+0  
 
+1
150
8
avatar+1117 

Joe flew 300 miles with the wind in two hours. After flying against the wind for 2 hours, he had made 270 miles of the return trip. Find the wind speed and the speed of the plane in still air.

 

with -              r      *      t      =      d

 

against-

 

 

 

This is a motion problem but I don't know how to do it.

 May 7, 2019
 #1
avatar+19914 
0

When flying WITH the wind his rate is   s +w        where s = plane speed  w = wind speed

 

so   d = r * t

       300 = (s+w) * 2

 

 

AGAINST the wind, his rate is   s - w

so  270 = (s-w)*2                                       Now you have two equations and two unknowns....solve the system of equations to find ' s '

 May 7, 2019
 #2
avatar+1117 
+2

I can't solve that, it has two variables.

SmartMathMan  May 7, 2019
edited by SmartMathMan  May 7, 2019
 #3
avatar+19914 
0

What box?

ElectricPavlov  May 7, 2019
 #4
avatar+1117 
+2

it is supposed to be the box way of solving it, nevermind. But I can't solve a problem with two variables.

SmartMathMan  May 7, 2019
edited by SmartMathMan  May 7, 2019
 #5
avatar+1117 
+1

I got plane speed is  142.5 mph and wind speed is 7.5 mph.

 May 7, 2019
 #7
avatar+19914 
0

Correct-a-mundo! cheeky

ElectricPavlov  May 7, 2019
 #6
avatar+19914 
0

It is WAY easier to solve this NOT using the clumsy box method:

 

You have two equations:

300 = 2(s+w)     or    150 = s+w      (by dividing both sides of the equation by 2)

270 = 2 (s-w)     or    135 = s-w           <------ Now ADD the two equations together

                                 285 + 2s       thus s = 142.5 = plane speed in still air

 

 

 

I havn't any idea how to make this into a quadratic to solve by the Box method

(this ' box method', I think is crazy....sorry you are being taught this method)...

 May 7, 2019
 #8
avatar+1117 
+2

sorry, thx i got it!!!!!!!! we have to use the box method sadly.

SmartMathMan  May 7, 2019

4 Online Users

avatar