SOLUTION: At a cruising speed,Marie's small family plane can fly at 100 km/h. On a recent vacation trip, it took 3 h to fly with the wind to a resort and 7 h to fly back against the wind. Wh

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Question 1046376: At a cruising speed,Marie's small family plane can fly at 100 km/h. On a recent vacation trip, it took 3 h to fly with the wind to a resort and 7 h to fly back against the wind. What was the rate of the wind? How far away was the resort?
Please help me.
I'm not understanding uniform motion problems.
Found 3 solutions by solver91311, josgarithmetic, ikleyn:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Let be the wind speed. With the wind, the airplane has an airspeed of . Against the wind, the airplane has an airspeed of . Distance equals rate times time. The distance to the resort is then . The distance from the resort back home is . Since we presume that neither home nor the resort moved during the time they were on vacation, it is safe to assume that the distance going to the resort is the same as the distance returning from the resort. Hence, the two expressions that represent the distance to and from the resort must be equal, to wit:



Solve for and then calculate either or

John

My calculator said it, I believe it, that settles it

Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
R is a constant rate.
D for Distance
T for Time
D%2FT=R ----------You can also include the units and see how they fit with the numbers for left member and right member.

Multiply the two members by 1%2FT, and you have....... D=RT.
The basic travel rate rule for constant rate of travel is RT=D.

The meaning of "speed" in your example problem can be assumed as "rate", because any angle is not really important for the example here.


SOLUTION______________________________

Analyzing and solving your example best is handled by assigning variables and putting data into a table.

The wind's speed and the trip distance are not known. The wind going in the same direction as the plane will ADD to the plane's speed,
but the same wind with its same speed will SUBTRACT from the plane's speed while flying against this wind.

Variables are assigned according to what is shown in this table:
                    SPEED            TIME             DISTANCE

GOING               100+w             3                  d 

FLY BACK            100-w             7                  d


SYSTEM OF EQUATIONS BASED ON THE UNIFORM TRAVEL RATE RULE:
system%28%28100%2Bw%29%2A3=d%2C%28100-w%29%2A7=d%29

Solve that system.
Notice you have two expressions for d, so substituting, you simply equate the two expressions for d. This is not the only way
but it is more efficient.

%28100%2Bw%293=%28100-w%297
300%2B3w=700-7w
3w%2B300=-7w%2B700
3w%2B300%2B7w-300=-7w%2B700%2B7w-300------just applying additive inverses,
10w=400
highlight%28w=40%29


You can evaluate d on your own.

Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
.
At a cruising speed,Marie's small family plane can fly at 100 km/h. On a recent vacation trip, it took 3 h to fly
with the wind to a resort and 7 h to fly back against the wind. What was the rate of the wind? How far away was the resort?
Please help me.
I'm not understanding uniform motion problems.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

The plane can fly at 100 km/h in still air (with no wind).
For the given problem, let "v" be the wind speed and D be the distance.


When the plane flies with the wind, its speed relative to the ground is 100+v km/h.

Therefore, your first equation is

D%2F3 = 100 + v.   (1)

As a reminder, when you divide the distance by the time, you get the speed.
It is what is written in the equation (1). 


When the plane flies against the wind, its speed relative to the ground is 100-v km/h.

Therefore, your second equation is

D%2F7 = 100 - v.   (2)

Again, when you divide the distance by the time, you get the speed.
It is what is written in the equation (2). 


Now all your problem is presented in equations (1) and (2). It is the system of two equations in two unknowns, D and v.


To solve it, let us first find D, the distance. For it, add the equations (1) and (2) (both sides). You will get a single equation for D:

D%2F3+%2B+D%2F7 = 200,  or  (multiply both sides by 3*7)

7D + 3D = 200*3*7,  or 

10D = 4200,  which gives you

D = 4200%2F10 = 420 km.

So, you just found the distance. It is D = 420 km.

Now you can easily find the speed "v", which is under the question. From (1), you have

v = D%2F3+-+100 = 420%2F3+-+100 = 140 - 100 = 40 km/h.

The wind speed is 40 km/h.

Check.  420%2F%28100%2B40%29 = 420%2F140 = 3 hours.

        420%2F%28100-40%29 = 420%2F60 = 7 hours.

Answer.  The speed of wind is 40 mph.

For many other solved Travel and Distance problems see the lessons
    - Travel and Distance problems
    - Wind and Current problems
    - More problems on upstream and downstream round trips
    - Wind and Current problems solvable by quadratic equations
    - Unpowered raft floating downstream along a river
    - Selected Travel and Distance problems from the archive
    - Selected problems from the archive on the boat floating Upstream and Downstream
    - Selected problems from the archive on a plane flying with and against the wind
in this site.