Question

You fly \(32.0{\rm{ km}}\) in a straight line in still air in the direction \(35.0^\circ \) south of west.

(a) Find the distances you would have to fly straight south and then straight west to arrive at the same point. (This determination is equivalent to finding the components of the displacement along the south and west directions.)

(b) Find the distances you would have to fly first in a direction \(45.0^\circ \) south of west and then in a direction \(45.0^\circ \) west of north. These are the components of the displacement along a different set of axes—one rotated \(45.0^\circ \).

Short answer

(a) The south component of the displacement is\(18.35\;{\rm{km}}\), and the west component of the displacement is\(26.21\;{\rm{km}}\).

(b) When the axis is rotated \(45.0^\circ \) counterclockwise, the southwest component of the displacement is \(31.51\;{\rm{km}}\), and the northwest component of the displacement is \(5.56\;{\rm{km}}\).

Step-by-step solution

Resolution of a vector

The splitting of a single vector into its two or more components in a particular direction, which, when added, will produce the same effect as produced by the single vector, is called the resolution of the vector.

Step 2: Given data

• Magnitude of the displacement vector\(S = 32.0\,{\rm{km}}\).
• Initial direction of the displacement vector is \(35.0^\circ \) south of west.
• New direction of the displacement vector is \(45.0^\circ \).

(a) Vector representation

The resultant vector is represented as,

Figure: Vector representation

South component of the displacement vector

The south component of the displacement vector is,

\({S_S} = S\sin \left( {35^\circ } \right)\)

Here\(S\)is the magnitude of the displacement vector\({\rm{S}}\).

Substitute\(32{\rm{ km}}\)for\(S\),

\(\begin{aligned}{}{S_S} = \left( {32{\rm{ km}}} \right) \times \sin \left( {35^\circ } \right)\\ = 18.35{\rm{ km}}\end{aligned}\)

West component of the displacement vector

The west component of the displacement vector is,

\({S_W} = S\cos \left( {35^\circ } \right)\)

Here\(S\)is the magnitude of the displacement vector\({\rm{S}}\).

Substitute\(32{\rm{ km}}\)for\(S\),

\(\begin{aligned}{}{S_W} = \left( {32{\rm{ km}}} \right) \times \cos \left( {35^\circ } \right)\\ = 26.21{\rm{ km}}\end{aligned}\)

Hence, the south component of the displacement is \(18.35{\rm{ km}}\) , and the west component of the displacement is \(26.21{\rm{ km}}\).

(b) When the axis is rotated

When the axis is rotated by \(45^\circ \), the displacement vector is represented as,

Figure: Vector representation

Angle between the displacement vector and southwest axis

The angle\(\theta \)between the displacement vector and southwest axis is,

\(\begin{aligned}{}\theta = 45^\circ - 35^\circ \\ = 10^\circ \end{aligned}\)

Southwest component of the displacement vector

The southwest component of the displacement vector is,

\({S_{SW}} = S\cos \theta \)

Here\(S\)is the magnitude of the displacement vector\({\rm{S}}\)and\(\theta \)is the angle between the displacement vector and the southwest axis.

Substitute\(32{\rm{ km}}\)for\(S\),\(10^\circ \)for\(\theta \),

\(\begin{aligned}{}{S_{SW}} = \left( {32{\rm{ km}}} \right) \times \cos \left( {10^\circ } \right)\\ = 31.51{\rm{ km}}\end{aligned}\)

Northwest component of the displacement vector

The northwest component of the displacement vector is,

\({S_{NW}} = S\sin \theta \)

Here\(S\)is the magnitude of the displacement vector\({\rm{S}}\)and\(\theta \)is the angle between the displacement vector and the southwest axis.

Substitute\(32{\rm{ km}}\)for\(S\),\(10^\circ \)for\(\theta \),

\(\begin{aligned}{}{S_{NW}} = \left( {32{\rm{ km}}} \right) \times \sin \left( {10^\circ } \right)\\ = 5.56{\rm{ km}}\end{aligned}\)

Hence, when the axis is rotated \(45.0^\circ \) counterclockwise, the southwest component of the displacement is \(31.51{\rm{ km}}\) , and the northwest component of the displacement is \(5.56{\rm{ km}}\).