Question

A plane travels at a speed of 205mph in still air. Flying with a tailwind, the plane is clocked over a distance of 1000 miles. Flying against a headwind, it takes 2 hours longer to complete the return trip. What was the wind velocity? (Round your answer to the nearest tenth.)

Answer 1

While plane is moving under tailwind condition it took time "t"

so here we will have

$t = \frac{d}{v_{net}}$

here net speed of the plane will be given as

$v_{net} = v + v_w$

$t = \frac{1000}{205 + v_w}$

similarly when it moves under the condition of headwind its net speed is given as

$v_{net} = v - v_w$

now time taken to cover the distance is 2 hours more

$t + 2 = \frac{1000}{205 - v_w}$

now solving two equations

$\frac{1000}{205 + v_w} + 2 = \frac{1000}{205 - v_w}$

solving above for v_w we got

$v_w = 40.4 mph$