1. ## Vector Intersection

Hi, I have a mechanics question here I can't quite get.

A destroyer sights a ship at a point with position vector 600(3i + j)m relative to it and moving with velocity 5j m/s. The destroyer alters course so that it moves with speed v m/s in the direction of the vector 4i + 3j. Find v so that the destroyer intercepts the ship and the time to the interception.

Any help would be greatly appreciated.
Thanks

2. Hello steve989
Originally Posted by steve989
Hi, I have a mechanics question here I can't quite get.

A destroyer sights a ship at a point with position vector 600(3i + j)m relative to it and moving with velocity 5j m/s. The destroyer alters course so that it moves with speed v m/s in the direction of the vector 4i + 3j. Find v so that the destroyer intercepts the ship and the time to the interception.

Any help would be greatly appreciated.
Thanks
Denote the velocity of the destroyer by $\vec{v_D}$ and the velocity of the ship by $\vec{v_S}$. Then:
$\vec{v_D}=\frac{v}{5}\Big(4\vec i + 3 \vec j\Big)$, since this vector has magnitude $v$.
and
$\vec{v_S} = 5\vec j$
So the velocity of the ship relative to the destroyer is
$\vec{v_S}-\vec{v_D} = -\frac{4v}{5}\vec i +\Big(5-\frac{3v}{5}\Big)\vec j$
Therefore after time $t$, the position of the ship relative to the destroyer is:
$600(3\vec i + \vec j) -\frac{4vt}{5}\vec i +\Big(5-\frac{3v}{5}\Big)t\vec j$
$=0\vec i + 0\vec j$ when the destroyer intercepts the ship
Solving for $v$ and $t$ gives:
$t = 390$ and $v = \frac{75}{13}$
But check my working!

3. Hello, steve989!

A destroyer sights a ship at a point with position vector $1800i + 600j$ relative to it
and moving with velocity $5j$ m/s.

The destroyer alters course so that it moves with speed $v$ m/s
in the direction of the vector $4i+3j$

Find $v$ so that the destroyer intercepts the ship and the time to the interception.
Code:
      |
|                     o P
|                   o |
|                 o   | 5t
|               o     |
|             o       |
|           o         * S
|         o           |
|       *             |
|  v  * |             | 600
|   *   |3            |
| * θ   |             |
D* - - - + - - - - - - *
:   4   :
: - - - -1800 - - - - :

The destroyer is at: $D(0,0)$

The ship is at: $S(1800, 600)$

The ship is moving north at 5 m/s.
In the next $t$ seconds, it moves $5t$ m to point $P.$

. . Its position vector is: . $\vec S \:=\:\langle 1800,\:600+5t\rangle$

The destroyer heads in direction $4i+3j$
Let $\theta$ represent its direction, where: $\cos\theta \,=\,\tfrac{4}{5},\;\sin\theta\,=\,\tfrac{3}{5}$

. . Its position vector is: . $\vec D \;=\;\langle (v\cos\theta)t,\:(v\sin\theta)t\rangle \;=\;\left\langle\tfrac{4}{5}vt,\:\tfrac{3}{5}vt\r ight\rangle$

The destroyer intercepts the ship at point $P.$
This happens when: $\vec D \,=\,\vec S$

We have: . $\begin{Bmatrix}\frac{4}{5}vt &=& 1800 & [1] \\ \\[-3mm] \frac{3}{5}vt &=& 600 + 5t & [2] \end{Bmatrix}$

Divide [2] by [1]: . $\frac{600+5t}{1800} \:=\:\frac{\frac{3}{5}vt}{\frac{4}{5}vt} \quad\Rightarrow\quad\frac{120+t}{360} \:=\:\frac{3}{4}$

. . $480 + 4t \:=\:1080 \quad\Rightarrow\quad 4t \:=\:600 \quad\Rightarrow\quad t \:=\:150$

Therefore, the destroyer intercepts the ship in 150 seconds.

Substitute into [1]: . $\tfrac{4}{5}v(150) \:=\:1800 \quad\Rightarrow\quad v \:=\:15$

Therefore, the destroyer must travel at 15 m/s.

4. Thanks for both of your replies! I've gone though them as well as I can.

Grandad - I tried your way and think I see how you're thinking there, I got a different formula for one stage though.

Therefore after time , the position of the ship relative to the destroyer is:
when the destroyer intercepts the ship
For that bit; wouldn't it be $600(3i + j) + (4/5)vti - (5 - (3/5)v)tj$ ? The signs switch as they are on different sides of the equation I assumed, but I may of miss-understood since I got a completely different answer anyway. Sorry about the layout too, I'm useless with this maths typing.

Soroban - I do like that approach and the answers seem very reasonable, I missed the fact that the ship was moving directly north. I do have one question though for both you and grandad; why does the velocity vector of the destroyer end up with $v/5$ as the scalar? I thought it was just $v$, I think I'm going to get a simple answer to that question though.

5. Hello everyone

I agree with Soroban's answer: I got a sign wrong in my working. My equation for the relative velocity was correct, though. The principle is:
The velocity of A relative to B is the velocity of A minus the velocity of B.
So, as I said:
$\vec{v_S}-\vec{v_D} = -\frac{4v}{5}\vec i +\Big(5-\frac{3v}{5}\Big)\vec j$
not as you suggest:
Originally Posted by steve989
...
For that bit; wouldn't it be $600(3i + j) + (4/5)vti - (5 - (3/5)v)tj$ ? The signs switch as they are on different sides of the equation I assumed, but I may of miss-understood since I got a completely different answer anyway. Sorry about the layout too, I'm useless with this maths typing.

However, I then got a sign wrong when I equated the $\vec j$ component of the displacement to zero. The equations should have been:
$\vec i$ component: $1800 - \frac{4vt}{5} = 0$

$\vec j$ component: $600 + 5t -\frac{3vt}{5}=0$
which, of course, are the same as Soroban's equations [1] and [2].
Originally Posted by steve989
...I do have one question though for both you and grandad; why does the velocity vector of the destroyer end up with $v/5$ as the scalar? I thought it was just $v$, I think I'm going to get a simple answer to that question though.

Because the magnitude of $4\vec i + 3\vec j$ is $5$. So $v(4\vec i + 3\vec j)$ will have magnitude $5v$.

6. Ahh I see, all makes perfect sense now, I knew there had to be simple reasoning behind that. Thanks a lot to you both.