Question

The acceleration, initial velocity, and initial position of a particle traveling through space are given below. a(t) = (2, −4, −8), v(0) = (−8, 1, 5), r(0) = (15, −2, 4) The particle's trajectory intersects the yz-plane exactly twice. Find these two intersection points. (Order your answers from smallest to largest x, then from smallest to largest y.) (x, y, z) = (x, y, z) =

Answer 1

Answer:

The intesection points are:

$\vec r_{2} = (0,19,-17)$

$\vec r_{1} = (0, 53,-71)$

Step-by-step explanation:

The acceleration vector is:

$\vec a = (2,-4,-8)$

The velocity vector is found by integrating the previous vector as a function of time:

$\vec v = (-8+2\cdot t, 1 +4\cdot t, 5 - 8\cdot t)$

Afterwards, the position vector is constructed by means of integrating velocity  vector:

$\vec r = (15-8\cdot t+t^{2},-2+t+2\cdot t^{2}, 4+5\cdot t - 4\cdot t^{2})$

The particle trajectory intersects the yz-plane when x = 0, then:

$t^{2}-8\cdot t +15 = 0$

The polynomial roots are:

$t_{1}=5, t_{2}= 3$

The intesection points are:

$\vec r_{2} = (0,19,-17)$

$\vec r_{1} = (0, 53,-71)$