HELP9) An aircraft carrier left Hawaii and traveledtoward dry dock at an average speed of 15mph. A submarine left two hours later andtraveled in the opposite direction with anaverage speed of 15 mph. Find thenumber of hours the submarine needs totravel before the vessels are 300 mi. apart.

Answer:The Time taken by submarines to travel before the vessel is 11 hours Step-by-step explanation:Given as :The average speed of aircraft carrier = 15 mphThe average speed of submarine = 15 mphThe distance between them = 300 mileThe time taken by aircraft = T hourThe time taken by submarine = T + 2  hourNow, Speed = [tex]\dfrac{\textrm Distance}{\textrm Time}[/tex]So, Distance =  Speed × TimeNow, 300 miles = 15 × T + 15 × ( T + 2)Or, 300 = 15 T + 15 T + 30or, 300 - 30 = 30 TOr, 270 = 30 T∴ T = [tex]\frac{270}{30}[/tex]I,e T = 9 hoursSo, T + 2 = 9 + 2 = 11 hoursHence The Time taken by submarines to travel before the vessel is 11 hours Answer