G&T Can u plz help me with a problem relating with 3Dvec

[Guest]

hello, this is my first time using this so can u please help me

Q:A US submarine is stationary at position (4.5i-3j-0.4k)km. It locates an enemy submarine at position (4i-5j-0.3k)km moving at velocity (3.6i+2.4j-1k)m/s. At the instant the US submarine fires a torpedo which collides with the enemy 8 seconds later. Determine the velocity the torpedo was fired at.

Thank you

 

[soroban]

Re: G&T Can u plz help me with a problem relating with 3

Hello, barule22!

A US submarine is stationary at position (4.5i - 3j - 0.4k)km.
It locates an enemy submarine at position (4i - 5j - 0.3k)km
moving at velocity (3.6i + 2.4j - 1k)m/s.
At that instant the US submarine fires a torpedo which collides with the enemy 8 seconds later.
Determine the velocity of the torpedo.

Change all units to meters.

The enemy sub is at: .\(\displaystyle E(4000,\ -5000,\ -300)\)

. . Its velocity is: .\(\displaystyle \vec{v} \ = \ \langle3.6,\ 2.4,\ -1\rangle\)

. . Its position at any time \(\displaystyle t\) is: .\(\displaystyle \begin{array}{ccc}x \ = \ 4000 + 3.6t\quad\\ \;y \ = \ -5000 + 2.4t \\ z \ = \ -300 - t\qquad\quad\end{array}\)


The US sub is at: .\(\displaystyle U(4500,\ -3000,\ -400)\)

. . The torpedo's velocity is: .\(\displaystyle \vec{v} \ = \ \langle a,b,c\rangle\)

. . Its position any time \(\displaystyle t\) is: .\(\displaystyle \begin{array}{ccc}x \ = \ 4500 + at\\ \quad y \ = \ -3000 + bt\\ z \ = \ -400 + ct\end{array}\)


At \(\displaystyle t = 8\), the positions are equal.

We have: . \(\displaystyle \begin{array}{ccc}4500 + 8a \ = \ 4000 + 3.6(8) \\ \;-3000 + 8b \ = \ -5000 + 2.4(8) \\ -400 + 8c \ = \ -300 - 8\qquad\qquad\end{array}\)

Therefore, the velocity of the torpedo is:

. . . \(\displaystyle \vec{v} \:= \:\langle a,b,c\rangle \:= \:\langle -58.9,\ -247.6,\ 11.5\rangle \ =\) .-58.9i - 247.6j + 11.5k