# Distance application

• May 8th 2008, 06:20 PM
lemontea
Distance application
Ship A leaves port sailing north at a speed of 25km/h. A half hour later, ship B leaves the same port sailing east at a speed of 20km/hr. Let t (in hours) be the time ship B has been at sea.
a) find an expression in terms of t giving the distance between the two ships

b) use the expression obtained in part a to find the distance between the two ships 2h after ship A has left port.
• May 8th 2008, 07:27 PM
Chris L T521
Quote:

Originally Posted by lemontea
Ship A leaves port sailing north at a speed of 25km/h. A half hour later, ship B leaves the same port sailing east at a speed of 20km/hr. Let t (in hours) be the time ship B has been at sea.
a) find an expression in terms of t giving the distance between the two ships

b) use the expression obtained in part a to find the distance between the two ships 2h after ship A has left port.

If ship B is traveling due east at a speed of $20\frac{km}{hr}$, then the total distance traveled after $t$ hours is $20t$ km. However, Ship A leaves a half hour before, traveling at a speed of $20\frac{km}{hr}$. The distance traveled by Ship A would have to be $25(t+.5)$ km, since Ship A has been traveling a half hour longer that Ship B. When drawn, these two distances can be represented as legs of a triangle, with the hypotenuse being the distance between the ships. Thus, using the Pythagorean Theorem, we find that distance to be:

$D^2=(25(t+.5))^2+(20t)^2$
$\longrightarrow D=\sqrt{(25(t+.5))^2+(20t)^2}$

Plug in 2 for t to find how far apart they are.

Hope my explanation was clear!