ship problem

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

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

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

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


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


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...

You are approaching it correctly, this is best solved by relative motion. Write the distance function of either ship with respect to the other, then minimize (calculus) that distance with respect to time.

Q

Q'


P

This is the graph, Q' is the new RELATIVE position of Q at some time delta T. The distance you are looking for is PQ', minimize that. PQ=14.4km
QQ'= relative veloicty*deltaTime

You can use the direction of the relative velocity to find the angle PQQ'. Write the equation of PQ (magnitude) using the law of cosines, then minimize PQ with respect to time.


There is a neat qraphical solution to this, used by all navigators. Leaving P stationary, draw the line representing the direction of relative velocity of Q with respect to P. That line represents the relative motion of Q. Draw a perpendicular from P to the relative motion line, the measure of that perpendicular is the CPA (closest point of approach). Normally this graph is done directly with a marker on a radar screen, as it is a built in relative motion plotter.


rav, you might want to think of the ship moving east as x(t)= (10-30t), the one moving south as y(t)=(10-40t); these are the conventional directions anyway.
You then have D(t) = root(x(t)^2 + y(t)^2). Find D' and solve for t = 0. You should get a "nice" answer, if I did this correctly.
I used the neg. signs above because they were approaching each other. If they had been moving apart we would have 10 + 30t and 10 + 40t. Check if I did this right.


yea that's exactly what i did before you posted it..so im guessing its the right answer...i got 7/25 thankz


Guessing?? hmm... what math class taught that? Just funnin' with you rav.
BTW, this problem is called the 2-ships one in related rates. You will probably see it again. A variation of this is 2 cars approaching/leaving an intersection.

After I made my last post I remembered how it might be set-up. Since we want a function of 1-var., we use time. You might think of each ship as being on it's own t-axis, with the 2-axis having 0 in common. At t = 0, we want both ships to be 10km from 0 and moving to the origin. Thus we have either:
x(t) = 30t - 10 or 10 - 30t, and
y(t) = 40t - 10 or 10 - 40t
Both should work. This is how I remember seeing it done in class.
I did work the problem and I think 7/25 is what I had too.

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