1. Wind and Current Word Problems
Math 8H
Wind and Current
Word Problems
Algebra Glencoe McGraw-Hill JoAnn Evans

2. Wind and current problems are just a variation of a type of problem you’ve worked with many times this year: =
Distance
Rate
Time

3. When traveling with the current, the speed of a boat is increased by the speed of the current.
An similar example is when you walk on a moving walkway at the airport.
When traveling against the current, the speed of a boat is reduced by the speed of the current.

4. Mr. Lepire went 27 miles downstream in a boat in three hours
Mr. Lepire went 27 miles downstream in a boat in three hours. The return trip upstream took three times as long (nine hours). Find the rate of the boat in still water and the rate of the current.
Let b = boat’s speed
Let c = current’s speed
Going downstream, the boat is traveling with the current. The boat’s speed will be:
b + c
Going upstream, the boat is traveling against the current. The boat’s speed
will be:
b - c

5. ÷ all terms by 3 ÷ all terms by 9 System of equations:
RATE · TIME = DISTANCE
with the current:
against the current:
÷ all terms by 3
÷ all terms by 9
Add the equations together.
The speed of the boat is 6 mph. Find the speed of the current by substitution.
Boat speed is 6 mph.
Current speed is 3 mph.

6. Rowboat speed was 11 km/h and current speed was 1 km/h.
Ms. Wasson took 5 hours to row 60 km downstream and 4 hours to row 40 km upstream. Find the rate of the boat in still water and the rate of the current.
Let b = rate of boat in still water
Let c = rate of current
rate · time = distance
with current: (b + c)5 = 60
against current: (b - c)4 = 40
b + c = 12 Divide by 5.
b – c = 10 Divide by 4.
2b = 22
b = 11
Rowboat speed was 11 km/h and current speed was 1 km/h.

7. Express the times in terms of hours.
Mrs. Ashmore can swim 4 km with the current in 24 minutes. The same distance would take her 40 minutes against the current. Find the rate of the current and Mrs. Ashmore’s speed.
Let s = rate swimming in still water
Let c = rate of current
Express the times in terms of hours.
rate · time = distance
with current:
against current:
Simplify:
Simplify:

8. Mrs. Ashmore’s speed is 8 km/h and current speed was 2 km/h.

9. The rate you are traveling when you go with the wind current is a combination of the speed possible by the plane’s engine and the speed of the wind current.
When you travel against the wind, the speed possible by the plane’s engine is reduced by the wind current, which is going in the opposite direction.

10. Plane speed was 960 km/h and wind speed was 160 km/h.
Flying into the wind, Mr. Wutkee flew 2800 km in 3.5 hours. The return trip, flying with the wind, took 2.5 hours. Find the rate of his plane in still air and the rate of the wind.
Let p = rate of plane in still air
Let w = rate of wind
rate · time = distance
with wind: (p + w)2.5 = 2800
against wind: (p - w)3.5 = 2800
p + w = Divide by 2.5.
p – w = Divide by 3.5.
2p = 1920
p = 960
Plane speed was 960 km/h and wind speed was 160 km/h.

11. Plane speed was 770 km/h and wind speed was 70 km/h.
The 4200 km trip from New York to San Francisco takes 6 hours flying against the wind, but only 5 hours returning. Find the speed of the plane in still air and the wind speed.
Let p = rate of plane in still air
Let w = rate of wind
rate · time = distance
with wind: (p + w)5 = 4200
against wind: (p - w)6 = 4200
p + w = 840 Divide by 5.
p – w = 700 Divide by 6.
2p = 1540
p = 770
Plane speed was 770 km/h and wind speed was 70 km/h.