Particle Motion Along a Straight Line

[jdawg]

Homework Statement

At time t=0, a particle is located at the point (1,2,3). It travels in a straight line to the point (4,1,4), has speed 2 at (1,2,3) and constant acceleration 3i-j+k. Find an equation for the position vector r(t) of the particle at time t.

Homework Equations

The Attempt at a Solution

So I'm not exactly sure what to do with this information. I was thinking maybe integrate the acceleration vector to get the velocity vector, and the integrate that to get the position. But if it were that simple, then why would they give me all that other information? I'm very confused by this problem, please give me a hint on how to start!

 

[LCKurtz]

Homework Statement

At time t=0, a particle is located at the point (1,2,3). It travels in a straight line to the point (4,1,4), has speed 2 at (1,2,3) and constant acceleration 3i-j+k. Find an equation for the position vector r(t) of the particle at time t.

Homework Equations

The Attempt at a Solution

So I'm not exactly sure what to do with this information. I was thinking maybe integrate the acceleration vector to get the velocity vector, and the integrate that to get the position. But if it were that simple, then why would they give me all that other information? I'm very confused by this problem, please give me a hint on how to start!

You have the right idea. You start by integrating the acceleration vector, after you figure out its length and direction to start. Remember you get vector constants of integration each time you integrate, so you will need the extra info. Try it and show us what you get.

 

[jdawg]

This is what I got for the integration:
v(t)=(3t)I-(t)j+(t)k
r(t)=([itex]\frac{2}{3}[/itex]t²)I-([itex]\frac{t²}{2}[/itex])j+([itex]\frac{t²}{2}[/itex])k

acceleration magnitude: [itex]\sqrt{11}[/itex]

How do you find the direction of a vector? Is it just [itex]\frac{1}{magnitude}[/itex]*the vector?
Or am I looking for an angle? Sorry, I'm a little fuzzy on my vectors.

 

[LCKurtz]

This is what I got for the integration:
v(t)=(3t)i-(t)j+(t)k

You need to add the constant vector of integration ##\vec v_0##. Figure out its components from the given before integrating again.

r(t)=([itex]\frac{2}{3}[/itex]t²)I-([itex]\frac{t²}{2}[/itex])j+([itex]\frac{t²}{2}[/itex])k

acceleration magnitude: [itex]\sqrt{11}[/itex]

How do you find the direction of a vector? Is it just [itex]\frac{1}{magnitude}[/itex]*the vector?
Or am I looking for an angle? Sorry, I'm a little fuzzy on my vectors.

Don't mix tex with sup icons. You can write the whole expression in tex like this:
[itex]\frac{2}{3}t^2i-(\frac{t^2}{2})j+(\frac{t^2}{2})k[/itex].

Quote this to see how to do it. We will get to that after you get the velocity right.