Answer:
(0,-26,-8) and (0,-12,-1)
Step-by-step explanation:
a(t) =
(v(t))
⇒
= ![\int\limits^t_0 {a} \, dt](https://tex.z-dn.net/?f=%5Cint%5Climits%5Et_0%20%7Ba%7D%20%5C%2C%20dt)
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) =
(r(t))
⇒
= ![\int\limits^t_0 {v} \, dt](https://tex.z-dn.net/?f=%5Cint%5Climits%5Et_0%20%7Bv%7D%20%5C%2C%20dt)
r(t) - r(0) = (
-5t, -3
+t, -2
+3t)
⇒r(t) = (
-5t+6, -3
+t-2, -2
+3t+1)
when the particle intersects the yz plane, it's x-coordinate is 0
⇒
-5t+6=0
⇒
-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) = (
-5·2+6, -3·
+2-2, -2·
+3·2+1)
<u>r(2) = (0,-12,-1)</u>
r(3) = (
-5·3+6, -3·
+3-2, -2·
+3·3+1)
<u>r(3) = (0,-26,-8)</u>
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)