Question 209818
Using {{{d = rt}}} where d = distance, r = rate(speed), and t = time of travel, we can express the two legs (out and return) of the airplane.
For the outbound leg, the speed of the airplane can be expressed as:
{{{r[1] = 180+w}}} The speed is faster because the wind speed is added to that of the air plane when flying with the wind.
For the return leg, the speed of the airplane can be expressed as:
{{{r[2] = 180-w}}} The speed is slower because the wind speed is subtracted from that of the airplane when flying against the wind.
Now the distance traveled is given as:
{{{d[1] = 7}}} and
{{{d[2] = 5}}} and the time of travel, t, is the same for both legs.
So, we have enough data to set up our two equations.
{{{d[1] = r[1]*t}}} and...
{{{d[2] = r[2]*t}}}
Making the appropriate substitutions, we get:
{{{7 = (180+w)*t}}} and...
{{{5 = (180-w)*t}}}
Solving both of the equations for t, we get:
{{{t = 7/(180+w)}}} and...
{{{t = 5/(180-w)}}}
Now since t = t we can set these two equations equal to each other to get...
{{{7/(180+w) = 5/(180-w)}}} Now we can solve for w, the wind speed.
Cross-multiply.
{{{7(180-w) = 5(180+w)}}} Simplify.
{{{1260-7w = 900+5w}}} Add 7w to both sides.
{{{1260 = 900+12w}}} Subtract 900 from both sides.
{{{360 = 12w}}} Finally, divide both sides by 12.
{{{30 = w}}} or {{{w = 30}}}
The wind speed is 30mph.