# Math Help - ship problem

1. ## ship problem

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. At t=0, they are both 10 kms from their intersection.

Think Pythagoras. Let D=square of distance between ships.

$D(t)=(10-30t)^{2}+(10-40t)^{2}$

$D'(t)=2(10-30t)(-30)+2(10=40t)(-40)$

$D'(t)=5000t-1400$

$5000t-1400=0$

$t=\frac{7}{25}$

3. Originally Posted by galactus
At t=0, they are both 10 kms from their intersection.

Think Pythagoras. Let D=square of distance between ships.

$D(t)=(10-30t)^{2}+(10-40t)^{2}$

$D'(t)=2(10-30t)(-30)+2(10=40t)(-40)$

$D'(t)=5000t-1400$

$5000t-1400=0$

$t=\frac{7}{25}$
thankz...but i was wondering if i did the first part right that i showed?(shown below) because im not too sure if i was meant to include negative signs with any of the answers? can u please just check that for me thankz

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

4. Originally Posted by dopi
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.
Hello, dopi,

with your problem you use automatically a coordinate system: The pos. x-axis points to East and the pos. y-axis points to North.

At the time t = 0 the ship P is at P(-10, 0) and the ship Q is at Q(0, 10). The movement of both ships is described by a straight line. The speed is described by a vector:
$\overrightarrow{v_P}=(30, 0)$
$\overrightarrow{v_Q}=(0, -40)$

Thus P is moving along the line:
$\overrightarrow{x_P}=(-10, 0)+t*(30, 0)$
and Q is moving along the line:
$\overrightarrow{x_Q}=(0, 10)+t*(0, -40)$

The distance between the ships is: $\vec{d}=\overrightarrow{x_P}-\overrightarrow{x_Q}$

Originally Posted by dopi
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
I don't understand why you want to calculate the relative coordinates. It isn't necessary with your problem.

Originally Posted by dopi
but im struggling to find the time at which P and Q are closest to each other...is there anyone that can help??? thankz
This was already done by galactus.

Greetings

EB

5. Hello, dopi!

You seem to be using a coordinate system and vectors
. . and I don't know what you plan to do with them.

galactus gave you the solution . . . so why struggle with your approach?

I'll make some diagrams to accompany his excellent solution.

Originally Posted by galactus

At t=0, they are both 10 kms from their intersection.
Code:
                        * Q
|
|
| 10
|
|
|
P * - - - - - - - - o
10

$t$ hours later, $P$ has moved $30t$ km east to point $A$
. . and $Q$ has moved $40t$ km south to point $B.$
Code:
                        *
|
| 40t
|
o B
|
| 10-40t
* - - - o - - - - *
30t  A  10-30t
And we want to minimize the distance $\overline{AB}.$

Originally Posted by galactus

Think Pythagoras. Let D = square of distance between ships.

$D(t) \:= \10-30t)^{2} + (10-40t)^{2}" alt="D(t) \:= \10-30t)^{2} + (10-40t)^{2}" />

$D'(t) \:= \:2(10-30t)(-30) + 2(10=40t)(-40)$

$D'(t) \:= \:5000t - 1400$

$5000t -1400\:=\:0$

$t \:= \:\frac{7}{25}$