(a) Do Exercise using analytical techniques and change the second leg of the walk to\(25.0{\rm{ m}}\)straight south. (This is equivalent to subtracting\({\rm{B}}\)from\({\rm{A}}\)—that is, finding\({\rm{R'}} = {\rm{A}} - {\rm{B}}\))

(b) Repeat again, but now you first walk \(25.0{\rm{ m}}\) north and then \(18.0{\rm{ m}}\) east. (This is equivalent to subtract \({\rm{A}}\) from \({\rm{B}}\) —that is, to find \({\rm{A}} = {\rm{B}} + {\rm{C}}\). Is that consistent with your result?)

The resultant vector is the combination of two or more single vector having their own direction and magnitude.

• Magnitude of displacement towards west \(A = 18\,{\rm{m}}\).
• Magnitude of displacement towards south \(B = 25\,{\rm{m}}\).
• Magnitude of displacement towards east \(B' = 18\,{\rm{m}}\).
• Magnitude of displacement towards north \(A' = 25\,{\rm{m}}\).

The displacement\({\rm{A}}\)and\({\rm{B}}\)is represented as,

Figure: Vector representation

The magnitude of the resultant vector is,

\(R = \sqrt {{A^2} + {B^2}} \)

Here\(A\)is the magnitude of the displacement vector\({\rm{A}}\)(towards west), and\(B\)is the magnitude of the displacement vector\({\rm{B}}\)(towards south).

Substitute\(18{\rm{ m}}\)for\(A\)and\(25{\rm{ m}}\)for\(B\),

\(\begin{array}{c}R = \sqrt {{{\left( {18{\rm{ m}}} \right)}^2} + {{\left( {25{\rm{ m}}} \right)}^2}} \\ = 30.8{\rm{ m}}\end{array}\)

The direction of the resultant vector is,

\(\theta = {\tan ^{ - 1}}\left( {\frac{B}{A}} \right)\)

Substitute\(18{\rm{ m}}\)for\(A\)and\(25{\rm{ m}}\)for\(B\),

\(\begin{array}{c}\theta = {\tan ^{ - 1}}\left( {\frac{{25{\rm{ m}}}}{{18{\rm{ m}}}}} \right)\\ = 54.2^\circ \end{array}\)

Hence, the distance from the starting point is \(30.8{\rm{ m}}\), and the compass direction of a line connecting starting point to the final point is \(54.2^\circ \) south of west.

The displacement\({\rm{A'}}\)and\({\rm{B'}}\)is represented as,

Figure: Vector representation

The magnitude of the resultant vector is,

\(R' = \sqrt {{{A'}^2} + {{B'}^2}} \)

Here \(A'\) is the magnitude of the displacement vector \({\rm{A'}}\) (towards north) and \(B'\) is the magnitude of the displacement vector \({\rm{B'}}\) (towards east).

Substitute \(25{\rm{ m}}\) for \(A'\) and \(18{\rm{ m}}\) for \(B'\),

\(\begin{array}{c}R' = \sqrt {{{\left( {25{\rm{ m}}} \right)}^2} + {{\left( {18{\rm{ m}}} \right)}^2}} \\ = 30.8{\rm{ m}}\end{array}\)

The direction of the resultant vector is,

\[\varphi = {\tan ^{ - 1}}\left( {\frac{{B'}}{{A'}}} \right)\]

Substitute \(25{\rm{ m}}\) for \(A'\) and \(18{\rm{ m}}\) for \(B'\),

\(\begin{array}{c}\varphi = {\tan ^{ - 1}}\left( {\frac{{18{\rm{ m}}}}{{25{\rm{ m}}}}} \right)\\ = 35.8^\circ \end{array}\)

The angle of the resultant vector from the east is,

\(\theta = 90^\circ - \varphi \)

Substitute\(35.8^\circ \)for\(\varphi \),

\(\begin{array}{c}\theta = 90^\circ - 35.8^\circ \\ = 54.2^\circ \end{array}\)

Hence, the distance from the starting point is \(30.8{\rm{ m}}\), and the compass direction of a line connecting starting point to the final point is \(54.2^\circ \) north of east.