The pilot of an airplane wishes to fly due north, but there is a 71 km/h wind blowing toward the east.
(a) In what direction should the pilot head her plane if its speed relative to the air is 314 km/h?
° west of north
Should the angle just be the opposite of the one created by the east wind? I'm not sure how to "undo" it
To determine the direction the pilot should head, we need to consider the effect of the wind on the plane's actual motion.
Let's break down the given information:
- The wind is blowing toward the east with a speed of 71 km/h.
- The plane's speed relative to the air is 314 km/h.
- The pilot wishes to fly due north.
To "undo" the effect of the wind, we can subtract the wind's velocity vector from the plane's velocity vector.
Given that the wind is blowing east, its velocity vector can be represented as 71 km/h east (or -71 km/h west).
To find the resulting velocity vector of the plane, we subtract the wind's velocity vector from the plane's velocity vector:
Plane's velocity vector = 314 km/h north
Wind's velocity vector = -71 km/h west
Resultant velocity vector = Plane's velocity vector + Wind's velocity vector
= 314 km/h north + (-71 km/h west)
To add these vectors, we can visualize them on a coordinate system. One vector points in the north direction, while the other points toward the west.
By using Pythagorean's theorem, we can calculate the magnitude of the resultant vector:
Magnitude = sqrt((314 km/h)^2 + (-71 km/h)^2)
= sqrt(98596 km^2/h^2 + 5041 km^2/h^2)
= sqrt(103637 km^2/h^2)
≈ 322.05 km/h
To determine the direction, we use trigonometry. The angle between the north direction and the resultant vector can be found using the inverse tangent function:
Angle = atan((71 km/h) / (314 km/h))
≈ 12.86 degrees
Since the wind is blowing towards the east, we need to subtract the angle from 90 degrees (north) to find the direction relative to north that the pilot should head:
Direction = 90 degrees - 12.86 degrees
≈ 77.14 degrees
Therefore, the pilot should head the plane in a direction approximately 77.14 degrees west of north.
To determine the direction the pilot should head her plane, we can consider the relative velocities of the plane and the wind.
First, let's break down the velocities into their north-south and east-west components.
The speed of the plane relative to the air is given as 314 km/h. Since the pilot wants to fly due north, we can say that the plane's north-south velocity (VP-NS) is 314 km/h.
The wind is blowing towards the east with a speed of 71 km/h. This means the wind's east-west velocity (VW-EW) is 71 km/h.
To find the resultant velocity (VR) of the plane in terms of its north-south and east-west components, we can use vector addition.
VR-NS = VP-NS
VR-EW = VW-EW
Now, to cancel out the effect of the eastward wind velocity, we need to find the opposite (or negation) of the wind's east-west velocity.
Negating the wind's east-west velocity, we get -71 km/h.
Since the plane's east-west component is now opposite to that of the wind, we need to add the negated east-west velocity to the plane's east-west component:
VR-EW = VP-EW + (-71 km/h)
Now, if we are assuming that the plane's east-west component relative to the air is zero (as the pilot wishes to fly due north), we can solve for VP-EW:
VP-EW + (-71 km/h) = 0
Rearranging the equation, we get:
VP-EW = 71 km/h
Therefore, the pilot should head her plane with an east-west component (VP-EW) of 71 km/h. Since the pilot wishes to fly due north, this means the pilot should head the plane west of north.
Hence, the direction the pilot should head her plane is west of north (° west of north).
Vp + Vw = 314i
Vp + 71 = 314i
Vp = -71 + 314i
Tan Ar = 314/-71 = -4.42254
Ar = -77.26o = Reference angle.
A = -77.26 + 180 = 102.74o CCW = 12.74o
W. of N.