Question 1127565

A plane flies into the wind (against the wind current) for 3.5 hours for a distance of 1,260 miles. The return trip with a tailwind (with the wind current) only takes 3 hours. Find the average speed of the plane in still air and the speed of the wind current.
<pre>Let the speed of the plane, in still air, and the speed of the wind, be S and W, respectively
Then we get the following TIME equations: {{{matrix(2,10, S - W, "=", "1,260"/3.5, "=======>", S - W, "=", 360, "----", eq, "(i)", S + W, "=", "1,260"/3, "=======>", S + W, "=", 420, "----", eq, "(ii)")}}}
2S = 780 ------- Adding eqs (ii) & (i)
S, or speed, in still air = {{{highlight_green(matrix(1,4, 780/2, "=", 390, mph))}}}
W, or windspeed: {{{highlight_green(matrix(1,4, 420 - 390, "=", 30, mph))}}}