   Chapter 12.2, Problem 34E

Chapter
Section
Textbook Problem

# The magnitude of a velocity vector is called speed. Suppose that a wind is blowing from the direction N45°W at a speed of 50 km/h. (This means that the direction from which the wind blows is 45° west of the northerly direction.) A pilot is steering a plane in the direction N60°E at an airspeed (speed in still air) of 250 km/h. The true course, or track, of the plane is the direction of the resultant of the velocity vectors of the plane and the wind. The ground speed of the plane is the magnitude of the resultant. Find the true course and the ground speed of the plane.

To determine

To find: The true course and ground speed of plane.

Explanation

Given data:

The speed of wind is 50kmh in direction of N45°W an speed of plane is 250kmh in direction of N60°E

Formula used:

Consider the two two-dimensional vectors such as a=a1,a2 and b=b1,b2 .

The vector sum of two vectors (a+b) is,

a+b=a1,a2+b1,b2=a1+b1,a2+b2

Here,

a1 and b1 are x-coordinates, and

a2 and b2 are y-coordinates.

Write the expression for magnitude of vector a (|a|) .

|a|=a12+a22 (1)

Write the expression for angle of vector a (θ) .

θ=tan1(a2a1) (2)

Consider the east direction is located on positive x-axis and north direction is located on positive y-axis.

Draw the vector diagram of velocities as shown in Figure 1.

From Figure 1, write the expression for velocity vector of wind (vwind) .

vwind=|vwind|(cosθwindisinθwindj)

Here,

|vwind| is speed of wind, and

θwind is angle of wind.

Substitute 50kmh for |vwind| and 45° for θwind ,

vwind=50(cos(45°)isin(45°)j)=50(12i12j) {cos(45°)=12,sin(45°)=12}=252i252j

From Figure 1, write the expression for velocity vector of plane (vplane) .

vplane=|vplane|(cosθplanei+sinθplanej) (3)

Here,

|vplane| is speed of plane, and

θplane is angle of plane

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