{vectors}find the position vector of this

[niekehecv]

Homework Statement

a particle traveling in a straight line is located at the point (4, -2, 3) with the speed of 2 m/s at time t=0. the particle moves toward the point (6, 0, 9) with constant acceleration 3i-j+k. find its position vector r(t) at time t

The Attempt at a Solution

well, i tried to integrate the acceleration vector to obtain the velocity vector, but the constant of integration couldn't be determined by means i know of.

v=3t i- t j + t k + c

where C is the constant of integration

I am stuck at trying to find the constant of integration, which none of my ways i know of could be applied at all...

 

[niekehecv]

if there isn't enough information provided please inform me so i could inform the lecturer as soon as possible

 

[HallsofIvy]

I'm not sure whether there "isn't enough information" or how to interpret we are given. Does "moves toward the point (6, 0, 9)" mean that at the instant when the particle is (4, -2, 3) the instantaneous velocity vector is in the direction of <6-4, -2-0, 3- 9>= <2, -2, -6> or does it mean the curving path of the particle passes through (6, 0, 9)?

If it is the former then, which I am inclined to believe just because it is easier, since $\vec{a}= <3, -1, 1>$, it follows that the velocity vector is <3t+ a, -t+ b+ t+ c> where a, b, and c are constants, where a, b, and c are the components of the "constant of integration vector" you call C. Taking t= 0 to be the instant when the particle is at (4, -2, 3), you must have $<a, b, c>= (2/\sqrt{11})<3, -1, 1>$ which gives you the constants (the "$2/\sqrt{11}$ is to make the speed 2).

If it is the latter you do not know the initial velocity vector so you will have to leave it as <3t+ a, -t+ b, t+ c>. Integrating again gives the position vector as $(3/2)t^2+ at+ 4 -(1/2)t^2+ bt- 2, (1/2)t^2+ ct+ 3>$ (the new "constants of integration" gives the initial point when t= 0). Now, choose a, b, c such that for some t, the particle is at (6, 0, 9).

 

[niekehecv]

hello there,
the particle is traveling at a straight line.
so it would be the former.

 

[niekehecv]

Taking t= 0 to be the instant when the particle is at (4, -2, 3), you must have $<a, b, c>= (2/\sqrt{11})<3, -1, 1>$ which gives you the constants (the "$2/\sqrt{11}$ is to make the speed 2).

hey, i do not understand how you found out that <a,b,c> =(2/\sqrt{11})<3, -1, 1>.
how can you elaborate more on this part?