In the absence of wind a pigeon flies between the two cities for t = 60 minutes and it flies against the wind for t2 = 75 minutes. How long does it take for the pigeon to overcome this distance with the wind?

###### Solution.

The first equation of motion of the pigeon is given by the formula

X = V * t ,

where X – the distance between the two cities, V – the velocity of the pigeon in the absence of wind and t – the time it takes the pigeon to move between the two cities.

In the case when the pigeon flies against the wind the second equation of motion of the pigeon is given by the formula

X = (V – V_{w}) * t2 ,

where V_{w} – the velocity of the wind and t2 – the time it takes the pigeon to move between the two cities against the wind.

In the case when the pigeon flies with the wind the third equation of motion of the pigeon is given by the formula

X = (V + V_{w}) * t1 ,

where t1 – the time it takes the pigeon to move between the two cities with the wind.

Let’s start with second and third equations

X = (V – V_{w}) * t2 = (V + V_{w}) * t1

Let’s remove the parentheses and find the new factors

V * t2 – V_{w} * t2 = V * t1 + V_{w} * t1

V * (t2 – t1) = V_{w} * (t1 + t2)

V_{w} = V * (t2 – t1) / (t1 + t2)

Substitute it in the second equation

X = (V – V * (t2 – t1) / (t1 + t2) ) * t2

and after expanding and simplifying

X = 2 * V * t1 * t2 / (t1 + t2)

Now, we have to substitute it in the first equation

V * t = 2 * V * t1 * t2 / (t1 + t2)

we got

t = 2 * t1 * t2 / (t1 + t2)

Hence,

t1 = t * t2 / (2 * t2 – t)

t1 = 60 * 75 / (2 * 75 – 60) = 50 minutes.