The 4200 km trip from new york to san francisco takes 7 h flying against the wind, but only 6 h returning. How do you find the speed of the plane in still air and the wind speed?

1 Answer
Apr 26, 2017

Answer:

Plane speed #=514 2/7" Km/h" "~~514.286# to 3 decimal places
Wind speed #=85 5/7" Km/h "~~85.714# to 3 decimal places

Explanation:

To solve these equations we need to manipulate so that we have just 1 unknown and its relationship to some values.

Assumption: the speed (velocity) of the wind is constant

Let the speed of the plane be #p#
Let the speed of the wind be #w#
Let time going be in hours be #t_g->7h#
Let the time returning in hours be #t_r->6h#

Note that speed = #("distance")/("time")#

So #"distance "="time "xx" speed"#

Consider going: relative to the ground, the actual speed is #p-w#
Consider returning: relative to the ground the actual speed is #p+w#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the ratio of wind speed to plane speed")#
We need this to change things so that, by substitution, we end up with just 1 unknown.

So for going we have:
#" "4200=7xx(p-w)" ".........Equation(1)#

And for returning we have:
#" "4200=6xx(p+w)" "............Equation(2)#

Equation both to each other through distance

#7xx(p-w)=4200=6xx(p+w)#

#7xx(p-w)=6xx(p+w)#

#7p-7w=6p+6w#

Subtract #6p# from both sides

#7p-6p=0+6w#

#p=6w" "...........Equation(3)#
#w=p/6" "...........Equation(4)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determining the actual speeds")#

Using Equation(3) substitute for #p# in Equation(1) giving

#4200=7(p+w)" "->" "4200=7(6w+w)#
#" "4200=49w#

#w=4200/49=85.71428.... = 85 5/7" Km/h"#

Using Equation(4) substitute for #w# in Equation(1) giving

#4200=7(p+w)" "->" "4200=7(p+p/6)#
#" "4200=49/6 p#

#p=(6xx4200)/49=514.2857... =514 2/7" Km/h"#