An airplane flying at 390 mph has a bearing of N 28°W. After flying 3 hours, how far north and how far west has the plane traveled from its point of departure?
V = 390mi/h[28o]N of W.
D = 390mi/h[28o] * 3h = 1170 mi[28o]N. of W.
Dn = 1170*Cos28 =
Dw = 1170*sin28 =
To find out how far the airplane has traveled north and west, we can use trigonometry and basic motion equations. Let's break down the problem step by step:
Step 1: Find the distance traveled (D) by multiplying the speed (S) of the airplane by the time (T) it has been flying. In this case, speed (S) is given as 390 mph, and the time (T) is 3 hours. Therefore, the distance traveled (D) can be calculated as follows:
D = S * T
D = 390 mph * 3 hours
D = 1170 miles
So, the airplane has traveled a total distance of 1170 miles from its point of departure.
Step 2: Determine the north and west components of the distance traveled. To do this, we need to use trigonometry.
Considering the bearing of N 28°W, we can divide it into two components: the north component and the west component.
The north component is given by D * sin(bearing), where "bearing" is the angle N 28°W. The west component is given by D * cos(bearing).
Calculating the north component:
North component = 1170 miles * sin(28°)
North component ≈ 543.49 miles (rounded to two decimal places)
Calculating the west component:
West component = 1170 miles * cos(28°)
West component ≈ 1038.80 miles (rounded to two decimal places)
Therefore, the airplane has traveled approximately 543.49 miles north and 1038.80 miles west from its point of departure.