# Ships when ill they meet?

• May 27th 2009, 02:20 AM
Nuke
Ships when ill they meet?
Startng point 0,0
1 ship has the time T=0 posision 0.5 miles east of a point and 0,4north
the boat is sailing with the speed 0.8/hour with the dirction 55 degrees

Another boat have this speed

(xt),y(t))=(-2,236t+5,5;4.472t+1,o)

-What is the speed of th lastboat and in wich direction( degrees)
-And will h boats ever meet? if so in wich point

Sry 4 my english
• May 27th 2009, 04:55 AM
earboth
Quote:

Originally Posted by Nuke
Startng point 0,0
1 ship has the time T=0 posision 0.5 miles east of a point and 0,4north
the boat is sailing with the speed 0.8/hour with the dirction 55 degrees

Another boat have this speed

(xt),y(t))=(-2,236t+5,5;4.472t+1,o)

-What is the speed of th lastboat and in wich direction( degrees)
-And will h boats ever meet? if so in wich point

Sry 4 my english

Make a rough sketch!

Construct a coordinate system with the y-axis pointing North and the x-axis pointing East. You must transform the courses into angles with respect to the x-axis!

The way of the ship is a straight line:

$\vec s = (0.5, 0.4)+t \cdot (0.8 \cdot \cos(35^\circ), 0.8 \cdot \sin(35^\circ))$

Expand the values in the direction vector and use rounded decimals:

$\vec s = (0.5, 0.4)+t \cdot (0.6553, 0.4589)$

The way of the boat (the 2nd vessel!) is:

$\vec b = (5.5, 1)+r\cdot(-2.236, 4.472)$

The norm of the direction vector represents the speed of the boat:

$v=\sqrt{(-2.236)^2+ ( 4.472)^2}=5.0\ kn$

The angle with respect to the x-axis is:

$\tan(\phi)=\dfrac{4.472}{-2.236} = -2~\implies~\phi\approx -63.435^\circ$

Since the direction vector is pointing into the 4th quadrant the boat is heading $270^\circ+63.435^\circ = 333.435^\circ$

To answer the second question calculate the point of intersection between the two lines:

$\left|\begin{array}{rcl}0.5+0.6553t&=&5.5-2.236r \\ 0.4+0.4589 t&=& 1+4.472r\end{array}\right.$

I've got t = 5.99 and r = 0.48

And what does that mean?