Question

8) 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:

Correct answer:  Vsub = 5 mph

Step-by-step explanation:

Given:

t₁ = 2 h

Va = 25 mph

t₂ = 9 h

d = 280 mile

distance traveled by aircraft carrier  for time t₂ = 9 h

d₃ = Va · t₂ = 25 · 9 = 225 miles

the total distance traveled by the submarine

d₁ + d₂ = 280 - 225 = 55 miles

d₁ + d₂ = Vsub · t₁ + Vsub · t₂ = Vsub · ( t₁ + t₂ ) = 55

Vsub · ( t₁ + t₂ ) = 55  ⇒  Vsub = 55 / 2 + 9 = 55 / 11 = 5 mph

Vsub = 5 mph

God is with you!!!