Question

It is a question about space vector.
At the time t=0, a particle start at point (1,2,3), it along a straight line to point (4,1,4). The speed is 2 at the point (1,2,3) and it has a constant acceleration which is (3i-j+k). Find the location of particle r(t) at time t.

Answer 1

s = ut + 0.5 a t2?

Answer 2

u got everythin...............so substitute

Answer 3

s = s2-s1

Answer 4

well i translate it from a chinese book...maybe it's my reason to let you hard understand it
the answer it gave is far than yours :(

Answer 5

oh.... :P
but this is the way :P
u got speed.....................u got to put velocity vector....
u kno a unit vector.....parallel to that

Answer 6

i was thinking of that approach too^
and think the answer would be (2/11)(3,-1,1)t + 0.5t(-3,-1,1)

Answer 7

um i think u shud add (1,2,3)

Answer 8

oh yah u're right
(1,2,3)+(2/11)(3,-1,1)t + 0.5t(-3,-1,1)

Answer 9

@AndrewNJ dude is that it?

Answer 10

the answer is (0.5t^2+2t/(Square root of 11))(3i-j+k)+(i+2j+3k)
terrible internet, i spent 20 min refreshing this page!

Answer 11

@Tushara
@A.Avinash_Goutham
upper is the answer

Answer 12

my bad it shud hav been sqrt(11)

Answer 13

ok so dat is the same as (i+2j+3k)+(0.5t^2)(-3i-j+k)+(2/sqrt11)(-3i-j+k)

do you understand how to get to dat answer?

Answer 14

i only feel puzzled about where to get 2t/sqrt11

Answer 15

other part is ok

Answer 16

well u're finding the unit vector in da direction of the velocity:

(4,1,4)-(1,2,3)=(3,-1,1)
sqrt(3^2+(-1)^2+1^2) = sqrt(11)
hence unit vector= (1/sqrt11)(3,-1,1) which is direction vector
and you have to multiply it by the speed which is 2

therefore you get (2/sqrt11)(3,-1,1)

Answer 17

and since you have to multiply velocity by time, you get (2t/sqrt11)(3,-1,1)

Answer 18

ya... got it
thanks!!