# Ship problem,related time change

#### wolfsprint

##### New member
A ship sailing west at 9am with velocity 20km/h. after one hour another ship sailed from the same port with velocity 40km/h at 60 north of west . find the rate of increase of the distance between the two ships at 11am

Can someone please show me how to solve this? i have no idea

Last edited:

#### CaptainBlack

##### Well-known member
Re: Ship problem , related time change

A ship sailing west at 9am with velocity 20km/h. after one hour another ship sailed from the same port with velocity 40km/h at 60 north of west . find the rate of increase of the distance between the two ships at 11am

Can someone please show me how to solve this? i have no idea
Write the equations of the position of the two ships as a function of time. Now write the square of the distance between them as a function of time, and differentiate...

CB

#### soroban

##### Well-known member
Hello, wolfsprint!

Did you make a sketch?

A ship sailing west at 9 am with velocity 20 km/h.
After one hour, another ship sailed from the same port with velocity 40 km/h at 60o north of west .
Find the rate of increase of the distance between the two ships at 11 am.
Code:
                 C
o
.   *
x  .       * 40t
.           *
.               *
.               60[SUP]o  [/SUP]*
B o  *  *  o  *  *  *  *  o P
20t   A     20
Ship #1 leaves port $P$ at 9 am at 20 km/hr.
In one hour, it reaches point $A:\A = 20.$
In the next $t$ hours, it reaches point $B:\:AB = 20t.$

In the same $t$ hours, ship #2 leaves point $P$ at 40 km/hr.
. . and reaches point $C:\C = 40t.$

Let $x = BC.$

Law of Cosines:

. . $x^2 \;=\;(40t)^2 + (20t+20)^2 - 2(40t)(20t+20)\cos60^o$

. . $x^2 \;=\;1600t^2 + 400t^2 + 800t + 400 -800t^2 - 800t$

. . $x^2 \;=\;1200t^2 + 400$

Differentiate with respect to time:

. . $2x\dfrac{dx}{dt} \;=\;2400t \quad\Rightarrow\quad \dfrac{dx}{dt} \;=\; \dfrac{1200t}{x}$

When $t=1,\:x = 40:$

. . $\dfrac{dx}{dt} \;=\;\dfrac{1200(1)}{40} \;=\;30$

The ships are separating at 30 km/hr.