an ultralight plane had been flying for 40 min when a change of wind direction doubled its ground speed. the entire trip of 160miles took 2h. how far did the plane travel during the first 40 min?

Let V be the initial ground speed and 2V be the final ground speed, in miles per minute.
The length of time travelled at the higher speed was 120 - 40 = 80 minutes
Total distance travelled =
160 = 40 V + 80*(2V) = 200 V
V = 0.8 miles/minute
Inital distance travelled at velocity V = 0.8 miles/min* 40 min = 32 miles
Final distance travelled at velocity 2V = 80*1.6 = 128 miles

To solve this problem, we can use the formula Distance = Speed * Time.

Let's assume that the initial ground speed of the plane is represented by V miles per minute. Since the change in wind direction doubled the ground speed, the final ground speed is 2V miles per minute.

We know that the entire trip of 160 miles took 2 hours, which is equivalent to 120 minutes.

Now, let's break down the trip into two parts:

1. The distance traveled during the first 40 minutes at the initial ground speed:
Distance = Speed * Time = V * 40 minutes = 40V miles.

2. The distance traveled during the remaining 80 minutes at the final ground speed:
Distance = Speed * Time = (2V) * 80 minutes = 160V miles.

Since the total distance traveled is 160 miles, we can write the equation:
Total Distance = Distance at initial speed + Distance at final speed
160 = 40V + 160V.

Simplifying the equation:
160 = 200V.

Dividing both sides of the equation by 200:
V = 0.8 miles per minute.

Now that we have the initial ground speed, we can calculate the distance traveled during the first 40 minutes:
Distance = Speed * Time = 0.8 miles per minute * 40 minutes = 32 miles.

Therefore, the plane traveled a distance of 32 miles during the first 40 minutes.