Find the following for path D in Figure

(a) the total distance traveled and

(b) the magnitude and direction of the displacement from start to finish.

In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition.

The various lines represent paths taken by different people walking in a city. All blocks are \(120{\rm{ m}}\) on a side.

Distance is the actual path traveled by a moving body. It is a non-zero and positive scalar quantity.

The shortest path by the object between the initial and final position is also called distance. The shortest path between two points is a straight line. It is a vector quantity, which can be zero or a negative quantity under certain conditions.

Length of blocks \(L = 120\,{\rm{m}}\).

The total distance traveled by a moving body following path D is,

\(d = n \times L\)

Here \(n\) is the number of blocks in path D and \(L\) is the length of one block.

Substitute \(13\) for \(n\) and \(120{\rm{ m}}\) for \(L\),

\(\begin{aligned}{}d = 13 \times \left( {120{\rm{ m}}} \right)\\ = 1560{\rm{ m}}\end{aligned}\)

Hence, the total distance traveled is\(1560{\rm{ m}}\).

The displacement of the moving object along \(x\) the direction (towards East from starting point) following path D is,

\(\begin{aligned}{}{s_x} = 6 \times \left( {120{\rm{ m}}} \right) - \left( {120{\rm{ m}}} \right)\\ = 600{\rm{ m}}\end{aligned}\)

The displacement of the moving object along \(y\) the direction (towards North from starting point) following path D is,

\(\begin{aligned}{}{s_y} = - 2 \times \left( {120{\rm{ m}}} \right) + 4 \times \left( {120{\rm{ m}}} \right)\\ = 240{\rm{ m}}\end{aligned}\)

The magnitude of the displacement is,

\[s = \sqrt {s_x^2 + s_y^2} \]

Substitute \({\rm{600 m}}\) for \({s_x}\) and \(240{\rm{ m}}\) for \({s_y}\),

\(\begin{aligned}{}s = \sqrt {{{\left( {600{\rm{ m}}} \right)}^2} + {{\left( {240{\rm{ m}}} \right)}^2}} \\ = 646.22{\rm{ m}}\end{aligned}\)

The direction of the displacement vector is,

\(\theta = {\tan ^{ - 1}}\left( {\frac{{{s_y}}}{{{s_x}}}} \right)\)

Substitute \({\rm{600 m}}\) for \({s_x}\) and \(240{\rm{ m}}\) for \({s_y}\),

\(\begin{aligned}{}\theta = {\tan ^{ - 1}}\left( {\frac{{240{\rm{ m}}}}{{600{\rm{ m}}}}} \right)\\ = 21.8^\circ \end{aligned}\)

Hence, the displacement following path D is \(646.22{\rm{ m}}\), along \(21.8^\circ \) the North of the East.