Speed of Plane & Wind 2

[mathdad]

George and Glen are frequent fliers on Fast-n-Go Airlines. They often fly between two cities that are a distance of 800 miles apart. On one particular trip, they flew into the wind and the trip took 2.5 hours. The return trip with the wind behind them, only took about 2 hours. Find the speed of the wind and the speed of the plane in still air.

Solution:

Let p = plane
Let w = wind

Total distance of trip is 800 miles.

Flying into the wind: p - w
Flying with the wind: p + w

Going time: 2.5 hours
Return time: 5 hours

Equation 1

2.5(p - w) = 800

Equation 2

5(p + w) = 800

Is the set up correct? This is my third and final Sunday question. Everyday 3 questions will be posted per ONE TOPIC.

 

[ksdhart2]

So long as by "Let p = plane" you mean "Let p stand for the speed of the plane," and similar for the variable \(w\), then yes. Good job.

Edit: Additionally you appear to have a small typo in your answer that I initially overlooked. Your second equation should read \({\color{red}{2}}(p + w) = 800\) as the return trip took 2 hours, not 5.

 

[mathdad]

So long as by "Let p = plane" you mean "Let p stand for the speed of the plane," and similar for the variable \(w\), then yes. Good job.

Edit: Additionally you appear to have a small typo in your answer that I initially overlooked. Your second equation should read \({\color{red}{2}}(p + w) = 800\) as the return trip took 2 hours, not 5.

Yes, I see the error. The number 5 is from another problem.