An airplane maintains a speed of 615 km/h relative to the air it is flying throu
ID: 1659318 • Letter: A
Question
An airplane maintains a speed of 615 km/h relative to the air it is flying through as it makes a trip to a city 759 km away to the north. (Assume north is the positive y-direction and east is the positive x-direction.)
a) What time interval is required for the trip if the plane flies through a headwind blowing at 38.1 km/h toward the south?
b) What time interval is required if there is a tailwind with the same speed?
c) What time interval is required if there is a crosswind blowing at 38.1 km/h to the east relative to the ground?
Explanation / Answer
If the plane's relative speed is 615, then it means that the sum of its actual speed and the wind speed = 615
If actual speed is V, then,
V + 38.1 = 615
V = 573.9
Since we don't have any information about the air drag, we can ignore it and calculate the time taken by just using the plane's actual speed...
So, time taken = 759/573.9 = 1.32 hours
b) In this case, since the wind is blowing in the same direction as the plane, the relative speed = difference of their actual speeds.
V - 38.1 = 615
V = 653.1
Time taken = 759/653.1 = 1.16 hours
c) In this case, since the wind is blowing perpendicular to the direction of the plane, there is no component of wind velocity vector relative to the plane's velocity... as a result, the plane's speed is same as its relative speed to the wind.
time taken = 759/615 = 1.23 hours