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44 votes
44 votes
A plane traveled 1560 miles to Ft. Worth and back. The trip there was with the wind. It took 13 hours. The trip back was into the wind. The trip back took 26 hours. Find the speed of the plane in still air and the speed of the wind.​

User Lokesh Kumar
by
2.7k points

2 Answers

23 votes
23 votes

Answer:

The speed of the plane in still air was 90 mph

The speed of the wind was 30 mph

Explanation:

Your first step would be to make two equations so you can cancel them out and find what each variable is. So, I divided 1560 by 13 and got 120 mph which represents the plane's speed plus the wind's speed. Next, to find the speed of the trip to go back, I divided 1560 by 26 which got me to 60. This represents the plane's speed minus the wind's speed to find the way back. Once, you have these equations, line them up and use either method of elimination or substitution. I used elimination because you could see the variable "w" cancelling out very easily. Solve the equation and substitute the new value of the variable to get the other value.

p+w=120

p-w=60

--------------

2p=180

p=90

p+w=120--> 90+w=120--> w=30

W: 30 mph

P: 90 mph

I hope this helps!!

User Robin Vessey
by
3.1k points
25 votes
25 votes
tbh i think it’s


















coco puffs bro
User Asudhak
by
3.2k points
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