1.206) + t^{2}\cdot \left(\frac{3}{2},-\frac{1}{2}, -\frac{1}{2} \right)[/tex].
Vectorially speaking, the vector acceleration of the particle ([tex]\vec a[/tex]) is:
[tex]\vec a = (3, -1, -1)[/tex] (1)
By vectorial calculus, definition of velocity and vector properties, we derive the expression for the vector velocity:
[tex]\vec v = v_{o}\cdot \frac{\vec r_{AB}}{\|\vec r_{AB}\|} + (3,-1,-1)\cdot t[/tex] (2)
Where:
- [tex]v_{o}[/tex] - Initial speed.
- [tex]\vec r_{AB}[/tex] - Vector between initial and final positions.
- [tex]\|\vec r_{AB}\|[/tex] - Magnitude of the vector between initial and final positions.
- [tex]t[/tex] - Time.
Then, the vector and its magnitude are shown below:
[tex]\vec r_{AB} = (8,1,6)-(5,2,7)[/tex]
[tex]\vec r_{AB} = (3, -1,-1)[/tex]
[tex]\|\vec r_{AB}\| = \sqrt{3^{2}+(-1)^{2}+(-1)^{2}}[/tex]
[tex]\|\vec r_{AB}\| =3.317[/tex]
If we know that [tex]v_{o} = 4[/tex], [tex]\vec r_{AB} = (3, -1,-1)[/tex] and [tex]\|\vec r_{AB}\| =3.317[/tex], then the vector velocity is:
[tex]\vec v = \frac{4}{3.317} \cdot (3,-1,-1)+(3,-1,-1)\cdot t[/tex]
[tex]\vec v = (3.617, -1.206, -1.206) + t\cdot (3, -1, -1)[/tex]
And the vector acceleration is described by this equation derived from vectorial calculus and definition of position:
[tex]\vec r = (5,2,7) + t\cdot (3.617,-1.206, -1.206) + t^{2}\cdot \left(\frac{3}{2},-\frac{1}{2}, -\frac{1}{2} \right)[/tex] (3)
The equation for the vector position of the particle at time [tex]t[/tex] is [tex]\vec r = (5,2,7) + t\cdot (3.617,-1.206, -1.206) + t^{2}\cdot \left(\frac{3}{2},-\frac{1}{2}, -\frac{1}{2} \right)[/tex].
We kindly invite to check this question on vectors: https://brainly.com/question/4579006
