Couple of Vector Functions questions.

[99.96]

Having a bit of trouble with these questions:

The acceleration of the particle is given by sin(2t)i + 2(2t-1)j
If the particle was initially traveling in the direction i + j at a speed of 2^3/2, the velocity at time t is given by?

and this question:

The position of a particle is given by (t²-6t)(i + 2j -2k)
The distance traveled in the first 4 seconds is?

Help would be nice

[b^3]

For the second question. To find the distance we can find the displacement at the extreme points and the endpoints. So at t=0 r(0)=0. t=4 r(4)=-8(i+2j-2k)
now there is a turning point, so diff the function (or at least the first part anyway)
2t-6=0
t=3
so at t=3, the particle turns around. r(3)=-9(i+2j-2k)
so the distance travelled will be from 0 to -9(i+2j-2k) and back to -8(i+2j-2k). from that find the magnitude. I'll let you do the rest.

For the first question. Find the velocity by diffing. add the +c and using the magnitude of the resulting equation you should be able to let it equal the velocity given and solve for c (and t=0)
Then antidiff it again and sub in t=0 and r=1+j

I hope that helps.

[99.96]

Ah ok, second one makes sense now

Didn't quite get your explanation for the first one.

[b^3]

For the second question. To find the distance we can find the displacement at the extreme points and the endpoints. So at t=0 r(0)=0. t=4 r(4)=-8(i+2j-2k)
now there is a turning point, so diff the function (or at least the first part anyway)
2t-6=0
t=3
so at t=3, the particle turns around. r(3)=-9(i+2j-2k)
so the distance travelled will be from 0 to -9(i+2j-2k) and back to -8(i+2j-2k). from that find the magnitude. I'll let you do the rest.

For the first question. Find the velocity by diffingantidiffing. add the +c and using the magnitude of the resulting equation you should be able to let it equal the velocity given and solve for c (and t=0)
Then antidiff it again and sub in t=0 and r=1+j

I hope that helps.

Sorry it was meant to be antidiffing, i'll post a sol in a min.

[b^3]

a=sin(2t)i + 2(2t-1)j
so v=1/2 cos(2t)i+(2t²-2t)j+c
(sorry the next line is where i went wrong, ignore how to do it in the posts above, follow the one below.)
at t=0, v=u(i+j) as it is in that direction but we don't know the magnitude in that direction
so u(i+j)=1/2*i+c (subbing in t=0 and v=a(i+j))
so c=i(u-1/2)+uj
so v=(u-1/2+1/2 cos(2t))i+(2t²-2t+u)j
now the magnitude of that will equal 2^3/2 (mag is too hard to type out, you know you swaure teh coefficients infrom of the i and j then add together and root them)
that solves u to be +-2 but since it is moving in the positive direction intiall, we take +2
so v =(3/2+1/2 cos(2t))i+(2t²-2t+2)j
sorry about the confusion

EDIT: tried to make it clearer, changed a to u to avoid confusion with a (acceleration)

[99.96]

Ahk makes sense now.
cheers.

[david10d]

Hijacking this thread.


How do you find the time that the particles are the closest?

[luffy]

Hijacking this thread.


How do you find the time that the particles are the closest?

Never seen a question like this (or at least not for a long time). But I assume you would find the vector resolute (as they will be perpendicular at this point and hence, closest to each other) and then using the position vector, sub it in for the i and j components to solve for t.

[tony3272]

Haven't read the question that this relates to, but i'm assuming that you could also work out the distance between two points and derive/solve for a minimum.

[david10d]

It's in the Essentials book btw haha

tony's right :D

[luffy]

It's in the Essentials book btw haha

tony's right :D

Both methods should logically work, unless you can think of a counter-argument to mine. If so, let me know because I would definitely be interested. In order to check, its probably best that you actually do the question and see if the answer obtained from my method is correct. Haha.

[tony3272]

My rule of thumb is: If it's a vector question, go as far out of your way as humanly possible to find a non-vector method to solve it.

[david10d]

Hijacking this thread.


How do you find the time that the particles are the closest?

Never seen a question like this (or at least not for a long time). But I assume you would find the vector resolute (as they will be perpendicular at this point and hence, closest to each other) and then using the position vector, sub it in for the i and j components to solve for t.

Will try.