# ship problem

• Jul 15th 2006, 06:36 AM
dopi
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
• Jul 15th 2006, 07:17 AM
galactus
At t=0, they are both 10 kms from their intersection.

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

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

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

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

$\displaystyle 5000t-1400=0$

$\displaystyle t=\frac{7}{25}$
• Jul 15th 2006, 07:46 AM
dopi
Quote:

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

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

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

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

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

$\displaystyle 5000t-1400=0$

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

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
• Jul 16th 2006, 08:50 PM
earboth
Quote:

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:
$\displaystyle \overrightarrow{v_P}=(30, 0)$
$\displaystyle \overrightarrow{v_Q}=(0, -40)$

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

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

Quote:

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.

Quote:

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
• Jul 17th 2006, 04:32 AM
Soroban
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.

Quote:

Originally Posted by galactus

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

Code:

                        * Q                         |                         |                         | 10                         |                         |                         |     P * - - - - - - - - o             10

$\displaystyle t$ hours later, $\displaystyle P$ has moved $\displaystyle 30t$ km east to point $\displaystyle A$
. . and $\displaystyle Q$ has moved $\displaystyle 40t$ km south to point $\displaystyle B.$
Code:

                        *                         |                         | 40t                         |                         o B                         |                         | 10-40t       * - - - o - - - - *         30t  A  10-30t
And we want to minimize the distance $\displaystyle \overline{AB}.$

Quote:

Originally Posted by galactus

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

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

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

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

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

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