Question

Two fishing boats depart a harbor at the same time, one traveling east, the other south. the eastbound boat travels at a speed 3 mi/h faster than the southbound boat. after 3 h the boats are 45 mi apart. find the speed of the southbound boat.

Answer 1

1) Let's call $V_S$ the speed of the southbound boat, and $V_E=V_s+3~mph$ the speed of the eastbound boat, which is 3 mph faster than the southbound boat. We can write the law of motion for the two boats:
$S_E(t)=V_E t=(V_S+3)t$
$S_S(t)=V_S t$


2) After a time $t=3~h$, the two boats are $45~mi$ apart. Using the laws of motion written at step 1, we can write the distance the two boats covered:
$S_E(3~h)=3(V_S+3)=3V_S+9$
$S_S(3~h)=3V_S$
The two boats travelled in perpendicular directions. Therefore, we can imagine the distance between them (45 mi) being the hypotenuse of a triangle, of which $S_E$ and $S_S$ are the two sides. Therefore, we can use Pythagorean theorem and write:
$45= \sqrt{(3V_S)^2+(3V_S+9)^2}$
Solving this, we find two solutions. Discarding the negative solution, we have $V_S=9~mph$, which is the speed of the southbound boat.