Vectors and ships question

Question

Port A is defined to be the origin of a set of coordinate axes and port B is located at the point $(70, 30)$, where distances are measured in kilometres. A ship $S_1$ sails from port A at 10:00 in a straight line such that its position $t$ hours after 10:00 is given by $r=t(10, 20)$ (I don't know how to make column vectors sorry, but that 20 should be under the 10)

A speedboat S2 is capable of three times the speed of S1 and is to meet S1 by travelling the shortest possible distance. What is the latest time that S2 can leave port B?

My working so far:

$r_1=t(10, 20)$

$r_2=(70, 30)+(t-k)(x, y)$ as the second ship sets of k minutes later. How do I know what the direction vectors x and y are though? Maybe I'm not visualising the problem correctly...

Answer 1

You want to minimise the distance that ship $S_2$ has to travel. Hence you want to set up an expression for the distance between ship $S_1$ and port $B$, and mnimise this (calculus).

Next, you can easily calculate how late ship $S_2$ can leave port $B$ to reach this point at the same time as ship $S_1$.

Answer 2

If you draw a picture, you'll see that the shortest possible distance that the speedboat can travel is if it goes perpendicularly to the path of the ship. If the ship's direction vector is $\mathbf v=(10,20)$ then we need a direction vector $\mathbf u=(x,y)$ such that $\mathbf{u.v}=0$, in other words $10x+20y=0$ or $x=-1/2 y$. If we define $\mathbf v$ to have the length three times $\mathbf v$ it also takes the speedboat's triple speed into account, so $(-60,30)$ is a good choice. Then if the speedboat has travelled for $q$ hours its position is $(70-60q,30+30q)$.

Can you do the rest yourself? You need to work out the point of intersection, solve for $q$ and then get the difference between $t$ and $q$.