Flying against the jetstream, a jet travels 2800 km in 5 hours.?

Question

Flying with the jetstream, the same jet travels 2820 km in 3 hours. What is the speed of the jet in still air, and what is the speed of the jetstream?

Answer 1

You have two equations with two unknowns. Let X be the speed of the plane iun still air.and Y be the speed of the jetstream. Going against the wind, it will go (X-Y) kph. With the stream, it will go (X+Y)kph. You know how far it goes in three and five hours. You get the following two equations:

5*(X-Y) = 2800

and

3*(X+Y) = 2820

A little algebra gives you:

5X - 5Y = 2800
3X + 3Y = 2820

Divide everything in the top equation by 5 and you get

X - Y = 560

Divide the second equation by 3 and you get

X + Y = 940

Add the two equations together and you get

2X = 1500

so X = 750.

Plug this back into any equation to get Y

Answer 2

As yet another poster suggested this question is incorrect. The raise and drag on the airplane are regarding the fee it travels at, so in certainty there is greater to be taken into consideration here, besides the incontrovertible fact that at a typical physics point not pondering countless the better order concerns we are in a position to come to a answer: we are going to outline the fee of the airplane through thrust as x and we are going to outline the fee of the airplane through wind as y. d=v(dot)t, or for our elementary diagnosis distance = velocity x time for this reason we've: 1600 miles = (x+y) x 5 hrs (the place x and y are in mph) 1600 miles = (x-y) x 8 hrs fixing the 2d equation for x we've 1600/8+y= x Substituting into the 1st equation we've 1600=(200+2y) x 5 fixing for y supplies us 320=200+2y a hundred and twenty=2y y=60 mph changing returned into our equation for x supplies: 200 mph +60 mph= 260 mph So the wind velocity is 60 mph on an identical time as the airplane velocity is 260 mph, in this simplified diagnosis.

Answer 3

This looks just like another question answered at the link below...