Please send help!

Question

Flying against the wind, an airplane travels 3050

 kilometers in 5

 hours. Flying with the wind, the same plane travels 6960

 kilometers in 8

 hours. What is the rate of the plane in still air and what is the rate of the wind? 

Answer 1

d = r * t

3050 = (p - w)5

6960 = (p + w)8

3050 = 5p - 5w

6960 = 8p + 8w

Can you solve for p or w and answer?

Answer 2

Hi, Essence!

so this is d=rt problem.

I would solve for r

So r= d/t

let’s call the plane with no wind r

so against the wind it would be r-w bc it will slow down the plane

with the wind it would be r+w bc it will make it go faster

so we will have two equations

r-w= 3060/5 Can go 3050 km in 5 hours AGAINST the wind

r+w= 6960/8 Can go 6960 km in 8 hours WITH the wind

here is the system after simplifying:

r-w = 610

r+w= 870

ADD to eliminate the w. We will find that next we want to solve for r the speed in still air first

2r= 1480

Divide by 2

r= 740 (this is part A)

Now find the wind speed w. We will use r+w= 870 to avoid negatives.

r+w= 870

replace r with 740

740+w=870

subtract 740 on both sides

w=130 (that’s the wind speed, the part B answer.)

I hope this helps. Have a great week at school!

Mrs. S

Answer 3

distance = rate * time

R is the speed of the plane

S is the spped of the wind

3050 = 5(R-s) = 5R - 5s

6960 = 8(R+s) = 8R + 8s

Multiplies top equation by 8

and the bottom by 5:

24400 = 40R - 40s

34800 = 40R + 40s

59200 = 80R

R=740

3050 = 5*740 - 5s

3050 = 3700 -5s

3050 + 5s = 3700

5s = 650

s = 130

6960 = 8(740) + 8s

6960 = 5920 + 8s

1040 = 8s

s=130