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

Answer:

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

Step-by-step explanation:

a(t) = $\frac{d}{dt}$(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) = $\frac{d}{dt}$(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)