Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

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.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Distance and displacement

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.

Step 2: Given data

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

Step 3: (a) Determine the total distance

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 \(1560{\rm{ m}}\).

Step 4: (b) Determine the displacement

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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College Physics (Urone)

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Short Answer

Step by Step Solution

Step 1: Distance and displacement

Step 2: Given data

Step 3: (a) Determine the total distance

Step 4: (b) Determine the displacement

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College Physics (Urone)

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Short Answer

Step by Step Solution

Step 1: Distance and displacement

Step 2: Given data

Step 3: (a) Determine the total distance

Step 4: (b) Determine the displacement

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