Lesson Wind and Current problems solvable by quadratic equations

Wind and Current problems solvable by quadratic equations

In this lesson some typical problems of the type "Wind and Current" are presented for a motorboat and an airplane making round trips.

The given values (input data) are the length of the trip, the total time spent for the round trip, and one of two speeds - the speed of the moving object (a motorboat or an airplane) or the speed of the moving media (the current or the wind).
The second speed is unknown.
The way to solve such problems is to reduce them to a quadratic equation and then to solve the equation.

At the first look, problems in this lesson seem to be similar to those from the lesson Wind and Current problems of this module.
They are similar from the point of view of processes under consideration (a moving object in a moving media goes back and forward).
But in fact, problems in this lesson differ from those of the above referred lesson. The difference is in the nomenclature of the input data.
This difference produces the difference in the solution strategy.

Problem 1. Motorboat moving upstream and downstream on a river

A motorboat makes a round trip on a river of 45 miles upstream and 45 miles downstream, maintaining the constant speed 12 miles per hour relative to the water.
The entire upstream and downstream trip takes 8 hours.
What is the speed of the current?

Solution
Denote the unknown current speed of the river as miles per hour.
When the motorboat moves upstream, its speed relative to the bank of the river is miles per hour, and the time spent moving upstream is hours.
When the motorboat moves downstream, its speed relative to the bank of the river is miles per hour, and the time spent moving downstream is hours.
So, the total time upstream and downstream is , and it is equal to 8 hours, according to the problem input.
This gives an equation .
To simplify the equation, multiply both sides by and collect like terms. Step by step, you get
,
,
(after dividing both sides by 8),
,
,
.

Answer. The speed of the current is 3 mile per hour.

Problem 2. Airplane flying into the wind and with the wind

An airplane makes a trip of 2400 miles long into the wind and 2400 miles back with the same tail wind, maintaining the constant speed of 440 miles per hour relative to the air.
The entire round trip takes 11 hours.
Find the speed of the wind.

Solution
Denote the unknown speed of the wind as miles per hour.
When the airplane flies into the wind, its speed relative to the earth is miles per hour, and the time spent moving upstream is hours.
When the airplane flies with the wind, its speed relative to the earth is miles per hour, and the time spent moving downstream is hours.
So, the total time for the round trip is , and it is equal to 11 hours, according to the problem input.
This gives an equation .
To simplify the equation, multiply both sides by and collect like terms. Step by step, you get
,
,
(after dividing both sides by 11),
,
,
.

Answer. The speed of the wind is 40 mile per hour.

Problem 3 below is modified Problem 1. The difference in time spent for the upstream and downstream trips is given as known input value instead of the total time spent for the the upstream and downstream trips. The solution strategy remains the same.

Problem 3. Motorboat moving upstream and downstream on a river

A motorboat makes a round trip on a river of 45 miles upstream and 45 miles downstream, maintaining the constant speed 12 miles per hour relative to the water.
The upstream trip takes two hours more time than the downstream trip.
What is the speed of the current?

Solution
Denote the unknown current speed of the river as miles per hour.
When the motorboat moves upstream, its speed relative to the bank of the river is miles per hour, and the time spent moving upstream is hours.
When the motorboat moves downstream, its speed relative to the bank of the river is miles per hour, and the time spent moving downstream is hours.
So, the difference in time spent for the upstream and the downstream trip is , and it is equal to 2 hours, according to the problem input.
This gives an equation .
To simplify the equation, multiply both sides by and collect like terms. Step by step, you get
,
,
,
.
You obtained the quadratic equation.
Now apply the quadratic formula to solve this equation (see the lesson Introduction into Quadratic Equations):
.
Thus, the quadratic equation has two roots, and .
The root is the only appropriate solution of the problem.
Easy check shows that value satisfies the problem condition.

Answer. The speed of the current is 3 mile per hour.