Respuesta :
Answer:
(0,-26,-8) and (0,-12,-1)
Step-by-step explanation:
a(t) = [tex]\frac{d}{dt}[/tex](v(t))
⇒[tex]\int\limits^t_0 dv(t)[/tex] = [tex]\int\limits^t_0 {a} \, dt[/tex]
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) = [tex]\frac{d}{dt}[/tex](r(t))
⇒[tex]\int\limits^t_0 dr(t)[/tex] = [tex]\int\limits^t_0 {v} \, dt[/tex]
r(t) - r(0) = ([tex]t^{2}[/tex]-5t, -3[tex]t^{2}[/tex]+t, -2[tex]t^{2}[/tex]+3t)
⇒r(t) = ([tex]t^{2}[/tex]-5t+6, -3[tex]t^{2}[/tex]+t-2, -2[tex]t^{2}[/tex]+3t+1)
when the particle intersects the yz plane, it's x-coordinate is 0
⇒[tex]t^{2}[/tex]-5t+6=0
⇒[tex]t^{2}[/tex]-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) = ([tex]2^{2}[/tex]-5·2+6, -3·[tex]2^{2}[/tex]+2-2, -2·[tex]2^{2}[/tex]+3·2+1)
r(2) = (0,-12,-1)
r(3) = ([tex]3^{2}[/tex]-5·3+6, -3·[tex]3^{2}[/tex]+3-2, -2·[tex]3^{2}[/tex]+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)
