*Solving for both velocities*

Velocity (against the wind) = 1650 ÷ 3 = 550 miles per hour.

Velocity (with the wind) = 1650 ÷ 2.75 = 600 miles per hour.
*"psa" means plane speed in still air; "ws" means wind speed*

psa -ws = 550

psa +ws = 600

Adding both equations we get

2*psa = 1150

Plane speed in still air = 575

Since psa +ws = 600

By substitution we get wind speed = 25

**
B) Solving for airplane velocity and distance ***given wind velocity and time.*

Example:

An airplane flying *with* a 40 mph wind takes 4 hours to make a trip.

On the return trip, the airplane flies *against* a 40 mph wind and takes 4.5 hours to make the trip.

What is the airplane velocity and the distance travelled (one way)?

**
***Using the calculator,* we click "B" then enter

Wind Velocity 40

Time 1 4

Time 2 4.5

(*entering Time 1 = 4.5 Time 2 = 4 will also work*)

**
Clicking "Calculate" we see the answers are:**

Airplane Velocity 680

Distance 2880

**
***Without using the calculator:*

Distance = velocity * time

Distance = (plane velocity + wind velocity) * time

Dist (out) = (plane vel + 40) * 4

Dist (return) = (plane vel - 40) * 4.5

**
Since this is a round trip, the distance is the ***same* so:

(plane vel + 40) * 4 = (plane vel - 40) * 4.5

4*plane vel + 160 = 4.5 * plane vel - 180

340 = .5 * plane velocity

plane velocity = 680

**
Putting this amount into this equation:**

Dist (out) = (plane vel + 40) * 4

Distance = (680 +40) * 4

Distance = 720 * 4

Distance = 2,880

**
**

**
**

C) Solving for Distance and Wind Velocity *given airplane velocity and time.*

**
Example:
You take a round-trip on an airplane, that has a velocity in ***still air* of 440 mph.

On your trip out, the plane flies for 6 hours *against the wind* and on your return trip, it flies for 5 hours *with the wind*.

What is the wind speed and the one-way distance it traveled?

**
***Using the calculator,* we click "C" then enter

Airplane Velocity 440

Time 1 5

Time 2 6

(*entering Time 1 = 6 Time 2 = 5 will also work*)

**
Clicking "Calculate" we see the answers are:**

Distance 2400

Wind Velocity 40

**
***Without using the calculator,*

Distance = velocity * time

Distance = (plane velocity + wind velocity) * time

Distance (out) = (440 - wind velocity) * 6

Distance (return) = (440 + wind velocity) * 5

**
Since it's a round trip, the distance is the ***same* so:

(440 - wind velocity) * 6 = (440 + wind velocity) * 5

-6* wv + 2,640 = 2,200 +5*wv

440 = 11*wv

wind velocity = 40

**
Putting this value into this equation:**

Distance (out) = (440 - wind velocity) * 6

Distance = (440 - 40) * 6

Distance = (400) * 6

Distance = 2,400

**
D) Solving for Wind Velocity ***given Airplane Velocity and Distance.*

**
Example:
A plane's velocity in still air is 210 miles per hour.**

It flies for 725 miles *with* the wind and in the same amount of time, it flies 675 miles *against* the wind.

What is the wind velocity?

**
***Using the calculator,* we click "D" then enter

**
Airplane Velocity 210**

Distance 1 725

Distance 2 675

*Entering Distance 1 = 675 and Distance 2 = 725 will also work*

**
Clicking "Calculate" we see the answer is:**

Wind Velocity 7.5

**
Without using the calculator**

*There are 2 ways to do this.*

If you want the complete explanation (the solution that algebra teachers like), then just keep reading.

Otherwise scroll to "the shorter method".

We are *not* given a specific amount of time, but we *do* know that time = distance ÷ rate

Since time is the *same for both cases* we can set up 2 *"distance ÷ rate"* fractions that equal each other.

