Quadratic Equations in Distance Problems Practice Problems

Set 1: Introduction to Quadratic Equations in Distance Problems - Traveling in One Direction or Stream

To review quadratic equations in distance problems, go to Quadratic Equations in Distance Problems Help

Practice

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?
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 hours. The distance between Houston and New Orleans is 300 miles. How long was the plane in flight from Houston to New Orleans?
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?
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 headwind. The distance between Denver and Indianapolis is 1000 miles and the plane was in the air a total of hours. What would have been the plane’s average speed without the wind?
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 hours. What would the plane’s average speed have been without the wind?

Solutions

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 LCD is ( r + 40)( r – 40).

The plane’s average speed without the wind would have been about 320 mph.
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 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: hour. The plane was in flight from Houston to New Orleans for one hour.
Let r represent the boat’s speed in still water. The average speed downstream is r + 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 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.
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 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.
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 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 hours. The equation to solve is . The LCD is ( r – 30)( r + 30).

The plane’s average speed without the wind was 330 mph.

Set 2: Traveling at Right Angles

To review problems where you need to find the distance between two bodies traveling at right angles away from each other, go to Quadratic Equations in Distance Problems Help

Practice

A car and plane leave an airport at the same time. The car travels eastward at an average speed of 45 mph. The plane travels southward at an average speed of 200 mph. After how long will they be 164 miles apart?
Two joggers begin jogging from the same point. One jogs south at the rate of 8 mph and the other jogs east at a rate of 6 mph. When will they be five miles apart?
A cross-country cyclist crosses a railroad track just after a train passed. The train is traveling southward at an average speed of 60 mph. The cyclist is traveling westward at an average speed of 11 mph. When will they be 244 miles apart?
A motor scooter and a car left a parking lot at the same time. The motor scooter traveled north at 24 mph. The car traveled west at 45 mph. How long did it take for the scooter and car to be 34 miles apart?
Two cars pass each other at 4:00 at an overpass. One car is headed north at an average speed of 60 mph and the other is headed east at an average speed of 50 mph. At what time will the cars be 104 miles apart? Give your solution to the nearest minute.

Solutions

Let t represent the number of hours each has traveled. The plane’s distance after t hours is 200t and the car’s distance is 45 t .

The car and plane will be 164 miles apart after 0.80 hours or 48 minutes.
Let t represent the number of hours after the joggers began jogging. The distance covered by the southbound jogger after t hours is 8t, and the distance covered by the eastbound jogger is 6 t .

The joggers will be five miles apart after hour or 30 minutes.
Let t represent the number of hours after the cyclist crosses the track. The distance traveled by the bicycle after t hours is 11 t and the distance traveled by the train is 60 t .

After four hours the cyclist and train will be 244 miles apart.
Let t represent the number of hours after the scooter and car left the parking lot. The car’s distance after t hours is 45 t . The scooter’s distance is 24 t .

The car and scooter will be 34 miles apart after of an hour or 40 minutes.
Let t represent the number of hours after the cars passed the overpass. The northbound car’s distance after t hours is 60 t and the eastbound car’s distance is 50 t .

The cars will be 104 miles apart after about 1.33 hours or 1 hour 20 minutes. The time will be about 5:20.

Set 3: Round Trips with Different Average Speeds

To review distance problems where people are making a round trip, go to Quadratic Equations in Distance Problems Help

Practice

A man rode his bike six miles to work. The wind reduced his average speed on the way home by 2 mph. The round trip took 1 hour 21 minutes. How fast was he riding on the way to work?
On a road trip a saleswoman traveled 120 miles to visit a customer. She averaged 15 mph faster to the customer than on the return trip. She spent a total of 4 hours 40 minutes driving. What was her average speed on the return trip?
A couple walked on the beach from their house to a public beach four miles away. They walked 0.2 mph faster on the way home than on the way to the public beach. They walked for a total of 2 hours 35 minutes. How fast did they walk home?
A family drove from Detroit to Buffalo, a distance of 215 miles, for the weekend. They averaged 10 mph faster on the return trip. They spent a total of seven hours on the road. What was their average speed on the trip from Detroit to Buffalo? (Give your solution accurate to one decimal place.)
Boston and New York are 190 miles apart. A professor drove from his home in Boston to a conference in New York. On the return trip, he faced heavy traffic and averaged 17 mph slower than on his way to New York. He spent a total of 8 hours 5 minutes on the road. How long did his trip from Boston to New York last?

Solutions

Let r represent the man’s average speed on the way to work. Then r – 2 represents the man’s average speed on his way home. The distance each way is 6 miles, so the time he rode to work is , and the time he rode home is . The total time is 1 hour 21 minutes = hours. The equation to solve is . The LCD is 20 r (r – 2).

The man’s average speed on his way to work was 10 mph.
Let r represent the saleswoman’s average speed on her return trip. Her average speed on the way to the customer is r + 15. The distance each way is 120 miles. She spent a total of 4 hours 40 minutes hours driving. The time spent driving to the customer is . The time spent driving on the return trip is . The equation to solve is . The LCD is 3 r ( r + 15).

The saleswoman averaged 45 mph on her return trip.
Let r represent the couple’s average rate on their way home, then r – 0,2 represents the couple’s average speed to the public beach. The distance to the public beach is 4 miles. They walked for a total of 2 hours 35 minutes = hours. The time spent walking to the public beach is . The time spent walking home is . The equation to solve is . The LCD is 12 r ( r – 0.2).

(Multiplying by 10 to clear the decimals would result in fairly large numbers for the quadratic formula.)

The couple walked home at the rate of mph.
Let r represent the average speed from Detroit to Buffalo. The average speed from Buffalo to Detroit is r + 10. The distance from Detroit to Buffalo is 215 miles and the total time the family spent driving is 7 hours. The time spent driving from Detroit to Buffalo is . The time spent driving from Buffalo to Detroit is . The equation to solve is . The LCD is r ( r + 10).

The family averaged 56.8 mph from Detroit to Buffalo.
Let r represent the average speed on his trip from Boston to New York. Because his average speed was 17 mph slower on his return trip, r – 17 represents his average speed on his trip from New York to Boston. The distance between Boston and New York is 190 miles. The time on the road from Boston to New York is and the time on the road from New York to Boston is . The time on the road from Boston to New York plus the time on the road from New York to Boston is 8 hours 5 minutes hours. The equation to solve is . The LCD is 12 r ( r –17).

The rate cannot be because the total round trip is only 8 hours 5 minutes. The professor’s average speed from Boston to New York is 57 mph. We want his time on the road from Boston to New York. His time on the road from Boston to New York is hours or 3 hours 20 minutes.