An aircraft carrier made a trip to dry dock and back. The trip there took five hours and the trip back took nine hours. It averaged 8 mph faster on the trip the

Question

An aircraft carrier made a trip to dry dock and back. The trip there took five hours and the trip back took nine hours. It averaged 8 mph faster on the trip there than on the return trip. What was the aircraft carrier's average speed on the outbound trip?

Answer 1

An aircraft carrier made a trip to dry dock and back. The trip there took five hours and the trip back took nine hours. It averaged 8 mph faster on the trip there than on the return trip. What was the aircraft carrier's average speed on the outbound trip?

 

Distance = Rate*Time  ...  d = rt

Let r = outgoing rate:

 

There:  d = 5(r+8)

Back:  d = 9r

 

Since the distance is the same set the equations equal to each other:

9r = 5(r+8)

9r = 5r + 40

 

Finish solving for r to determine the outgoing rate of speed