141k views
1 vote
A businessperson took a small airplane for a quick flight up the coast for a lunch meeting and then returned home. The plane flew a total of 4 hours, and each way the trip was 200 miles. What was the speed of the wind that affected the plane, which was flying at a speed of 120mph? Round your answer to the nearest whole number.

User PmanAce
by
7.6k points

2 Answers

0 votes

Answer:

Step-by-step explanation:

Type Distance Rate Time

Headwind 200 120-r 200/120-r

Tailwind 200 120 - r 200/120 - r

We know the times add to 4, so we write the equation:

200/120−r + 200/120 + r = 4

We multiply both sides by the LCD and simplify to get:

(120−r)(120+r) ((200/120 -r ) + 200/120+r) = 4(120 -r) (120 +r)

200(120−r)+200(120+r)=4(120−r)(120+r)

Factor the 200 and simplify inside the parentheses to find:

200(120−r+120+r)=4(1202−r2)

200(240)=4(1202−r2)

200(60)=120^2−r^2

12,000=14,400−r^2

−2,400= −r^2

49 ≈ r

The speed of the wind is 49mph.

User Jim Horn
by
7.7k points
0 votes

Answer:

Speed of the wind is 48.989 mph

Step-by-step explanation:

We have given each trip is of 200 miles

So total distance = 200 +200 = 400 miles

Speed of the airplane = 120 mph

Let the speed of the wind = x mph

So the speed of the airplane with wind = 120+x

So time taken by airplane with wind =
(200)/(120+x)

Speed of the airplane against the wind = 120 - x

So time taken by the airplane against the wind
=(200)/(120-x)

Total time is given as t= 4 hour

So
(200)/(120+x)+(200)/(120-x)=4


(200(120-x)+200(120+x))/((120+x)(120-x))=4


48000=57600-4x^2


4x^2=9600

x = 48.989 mph

User Markt
by
8.5k points