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# at maximum speed an airplane can travel 2460 miles against the wind in 6 hours. The plane can go the same distance, but with the wind in 5 hours. What is the speed of the plane with no wind?

Dear Allison Lee,

I think there is some information that is needed and it is not given to you... having to do with wind resistance, inertia, friction...

1. Can you just say, well, since it takes the plane 6 hours with a headwind and 5 hours with a tailwind, then it can fly the distance in 5.5 hours if there is no wind? If so, then your answer is 2460/5.5 = 447.3 mph.

2. Can you say, accurately, the same factor f will slow the plane with a headwind, and speed the plane in a tailwind? Then if S is the speed of the plane with no wind, you would get two equations:

headwind: S - f = 2460/6

tailwind: S + f = 2460/5. Then solving for S,

2S = 902

S = 451 mph.

As suggested you can find your homework answer if you do the work, your math book as hundreds of examples, work on several for a couple of hours, then work your question to a solution.

Consider this line of reasoning.

Against the wind, it takes 6 hours to go 2460 miles. With the wind, the plane takes 5.5 hours to go 2460 miles.

Distance = (speed) * (time)

2 * 2460 = total distance or 4920 miles flown by plane in (how long?) 11.5 hours.

4920 miles / 11.5 hours = 427.8260869565217 miles / hour.

Did I eliminate the force of the wind as a factor or was it ever a factor *in this particular problem *?

And this particular problem is at least a *slightly *tricky one.

**let speed of wind be y mphlet speed of plane in still air be x..Against wind the speed = x-ydistance traveled = 24602460/x-y= 66x-6y= 2460x-y=410...........1..with wind speed = x+ydistance is same 24602460/x+y=5**

5x+5y=2460

x+y=492............2

..

Add equation 1 & 2

x-y+x+y=410+492

2x=902

x= 451 mph speed of plane in still air.

..

Plug the value of x in equation 1

451-y=410

y=451-410

y=40 mph the speed of wind