Quadratic Equations in Distance Problems Help (page 2)
Introduction to Quadratic Equations in 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.
Traveling in One Direction or Stream
“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 traveling 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.
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 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 from 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 hours. The equation to solve is . The LCD is ( r –50)( r + 50).
( is not a solution.) The plane’s average speed without the wind is 450 mph.
Find practice problems and solutions at Quadratic Equations in Distance Problems Practice Problems - Set 1.
Traveling at Right Angles
If you need to find the distance between two bodies traveling at right angles away from each other, you must use the Pythagorean Theorem: a 2 + b 2 = c 2 in addition to D = RT .
A car passes under a railway trestle at the same time a train is crossing the trestle. The car is headed south at an average speed of 40 mph. The train is traveling east at an average speed of 30 mph. After how long will the car and train be 10 miles apart?
Let t represent the number of hours after the train and car pass each other. (Because the rate is given in miles per hour, time must be given in hours.) The distance traveled by the car after t hours is 40t and that of the train is 30 t .
After of an hour (or 12 minutes) the car and train will be 10 miles apart.
Find practice problems and solutions at Quadratic Equations in Distance Problems Practice Problems - Set 2.
Round Trips with Different Average Speeds
In the following problems people are making a round trip. The average speed in each direction will be different and the total trip time will be given. The equation to solve is
Time to destination + Time on return trip = Total trip time.
To get the time to and from the destination, use D = RT and solve for T . The equation to solve becomes
A jogger jogged seven miles to a park then jogged home. He jogged 1 mph faster to the park than he jogged on the way home. The round trip took 2 hours 34 minutes. How fast did he jog to the park?
Let r represent the jogger’s average speed on the way home. He jogged 1 mph faster to the park, so r + 1 represents his average speed to the park. The distance to the park is 7 miles, so D = 7.
Time to the park + Time home = 2 hours 34 minutes
The time to the park is represented by . The time home is represented by . The round trip is 2 hours 34 minutes hours. The equation to solve becomes
The LCD is 30 r ( r + 1).
The jogger’s average speed to the park was 5 + 1 = 6 mph.
Find practice problems and solutions at Quadratic Equations in Distance Problems Practice Problems - Set 3.
More practice problems for this concept can be found at: Algebra Quadratic Applications Practice Test.
Today on Education.com
- Coats and Car Seats: A Lethal Combination?
- Kindergarten Sight Words List
- Signs Your Child Might Have Asperger's Syndrome
- Child Development Theories
- 10 Fun Activities for Children with Autism
- Social Cognitive Theory
- Why is Play Important? Social and Emotional Development, Physical Development, Creative Development
- GED Math Practice Test 1
- Problems With Standardized Testing
- The Homework Debate