Question

The acceleration, initial velocity, and initial position of a particle traveling through space are given by by a(t) = (2, −6, −4), v(0) = (−5, 1, 3), r(0) = (6, −2, 1). 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.)

Answer 1

(0,-26,-8) and (0,-12,-1)

Explanation:

a(t) =
`(d)/(dt)`(v(t))

⇒
`\int\limits^t_0 dv(t)` =
`\int\limits^t_0 {a} \, dt`

Integration of a vector is simultaneous integration of each of its components:

⇒v(t) - v(0) = (2t, -6t, -4t)

⇒v(t) = (2t-5, -6t+1, -4t+3)

v(t) =
`(d)/(dt)`(r(t))

⇒
`\int\limits^t_0 dr(t)` =
`\int\limits^t_0 {v} \, dt`

r(t) - r(0) = (
`t^(2)`-5t, -3
`t^(2)`+t, -2
`t^(2)`+3t)

⇒r(t) = (
`t^(2)`-5t+6, -3
`t^(2)`+t-2, -2
`t^(2)`+3t+1)

when the particle intersects the yz plane, it's x-coordinate is 0

⇒
`t^(2)`-5t+6=0

⇒
`t^(2)`-2t-3t+6=0

⇒t·(t-2)-3·(t-2)=0

⇒(t-2)·(t-3)=0

∴Particle hits the yz plane at t=2 and t=3.

r(2) = (
`2^(2)`-5·2+6, -3·
`2^(2)`+2-2, -2·
`2^(2)`+3·2+1)

r(2) = (0,-12,-1)

r(3) = (
`3^(2)`-5·3+6, -3·
`3^(2)`+3-2, -2·
`3^(2)`+3·3+1)

r(3) = (0,-26,-8)

x(2) = x(3) =0

y(3) = -26 is less than -12 = y(2)

∴The two intersection points with yz plane are (0,-26,-8) and (0,-12,-1)