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Q3.3-18 PE

Expert-verifiedFound in: Page 122

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 \(7.24{\rm{ km}}\).

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

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

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

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 \(\theta \),

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

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 \(7.24{\rm{ km}}\).

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

Vector representation

The east component of the displacement vector is,

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

Substitute \(7.5{\rm{ km}}\) for \(S\),

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

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.

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