A. A small plane departs from point A heading for an airport at point B 480 km due north. The airspeed of the plane is 200 km/h and there is a steady wind of 50 km/h blowing northwest to southeast. Determine the proper heading for the plane.
B. How long will the trip take?
To determine the proper heading for the plane in situation A, we need to consider the effect of the wind on the plane's trajectory. Since the wind is blowing from the northwest to the southeast, it will exert a force on the plane, pushing it slightly to the east as it flies north.
To find the proper heading, we can use vector addition. Let's break down the motion into two components: one in the north direction (plane's airspeed) and the other in the east direction (wind's speed and direction).
The north component of the plane's velocity is 200 km/h (its airspeed) since it is heading due north. The east component of the wind's velocity is 50 km/h multiplied by the cosine of 45 degrees (the angle it makes with the east-west axis), which is approximately 35.4 km/h.
Adding these two vectors together gives us:
Resultant velocity = √[(200 km/h)^2 + (35.4 km/h)^2]
= √[40000 km^2/h^2 + 1255.16 km^2/h^2]
= √41255.16 km^2/h^2
≈ 203 km/h
To find the proper heading, we can use trigonometry (specifically, the inverse tangent function) to determine the angle between the north axis and the resultant velocity vector:
Angle = tan^(-1) (35.4 km/h ÷ 200 km/h)
≈ 9.98 degrees
Therefore, the proper heading for the plane is a direction of approximately 9.98 degrees east of due north.
To calculate the time it will take for the plane to complete the trip in situation B, we can divide the distance between point A and point B by the speed of the plane:
Time = Distance ÷ Speed
= 480 km ÷ 200 km/h
= 2.4 hours
Therefore, the trip will take approximately 2.4 hours or 2 hours and 24 minutes.