Find the speed of the riverboat in still water and the speed of the current.

Question

A Mississippi riverboat travels 30 miles downstream in three hours and then makes the return trip upstream in five hours.

 

Rate x time = distance

 

please show the solution and steps! Thank you!

Answer 1

Hi Berryl.

 

If you let the current of the river be C, then when the

boat is travelling with the current, its speed is its own speed + the speed of the current.

When the boat is travelling against the current, its speed is its own speed - the speed of the current.

 

So we need a variable for the speed of the boat in still water, how about "s".

 

Rate going upstream then is "s - C"

Rate going downstream is "s + C"

 

The distance travelled is the same, so

 

Dist upstream = Dist downstream = 30 miles.

 

Rate * time = distance so 

 

UPSTREAM:  (s-C)(5)=30

DOWNSTREAM:  (s+C)(3)=30

 

Distribute and get

 

5s - 5C = 30

3s + 3C = 30

 

Solve one of these then substitute it into the other.

 

Solve the top equations, we get

 

s - C = 6   or that s = 6 + C

 

Now put that expression for "s" into the bottom equation

 

3s + 3C = 30

3(6+C) + 3C = 30

 

Distribute

18 + 3C + 3C = 30

 

18 + 6C = 30

 

Subtract 18 from both sides

 

6C = 12

 

C = 2   The current is moving 2 miles per hour.

 

Know that s = 6 + C, s = 6 + 2 = 8

 

The boat in still water is moving 8 miles per hour.