Question

A Coast Guard helicopter patrolled the

shoreline at an air speed of 50 km/h. With
the wind, it traveled 121 km in the same time
that it took to go 99 km against the wind.
Find the speed of the wind.

Answer 1

Let's assume that the speed of the wind is "w" km/h.

Against the wind, the effective speed of the helicopter is its air speed minus the speed of the wind, which is 50 - w km/h.

With the wind, the effective speed of the helicopter is its air speed plus the speed of the wind, which is 50 + w km/h.

We can use the formula:

time = distance / speed

Let's call the time it took to travel both distances "t".

Against the wind: t = 99 / (50 - w)

With the wind: t = 121 / (50 + w)

Since both expressions for "t" are equal, we can set them equal to each other and solve for "w":

99 / (50 - w) = 121 / (50 + w)

4950 + 99w = 6050 - 121w

220w = 1100

w = 5

Therefore, the speed of the wind is 5 km/h.

Answer 2

5 km/h

Explanation:

Let's denote the speed of the wind as "w" in km/h.

When the helicopter flies with the wind, its effective ground speed is increased by the speed of the wind, so we can express its ground speed as 50+w km/h. When it flies against the wind, its effective ground speed is reduced by the speed of the wind, so we can express its ground speed as 50-w km/h.

According to the problem, the helicopter traveled 121 km with the wind and 99 km against the wind in the same amount of time. Let's call this time "t" in hours. Then we can use the formula:

distance = speed x time

to write two equations:

121 = (50+w) x t

99 = (50-w) x t

We want to solve for the wind speed, so let's rearrange the second equation to get:

t = 99 / (50-w)

Now we can substitute this expression for "t" into the first equation and solve for "w":

121 = (50+w) x (99 / (50-w))

121(50-w) = (50+w) x 99

6050 - 121w = 4950 + 99w

220w = 1100

w = 5

Therefore, the speed of the wind is 5 km/h.