SOLUTION: An airplane travels 300 km from Manila and back in 4 hours and 30 minutes. The wind speed is 10 kph, what is the speed of the plane in still air?

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Question 1058973: An airplane travels 300 km from Manila and back in 4 hours and 30 minutes. The wind speed is 10 kph, what is the speed of the plane in still air?
Found 2 solutions by ikleyn, josmiceli:
Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
.
An airplane travels 300 km from Manila and back in 4 hours and 30 minutes. The wind speed is 10 kph, what is the speed of the plane in still air?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

Let "u" represents the airspeed of the plane, in miles per hour.
Then its speed when flying with the wind is (u+10).
Its speed when flying against the wind is (u-10).

The governing equation is

300%2F%28u%2B10%29+%2B+300%2F%28u-10%29 = 4.5.

The first addend on the left side is the time for flight with the wind.
The second addend on the left side is the time for flight against the wind. 

Solve it and find "u".

As the first step, multiply both sides of the equation by (u+10)*(u-10).

You will get a quadratic equation.

Simplify it, reduce to the standard form and then apply the quadratic formula.

It is a typical "tailwind and headwind" word problem.

See the lessons
    - Wind and Current problems
    - Wind and Current problems solvable by quadratic equations
    - Selected problems from the archive on a plane flying with and against the wind
in this site.

In these lessons you will find the detailed solutions of many similar problems.


Also, you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this textbook under the section "Word problems", the topic "Travel and Distance problems".


Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +s+ = the speed of the plane in km/hr in still air
+s+%2B+10+ = the speed of the plane flying with the wind
+s+-+10+ = the speed of the plane flying against the wind
Let +t+ = time in hrs flying with the wind
The one-way distance is +300%2F2+=+150+ km
-------------------------------------------
Equation for flying with the wind:
(1) +150+=+%28+s+%2B+10+%29%2At+
Equation for flying against the wind:
(2) +150+=+%28+s+-+10+%29%2A%28+4.5+-+t+%29+
---------------------------------
(1) +t+=+150%2F%28+s+%2B+10+%29+
and
(2) +150+=+4.5s+-+45+-+s%2At+%2B+10t+
(2) +150+%2B+45+=+4.5s+-+t%2A%28+s+-+10+%29+
Plug (1) into (2)
(2) +150+%2B+45+=+4.5s+-+%28+150%2F%28+s%2B10+%29+%29%2A%28+s+-+10+%29+
(2)
(2) +195%2A%28+s%2B+10+%29+=+4.5s%2A%28+s+%2B+10+%29+-+150s+%2B+1500+
(2) +195s+%2B+1950+=+4.5s%5E2+%2B+45s+-+150s+%2B+1500+
(2) +4.5s%5E2+-+195s+%2B+45s+-+150s+%2B+1500+-+1950+=+0+
(2) +4.5s%5E2+-300s+-+450+=+0+
(2) +45s%5E2+-+3000s+-+4500+=+0+
(2) +9s%5E2+-+600s+-+900+=+0+
--------------------------------
Use quadratic formula
+s+=+%28-b+%2B-+sqrt%28+b%5E2+-+4%2Aa%2Ac+%29%29+%2F+%282%2Aa%29+
+a+=+9+
+b+=+-600+
+c+=+-900+
+s+=+%28-%28-600%29+%2B-+sqrt%28+%28-600%29%5E2+-+4%2A9%2A%28-900%29+%29%29+%2F+%282%2A9%29+
+s+=+%28+600+%2B-+sqrt%28+360000+%2B+32400+%29%29+%2F+18+
+s+=+%28+600+%2B-+sqrt%28+392400+%29++%29+%2F+18+
+s+=+%28+600+%2B+626.418+%29+%2F+18+
+s+=+1226.418%2F18+
+s+=+68.134+
----------------
The speed of the plane in still air is 68.134 km/hr
--------------------
check:
(1) +150+=+%28+s+%2B+10+%29%2At+
(1) +150+=+%28+68.134+%2B+10+%29%2At
(1) +t+=+150+%2F+78.134+
(1) +t+=+1.92+ hrs
and
(2) +150+=+%28+s+-+10+%29%2A%28+4.5+-+t+%29+
(2) +150+=+%28+68.134+-+10+%29%2A%28+4.5+-+t+%29+
(2) +150+%2F+58.134+=+4.5+-+t+
(2) +2.5802+=+4.5+-+t+
(2) +t+=+4.5+-+2.5802+
(2) +t+=+1.92+ hrs
OK
Pretty strange numbers -ususally they
come out simple decimals. You might
want a 2nd opinion, too.