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Q3.3-13 PE

Expert-verifiedFound in: Page 121

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Find the following for path C 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.**

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

(a) The total distance traveled is \(1560\;{\rm{m}}\).

(b) The magnitude of displacement is \(120{\rm{ m}}\), and is directed towards the East.

**A moving object's entire route is defined as a scalar quantity called distance. It is a nonzero and positive quantity. The shortest path between a moving object's initial and ending positions is defined as displacement, which is a vector quantity. Under certain conditions, the displacement of a moving body is zero or a negative quantity.**

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

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

\(d = n \times L\)

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

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

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

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 C is,

\(\begin{array}{c}{s_x} = 5 \times \left( {120{\rm{ m}}} \right) - \left( {120{\rm{ m}}} \right) - 3 \times \left( {120{\rm{ m}}} \right)\\ = 120{\rm{ m}}\end{array}\)

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

\(\begin{array}{c}{s_y} = \left( {120{\rm{ m}}} \right) - 2 \times \left( {120{\rm{ m}}} \right) + \left( {120{\rm{ m}}} \right)\\ = 0\end{array}\)

The magnitude of the displacement is,

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

Substitute \(120{\rm{ m}}\) for \({s_x}\) and \(0\) for \({s_y}\),\({s_y}\)

\(\begin{array}{c}s = \sqrt {{{\left( {120{\rm{ m}}} \right)}^2} + {{\left( 0 \right)}^2}} \\ = 120{\rm{ m}}\end{array}\)

The direction of the displacement vector is,

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

Substitute \(120{\rm{ m}}\) for \({s_x}\) and \(0\) for ,

\(\begin{array}{c}\theta = {\tan ^{ - 1}}\left( {\frac{0}{{120{\rm{ m}}}}} \right)\\ = 0^\circ \end{array}\)

Hence, the magnitude of displacement is \(120{\rm{ m}}\), and is directed towards the East.

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