Question

A motorboat takes 3 hours to travel 108 mi going upstream. the return trip takes 2 hours going downstream. What is the rate of the boat in still water and what is the rate of the current?

1.Rate of the boat in still water is_________.
2.Rte of the current is__________.

Answer 1

Answer: The speed of the motorboat in still water is 45mph

The current rate is 9mph.

Explanation:

Let u be the motorboat speed in still water and v be the current rate.

The effective speed going upstream is

Distance÷Time = 108÷3 = 36mph

It is the DIFFERENCE of the motorboat speed in still water and the rate of the current. It gives you your first equation

u - v = 36. (1)

The effective speed going downstream is

Distance÷Time = 108÷2 = 54mph

It is the SUM of the motorboat speed in still water and the rate of the current. It gives you your second equation

u + v = 54. (2)

Thus you have this system of two equations in 2 unknowns

u - v = 36, (1) and

u + v = 54. (2)

Add the two equations. You will get

2u = 36 + 54 = 90 ====> u = 90÷2 = 45mph

So, you just found the speed of the motorboat in still water. It is 45mph

Then from the equation (2) you get v = 54 = 45 = 9mph is the current rate.

Answer. The speed of the motorboat in still water is 45mph

The current rate is 9mph