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Q5PE

Expert-verifiedFound in: Page 121

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Suppose you first walk 12.0 m in a direction 20ΒΊ west of north and then 20.0 m in a direction 40.0ΒΊ south of west. How far are you from your starting point and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A and B , as in Figure 3.56, then this problem finds their sum R = A + B.)**

You are at \(19.5\;{\rm{m}}\) from the starting point and direction of compass is \(4.65^\circ \) from west of south.

The displacement along west of north is \(A = 12\;{\rm{m}}\)

The direction of displacement along west of north is\(\alpha = 20^\circ \).

The displacement along south of west is \(B = 20\;{\rm{m}}\)

The direction of displacement along south of west is\(\beta = 40^\circ \).

**The displacement is resolved along horizontal and vertical direction then sum of all the horizontal components gives horizontal component of net displacement. The sum of vertical components gives vertical component of net displacement.**

The horizontal component of the net displacement is given as:

\(\begin{align}{}{R_x} &= - A\cos \left( {90^\circ - \alpha } \right) - B\cos \beta \\{R_x} &= - \left( {A\sin \alpha + B\cos \beta } \right)\end{align}\)

Substitute all the values in the above equation.

\(\begin{align}{}{R_x} &= - \left( {\left( {12\;{\rm{m}}} \right)\left( {\sin 20^\circ } \right) + \left( {20\;{\rm{m}}} \right)\left( {\cos 40^\circ } \right)} \right)\\{R_x} &= - 19.43\;{\rm{m}}\end{align}\)

The vertical component of the net displacement is given as:

\(\begin{align}{}{R_y} &= A\sin \left( {90^\circ - \alpha } \right) - B\sin \beta \\{R_y} &= A\cos \alpha - B\sin \beta \end{align}\)

Substitute all the values in the above equation.

\(\begin{align}{}{R_y} &= \left( {12\;{\rm{m}}} \right)\left( {\cos 20^\circ } \right) - \left( {20\;{\rm{m}}} \right)\left( {\sin 40^\circ } \right)\\{R_y} &= - 1.58\;{\rm{m}}\end{align}\)

The distance from starting point is given as:

\(R = \sqrt {R_x^2 + R_y^2} \)

Substitute all the values in the above equation.

\(\begin{align}{}R &= \sqrt {{{\left( { - 19.43\;{\rm{m}}} \right)}^2} + {{\left( { - 1.58\;{\rm{m}}} \right)}^2}} \\R &= 19.5\;{\rm{m}}\end{align}\)

The direction of compass from starting point is given as:

\(\tan \theta = \frac{{{R_y}}}{{{R_x}}}\)

Substitute all the values in the above equation.

\(\begin{align}{}\tan \theta = \frac{{ - 1.58\;{\rm{m}}}}{{ - 19.43\;{\rm{m}}}}\\\theta = 4.65^\circ \end{align}\)

Therefore, you are at \(19.5\;{\rm{m}}\) from the starting point and direction of compass is \(4.65^\circ \) from west of south.

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