Ship problem

1. Jul 15, 2006

dopey9

A ship P is travelling due East at 30 km/h and a Ship Q is travelling due South at 40 km/h.
Both ships keep constant speed and course. At t=0 they are each 10 km from the point of intersection of their courses and moving towards the point.

Iv found the co-ordinates of Q relative to P at t=0
---->X=(10,10)km

iv also found the velocity of Q relative to P----> V= V[Q] -V[P]
-----.V= (0,40) - V(30,0)
----- V= (-30,40)km/h

but im struggling to find the time at which P and Q are closest to each other...is there anyone that can help??? thankz

2. Jul 15, 2006

neutrino

Find an expression for the distance between the ships as a function of time and minimise that with respect to time.

3. Jul 15, 2006

Staff: Mentor

OK.

V[Q] = (0, -40)

The trajectory of Q with respect to P is just a straight line. Write the equation of that line and then find the point on that line closest to point P. (There are several ways to do that.)

4. Jul 15, 2006

dopey9

can i just confirm that are these parts right that i answered before...because i wasnt too show whether a negitive sign is requied with the 10 coz of the direction?:
"Iv found the co-ordinates of Q relative to P at t=0
---->X=(10,10)km

iv also found the velocity of Q relative to P----> V= V[Q] -V[P]
-----.V= (0,40) - V(30,0)
----- V= (-30,40)km/h "

5. Jul 15, 2006

Mindscrape

Maybe I am misunderstanding your use of minimize, but there is no need for calculus here.

For the topic creator, you are pretty close and really just need to resolve your new vectors in a single linear equation. Then once you have your distance and velocity you know that a change in distance over a change in velocity is time.

6. Jul 15, 2006

Staff: Mentor

Right.

Wrong. (As I pointed out earlier.)