SOLUTION: A light plane flies for 2h with a 20km/h tail wind. The return trip against the same wind takes 3h. Write equations to determine the speed of the plane when there is no wind and th

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: A light plane flies for 2h with a 20km/h tail wind. The return trip against the same wind takes 3h. Write equations to determine the speed of the plane when there is no wind and th      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 72815: A light plane flies for 2h with a 20km/h tail wind. The return trip against the same wind takes 3h. Write equations to determine the speed of the plane when there is no wind and the distance travelled
Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
This is a classic type of problem that uses the equation:
.
Distance = Rate * Time
.
or
.
D= R*T
.
for short.
.
All you have to figure out is what to put into the equation.
.
You are told that on one leg of a trip a plane flies for 2 hours (that means T = 2) and it
has a tail wind of 20 km per hour. Therefore the rate that it is going at is the speed of
the airplane (call it S) + 20 km per hour that the wind is pushing it. This rate can be
written as (S+20).
.
So on this leg of the trip the distance traveled is given by the equation:
.
D+=+%28S%2B20%29%2A2+=+2S+%2B+40
.
On the return home the plane flies for 3 hours so T = 3. But this time instead of boosting
the plane's speed the wind is slowing the plane down because the plane is flying into the wind.
So this time the rate at which the plane is flying is (S - 20) which is the speed of the
plane minus the speed of the wind. So the distance equation for the trip back home can
be written as:
.
D+=+%28S-20%29%2A3
.
But in both cases the distance covered is the same amount. Therefore the right sides of
these two equations have to be equal. This equality can be written as:
.
%28S%2B20%29%2A2+=+%28S-20%29%2A3
.
Multiply both sides out to get:
.
2S%2B40+=+3S-60
.
Subtract 3S from both sides to eliminate the 3S on the right side. The result is:
.
-S+%2B+40+=+-60
.
Subtract 40 from both sides to get rid of the +40 on the left side. When you do the equation
becomes:
.
-S+=+-100
.
And as a final step multiply both sides by -1 to get that S, the speed of the plane in
still air is:
.
S+=+100+km%2Fhr
.
So we have that the speed of the plane is 100 km/hr. When the wind pushes the plane it
boosts the speed of the plane by 20 km/hr so the plane is going 120 km/hr and it goes at
that speed for 2 hours. So multiplying the rate times the time gives us:
.
120%2A2+=+240 km
.
For grins, let's check the flight back home. This time the plane is flying at the speed
it would be at in still or calm air (100 km/hr) but because it is going into a 20 km/hr head
wind it is slowed to 100-20 = 80 km/hr. So it's true ground speed is 80 km/hr and it flies
this leg for 3 hrs before it gets home. Using the distance formula, the distance covered
is this rate times the time of 3 hrs or:
.
D+=+80%2A3+=+240 km.
.
The problem checks out because the distance is the same amount both going and coming.
.
Note that this plane is not very fast. In still air it flies at about 62 miles per hour,
and even with a 20 km/hr tail wind pushing it it is only going about 72 miles per hour.
.
Hope this helps you to understand the basics of this problem. The most difficult thing
about the problem is determining what the words are telling you about the performance
of the plane relative to the wind.