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Answers without the blur. Sign up and see all textbooks for free! Q3.3-18 PE

Expert-verified Found in: Page 122 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # You drive $$7.50{\rm{ km}}$$ in a straight line in a direction $$15^\circ$$ east of north. (a) Find the distances you would have to drive straight east and then straight north to arrive at the same point. (This determination is equivalent to find the components of the displacement along the east and north directions.) (b) Show that you still arrive at the same point if the east and north legs are reversed in order.

(a) The component of displacement along the east is $$1.94{\rm{ km}}$$, and the component of the displacement along the north is $$7.24{\rm{ km}}$$.

(b) When the order is reversed, the component of displacement along the east is $$1.94{\rm{ km}}$$, and the component of the displacement along the north is $$7.24{\rm{ km}}$$.

See the step by step solution

## Step 1: Components of a vector

A vector quantity's components are scalars that help us visualize the vector's influence in a certain direction.

## Step 2: (a) Vector representation

When vector representation of the displacement, when we drive straight to east and then straight to the north is represented as, Vector representation of the motion

## Step 3: Given data

• Magnitude of vector $${\rm{S}}$$ $$S = 7.5\,{\rm{km}}$$.
• Direction of the vector is $$15^\circ \,{\rm{EN}}$$.

## Step 4: Component of the displacement along the east

The angle formed between the east and displacement vector $${\rm{S}}$$ is,

$$\begin{array}{c}\theta = 90^\circ - 15^\circ \\ = 75^\circ \end{array}$$

The east component of the displacement vector is,

$${S_E} = S\cos \theta$$

Substitute $$7.5{\rm{ km}}$$ for $$S$$, and $$75^\circ$$ for $$\theta$$,

$$\begin{array}{c}{S_E} = \left( {7.5{\rm{ km}}} \right) \times \cos \left( {75^\circ } \right)\\ = 1.94{\rm{ km}}\end{array}$$

## Step 5: Component of the displacement vector along the north

The north component of the displacement vector is,

$${S_N} = S\sin \theta$$

Substitute $$7.5{\rm{ km}}$$ for $$S$$, and $$75^\circ$$ for $$\theta$$,

$$\begin{array}{c}{S_N} = \left( {7.5{\rm{ km}}} \right) \times \sin \left( {75^\circ } \right)\\ = 7.24{\rm{ km}}\end{array}$$

Hence, the component of displacement along the east is $$1.94{\rm{ km}}$$, and the component of the displacement along the north is $$7.24{\rm{ km}}$$.

## Step 6: (b) Vector representation when the orders of the vector are reversed

When the order of the vectors is reversed, the vectors are represented as, Vector representation

## Step 7: Component of the displacement along the east

The east component of the displacement vector is,

$${S_E} = S\sin \left( {15^\circ } \right)$$

Substitute $$7.5{\rm{ km}}$$ for $$S$$,

$\begin{array}{c}{S_E} = \left( {7.5{\rm{ km}}} \right) \times \sin \left( {15^\circ } \right)\\ = 1.94{\rm{ km}}\end{array}$

## Step 8: Component of the displacement vector along the north

The north component of the displacement vector is,

$${S_N} = S\cos \left( {15^\circ } \right)$$

Substitute $$7.5{\rm{ km}}$$ for $$S$$,

$$\begin{array}{c}{S_N} = \left( {7.5{\rm{ km}}} \right) \times \cos \left( {15^\circ } \right)\\ = 7.24{\rm{ km}}\end{array}$$

When the order is reversed, the component of displacement along the east is $$1.94{\rm{ km}}$$, and the component of the displacement along the north is $$7.24{\rm{ km}}$$.

Hence, you still arrive at the same point if the east and north legs are reversed in order. ### Want to see more solutions like these? 