word problem,

Question

a plane flew 70 miles with the wind from city A tto City B. He flew back to city A aginst the wind. If the wind was a constant of 30 miles per hour and the total going and reuturing time was 1.2 hours. Find the speed of the plane in still air

Answer 1

For this type of problem, it's often useful to make a table. Remember that distance = rate×time.

 

Let x = speed of the plane in still air. Then, on the way from A to B with the 30 mph wind, the plane's rate was x + 30 mph. Similarly, against the wind on the way from B to A, the plane's rate was x - 30 mph.

 

The total flight time was 1.2 hours, so let's say t = time from A to B. Then the time from B to A is whatever amount of the 1.2 hours remains, or 1.2 - t.

 

Now we have enough information to set up a table where the first row is the flight from A to B and the second is the flight from B to A:

 

DISTANCE                RATE                    TIME

70                                x+30                      t

70                                x-30                      1.2-t

 

Now we can write two equations using d = rt:

 

70 = (x+30)t

70 = (x-30)(1.2-t)

 

We'll solve both equations for t:

 

70 = (x+30)t

70 / (x+30) = t

 

70 = (x-30)(1.2-t)

70 = 1.2x - xt - 36 +30t

70 + 36 - 1.2x = 30t - xt

106 - 1.2x = t(30-x)

(106-1.2x) / (30-x) = t

 

Now, substitute 70 / (x+30) = t into the second equation and rearrange into quadratic form:

 

(106-1.2x) / (30-x) = 70 / (x+30)

(106-1.2x)(x+30) = 70(30-x)

106x + 3180 - 1.2x² - 36x = 2100 - 70x

0 = 2100 - 70x - 106x - 3180 + 1.2x² + 36x

0 = 1.2x² - 140x - 1080

 

Finally, solve for x using the quadratic formula where a = 1.2, b = -140, and c = -1080:

 

x = (140 ± √((-140)² - 4(1.2)(-1080))) / (2(1.2))

x = (140 ± √(19600 + 5184)) / 2.4

x = (140 ± √24780) / 2.4

x = (140 ± 157.42) / 2.4

 

So, x = (140 + 157.42) / 2.4 = 123.93 mph OR x = (140 - 157.42) / 2.4 = -7.26 mph.

 

Since miles per hour can't be negative, the plane's speed in still air is x = 123.93 mph.