Distance Problems
There are several distance problems that quadratic equations can solve. One of these types is ‘‘stream’’ problems: a vehicle travels the same distance up and back where in one direction, the ‘‘stream’s’’ average speed is added to the vehicle’s speed and in the other, the ‘‘stream’s’’ average speed is subtracted from the vehicle’s speed. Another type involves two bodies moving away from each other where their paths form a right angle (for instance, one travels north and the other west). Finally, the last type is where a vehicle makes a round trip that takes longer in one direction than in the other. In all of these types, the formula D = RT is key.
‘‘Stream’’
distance problems usually involve boats (traveling upstream or downstream) and
planes (traveling against a headwind or with a tailwind). The boat or plane
generally travels in one direction then turns around and travels in the
opposite direction. The distance upstream and downstream is usually the same.
If r represents the boat’s or plane’s
average speed travel- ing without the ‘‘stream,’’ then r stream’s speed represents the boat’s or plane’s average speed
traveling with the stream, and r stream’s
speed represents the boat’s or plane’s average speed traveling against the
stream.
Example
Miami and Pittsburgh are 1000 miles apart. A
plane flew into a 50-mph headwind from Miami to Pittsburgh. On the return flight
the 50-mph wind became a tailwind. The plane was in the air a total of 4 ½ hours
for the round trip. What would have been the plane’s average speed without the
wind?
Let r represent the plane’s average speed
(in mph) without the wind. The plane’s average speed against the wind is r 50 (from Miami to Pittsburgh) and the
plane’s average speed with the wind is r –
50 (from Pittsburgh to Miami). The distance from Miami to Pittsburgh is
1000 miles. With this information we can use 
to compute the time in the air in each
direction. The time in the air Miami to
Pittsburgh is
. The time in the air from Pittsburgh to Miami is 
. The time in the air from Miami to Pittsburgh plus the time in the air
from Pittsburgh to Miami is 4 ½ = 4.5 hours. The equation to solve is 
. The LCD is (r – 50)(r + 5).
Practice
1. A flight from Dallas to Chicago is 800 miles.
A plane flew with a 40- mph tailwind from Dallas to Chicago. On the return trip,
the plane flew against the same 40-mph wind. The plane was in the air a total of
5.08 hours for the flight from Dallas to Chicago and the return flight. What
would have been the plane’s speed without the wind?
2. A flight from Houston to New Orleans faced a
50-mph headwind, which became a 50-mph tailwind on the return flight. The total
time in the air was 13 hours. The distance between Houston and New
Orleans is 300 miles. How long was the plane in flight from Houston to New
Orleans?
3. A small motorboat traveled 15 miles
downstream then turned around and traveled 15 miles back. The total trip took 2
hours. The stream’s speed is 4 mph.
How fast would the boat have traveled in still water?
4. A plane on a flight from Denver to
Indianapolis flew with a 20-mph tailwind. On the return flight, the plane flew
into a 20-mph head- wind. The distance between Denver and Indianapolis is 1000
miles and the plane was in the air a total of 51 hoburs. What would
have been the plane’s average speed without the wind?
5. A plane flew from Minneapolis to Atlanta, a distance
of 900 miles, against a 30 mph-headwind. On the return flight, the 30-mph wind
became a tailwind. The plane was in the air for a total of 5 hours. What would
the plane’s average speed have been without the wind?
Solutions
1. Let r represent the plane’s average speed (in
mph) with no wind. Then the average speed from Dallas to Chicago (with the
tailwind) is r + 40, and the average speed from Chicago to Dallas is r – 40
(against the headwind). The distance between Dallas and Chicago is 800 miles.
The time in the air from Dallas to Chicago plus the time in the air from
Chicago to Dallas is 5.08 hours. The time in the air from Dallas to Chicago is 
. The time in the air from Chicago
to Dallas is 
. The equation to solve is 
.
The plane’s average speed without the wind would have been about 320 mph.
2. Let r represent the plane’s average speed
without the wind. The average speed from Houston to New Orleans (against the
headwind) is r – 50, and the average speed from New Orleans to Houston (with
the tailwind) is r + 50. The distance between Houston and New Orleans is 300 miles.
The time in the air from Houston to New Orleans is 
and the time in the air from New
Orleans to Houston is 
. The time in the air from Houston to New Orleans plus the time in the
air from New Orleans to Houston is 1 3/4 = 7/4
hours. The equation to solve is 
. The LCD is (4(r – 50)(r + 50).
The average speed of the plane without the wind was 350 mph. We want the
time in the air from Houston to New Orleans: 
. The plane was in flight from Houston to New Orleans for one hour.
3. Let r represent the boat’s speed in still
water. The average speed downstream is r þ 4 and the average speed upstream is
r _ 4. The boat was in the water a total of 2 hours. The distance traveled in each
direction is 15 miles. The time the boat traveled downstream is 
hours,
and it traveled upstream .png){kind=small}
hours. The time the boat traveled
upstream plus the time it traveled downstream equals 2 hours. The equation to
solve is 
The
LCD is ( r + 4)(r – 4).
The boat’s average speed in still water is 16 mph.
4. Let r represent the plane’s average speed
without the wind. The plane’s average speed from Denver to Indianapolis is r +
20, and the plane’s average speed from Indianapolis to Denver is r – 20. The
total time in flight is 5 ½ hours and the distance between Denver and
Indianapolis is 1000 miles. The time in the air from Denver to Indianapolis is 
hours and the time in the air from
Indianapolis to Denver is
hours. The time in the air from
Denver to Indianapolis plus the time in the air from Indianapolis to Denver is 5.5
hours. The equation to solve is 
. The LCD is (r + 20)(r – 20).
The plane would have averaged about 365 mph without the wind.
5. Let r represent the plane’s average speed
without the wind. The plane’s average speed from Minneapolis to Atlanta
(against the headwind) is r – 30. The plane’s average speed from Atlanta to Minneapolis
(with the tailwind) is r + 30. The total time in the air is 51 2 hours and the
distance between Atlanta to Minneapolis is 900 miles. The time in the air from
Minneapolis to Atlanta is 
hours,
and the time in the air from Atlanta to Minneapolis is 
hours.
The time in the air from Minneapolis to Atlanta plus the time in the air from
Minneapolis to Atlanta is 5 ½ = 5.5 hours. The equation to solve is 
. The LCD is (r – 30)( r + 30).
The plane’s average speed without the wind was 330 mph.
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