Question

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 position vector r(t) at t.
I got v(t) = 3ti -tj +tk +C, and |v(0)| =2, the direction would be 3i-j+k. I'm stuck at this point. Help!

Answer 1

Given r(0) = (1, 2, 3)
and speed |v(0)| = 2
at
\[a(t) = \frac{ dv }{ dt } = 3î - j + k\]
integrate on both sides.
where the unit vector is
\[\frac{ u }{ |u| }\]
and
\[|u| = \sqrt{a^2 + b^2 +c^2}\]
your answer should be \[\sqrt{11}\] for this part.

Answer 2

I'm confused. Is sqrt(11) the length of acceleration vector ? I think the question ask for r(t)

Answer 3

This you'r unit vector(s) are:
\[(\frac{ 3 }{ \sqrt{11} }), (-\frac{ 1 }{ \sqrt{11} }), (\frac{ 1 }{ \sqrt{11} }) \]
now unit vector multiplied by speed gives
\[\frac{ u }{ |u| }\times v\]
which is the same thing as v(0).

Answer 4

r(t) is your position vector. And it will be a function in terms of t.

Answer 5

You have to get your velocity function first so that you can integrate to get your position function, which is why I am doing this.

Answer 6

Can you take it from here or did you still need more help?

Answer 7

So, v(t) = 3ti -tj+tk +C

Answer 8

Correct.

Answer 9

i think I'm stuck right here :D

Answer 10

Well, v(0) = A
(I used a different constant. A for acceleration)
therefore, your velocity vector is:
\[v(t) = (3t)i - (t)j + (t)k + v(0) \]
and you already have v(0) which I gave to you before. it was your \[\frac{ u }{ |u| }\times v\]

Answer 11

Just a hint, your x component (your "i" vector) should start with
\[v(t) = (3t + \frac{ 6 }{ \sqrt{11} })i - ...\]

Answer 12

Once you have your v(t)
simply
\[\int\limits v(t) dt\]

Answer 13

I know this part, but still thinking about how u get r(t) unit vectors

Answer 14

\[\int\limits v(t) dt = r(t) + B\]

Answer 15

oh I got it. you indicate u for acceleration, that's why i'm confused. Thanks!.

Answer 16

v(t) = 3/sqrt(11)ti - 1/sqrt(11)tj +1/sqrt(11)tk

Answer 17

Your answer should be:
\[r(t) = (\frac{ 3t^2 }{ 2 }+\frac{ 6 }{ \sqrt{11}}t+1)i - (\frac{ t^2 }{ 2 }+\frac{ 2 }{ \sqrt{11}}t-2)j+(\frac{ t^2 }{ 2 }+\frac{ 2 }{ \sqrt{11}}t+3)k\]