The acceleration initial velocity and initial position of a

The acceleration, initial velocity, and initial position of a particle traveling through space are given below. a(t) = (2, 4, 8), v(0) = (7, 2, 2), r(0) = (12, 3, 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.)

(x,y,z)=( )

(x,y,z)=( )

Solution

a(t) is constant along all directions

So,

v(t)=(2t+v1,-4t+v2,-8t+v3)

v(0)=(v1,v2,v3)=(-7,2,2)

v(t)=(2t-7,-4t+2,-8t+2)

r(t)=(t^2-7t+r1,-2t^2+2t+r2,-4t^2+2t+r3)

r0=(12,-3,1)=(r1,r2,r3)

r(t)=(t^2-7t+12,-2t^2+2t-3,-4t^2+2t+1)

When particle intersections yz plane its x coordinate goes to 0. This happens at

t^2-7t+12=0

t^2-4t-3t+12=0

t=3,4

r(t)=(t^2-7t+12,-2t^2+2t-3,-4t^2+2t+1)

t=3

r(3)=(0,-18+6-3,-36+6+1)=(0,-15,-29)

r(4)=(0,-32+8-3,-64+8+1)=(0,-27,-55)

So in order of smallest to largest y points are

(0,-27,-55)

(0,-15,-29)