Question

The position vector of a ship (MB) from its starting position at a port is given by (x,y)=(5,20)+ t (12,16). Distances are in kilometres and speeds are in km/h. t is time after 00 hour.

a) Another ship (LW) is at sea in a location (41,68) relative to the same port. LW has stopped for some reason. Show that if LW does not start to move, the two ships will collide. Find the TIME of the potential collision.

Answer 1

did you have any ideas on this so far?

Answer 2

(x,y)=(41,68)+ t (12,16)
im not really sure what im doing

Answer 3

i believe you want to show that the position vector (x,y) = (5, 20) + t (12, 16) will at some point equal (41, 68). that would mean they have intersected, right?

Answer 4

yes , i need to find t when they intersect

Answer 5

so (5, 20) + t (22, 16) = (41, 68)
do you know how to proceed from here?

Answer 6

recall that the SUM of both of those vectors
that is, the resultant vector, due to the "t" value, will end up at (41, 68)

Answer 7

so would t=3hours

Answer 8

(5, 20) + 3 (12, 16) = (41, 68)
i believe this works out. you can check your work here to make sure. :)

Answer 9

\(\bf <5,20>+t<12,16>\implies <5,20>+<t12,t16>\implies <\square ,\square >
\\ \quad \\
\textit{and eventually due to "t"}\quad <\square ,\square > = <41,68>\)

Answer 10

\(\large { <5,20>+t<12,16>\implies <5,20>+<t12,t16>
\\ \quad \\
<5+12t, 20+16t>=<41,68>\implies
\begin{cases}
5+12t=41
\\ \quad \\
20+16t=68
\end{cases} }\)

Answer 11

ohhh thankss

Answer 12

yw