Question

A submarine left Hawaii two hours before an aircraft carrier. The vessels traveled in opposite directions. The aircraft carrier traveled at 25 mph for nine hours. After this time the vessels were 280 mi. apart. Find the submarine's speed.

Answer 1

Answer:

V  =  5 m/h

Step-by-step explanation:

An aircraft carrier, traveled 9 hours at speed of 25 mph, that means the aircraft position from Hawaii (the start point of the journey) is:

25 m/h *  9 h  = 225   miles far away from hawaii

When the aircraft has traveled 9 hours, the submarine has traveled 11 hours ( since its departure was 2 hours earlier ). They both traveled in opposite directions.

Then distance between the two vessels  =  d

d =  280  miles      

280   =   225 miles (from the aircraft to hawaii )  +  traveled distance for the submarine

And

Traveled distance by submarine  d(s)   =  V* 11   miles/hour. where V is speed of submarine

Then

280    =  225   +  11*V    ⇒  11*V =  280-225    ⇒    11*V =  55/11 mph

V  =  5 m/h