**
Distance1 ÷ (plane velocity + wind velocity) = Distance2 ÷ (plane velocity - wind velocity)**

**
725 ÷ (210 + wind velocity) = 675 ÷ (210 - wind velocity)**

(725 ÷ 675) = (210 + wind velocity) ÷ (210 - wind velocity)

(725 × 210) - (725 × wind velocity) = (675 × 210) + (675 × wind velocity)

(725 × 210) - (675 × 210) = (725 × wind velocity) + (675 × wind velocity)

152,250 - 141,750 = 1,400 wind velocity

10,500 = 1,400 wind velocity

wind velocity = 7.5 miles per hour

**
The shorter method:**

When using this method, make sure *distance1* is greater than *distance2*.

We'll also abbreviate plane velocity and wind velocity as *pv* and *wv*.

Insert the numbers from the problem into this equation:

[(725 × 210) -(675 × 210)] ÷ (725 + 675) = wind velocity

(152,250 -141,750) ÷ 1,400 = wind velocity

10,500 ÷ 1,400 = wind velocity

wind velocity = 7.5 miles per hour

**
**

**
E) Solving for Plane Velocity ***given Wind Velocity and Distance.*

**
Example:
The wind velocity is 12 miles per hour.**

A plane flies for 850 miles *with* the wind and in the same amount of time, it flies 750 miles *against* the wind.

What is the airplane velocity?

**
***Using the calculator,* we click "E" then enter

**
Wind Velocity 12**

Distance 1 850

Distance 2 750

*Entering Distance 1 = 750 and Distance 2 = 850 will also work*

**
Clicking "Calculate" we see the answer is:**

Airplane Velocity 192

**
Without using the calculator**

*There are 2 ways to do this.*

If you want the complete explanation (the solution that algebra teachers like), then just keep reading.

Otherwise scroll to "the shorter method".

We are *not* given a specific amount of time, but we *do* know that time = distance ÷ rate

Since time is the *same for both cases* we can set up 2 *"distance ÷ rate"* fractions that equal each other.

**
Distance1 ÷ (plane velocity + wind velocity) = Distance2 ÷ (plane velocity - wind velocity)**

**
850 ÷ (plane velocity + 12) = 750 ÷ (plane velocity - 12)**

(850 ÷ 750) = [(plane velocity + 12) ÷ (plane velocity - 12)]

(850 × -12) + 850 pv = (750 × 12) + 750 pv

-10,200 + 850 pv = 9,000 + 750 pv

100 pv = 19,200

plane velocity = 192

**
The shorter method:**

When using this method, make sure *distance1* is greater than *distance2*.

We'll also abbreviate plane velocity and wind velocity as *pv* and *wv*.

Insert the numbers from the problem into this equation:

[(850 × 12) + (750 × 12)] ÷ (850 -750) = plane velocity

(10,200 + 9,000) ÷ 100 = plane velocity

plane velocity = 192

**
F) Solving for Boat Velocity when ***given Current Velocity, Distance and Time.*

**
Example:
Each day, a boat makes a 30 mile trip upstream and 30 miles downstream, taking 8 hours for the round trip.**

The velocity of the current is 5 miles per hour.

What is the boat velocity?

**
**

**
***Using the calculator,* we click "F" then enter

**
Time 8**

Distance 30

Wind (or Current) Velocity 5

**
Clicking "Calculate" we see the answer is:**

Airplane (or Boat) Velocity 10

**
***Without using the calculator,*

We know that time = distance ÷ velocity.

We only know the *total time*, so any equations we write must take that into account.

**
Going ***upstream*, the time would be:

time1 = 30 ÷ (bv - 5)

**
Going ***downstream*, the time would be:

time2 = 30 ÷ (bv + 5)

**
One thing we ***do* know is that the *total time* equals 8 hours.

Since time1 + time2 = 8 hours, then we can say:

**
[ 30 ÷ (bv - 5) ] + [ 30 ÷ (bv + 5) ] = 8**

Multiplying both sides by (bv -5)

30 + [ 30 • (bv -5) ÷ (bv + 5) ] = 8bv -40

[ 30 • (bv -5) ÷ (bv + 5) ] = 8bv -70

Multiplying both sides by (bv +5)

30bv -150 = 8bv² -70bv +40bv -350

Collecting the terms

8bv² -60bv -200 = 0

Using the quadratic equation calculator

boat velocity = 10 miles per hour