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Q3.3-13 PE

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Found in: Page 121

### College Physics (Urone)

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.

See the step by step solution

## 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.