Chapter 3, Problem 14P

A hiker starts at his camp and moves the following distances while exploring his surroundings: 75.0 m north, 2.50 × 10² m east, 125 m at an angle 30.0° north of east, and 1.50 × 10² m south. (a) Find his resultant displacement from camp. (Take east as the positive x-direction and north as the positive y-direction.) (b) Would changes in the order in which the hiker makes the given displacements alter his final position? Explain.

To determine

The resultant displacement.

Answer to Problem 14P

Solution: The magnitude and direction of resultant displacement is $358\text{m}$ and $-2.00°$ respectively.

Explanation of Solution

Given Info: The distance travelled in north direction is $75.0\text{m}$, distance travelled in east direction is $2.50\times {10}^2\text{m}$,distance travelled in north east direction at angle $30.0°$ is $125\text{m}$ and the distance travelled in south direction is $1.50\times {10}^2\text{m}$.

Let the west, east, north and south directions corresponds to negative x-axis, positive x-axis, positive y-axis and negative y-axis respectively. There will be no x-component for displacements in north and south direction. The y-component is absent for displacement in east direction. There will be both and y-components for displacement in north east direction.

Write the formula to calculate the resultant displacement.

$R=\sqrt{R_{\text{x}}{}^2+R_{\text{y}}{}^2}$ (I)

• $R$ is the magnitude of resultant displacement
• $R_{\text{x}}$ is the x-component of resultant displacement
• $R_{\text{y}}$ is the y-component of resultant displacement

Write the formula to calculate $R_{\text{x}}$.

$R_{\text{x}}=A_{\text{x}}+B_{\text{x}}+C_{\text{x}}+D_{\text{x}}$      (II)

• $A_{\text{x}}$ is the x-component of displacement in north direction
• $B_{\text{x}}$ is the x-component of displacement in east direction
• $C_{\text{x}}$ is the x-component of displacement in north east direction at angle $30.0°$
• $D_{\text{x}}$ is the x-component of displacement in south direction

Write the formula to calculate $R_{\text{y}}$.

$R_{\text{y}}=A_{\text{y}}+B_{\text{y}}+C_{\text{y}}+D_{\text{y}}$ (III)

• $A_{\text{y}}$ is the y-component of displacement in north direction
• $B_{\text{y}}$ is the y-component of displacement in east direction
• $C_{\text{y}}$ is the y-component of displacement in north east direction at angle $30.0°$
• $D_{\text{y}}$ is the y-component of displacement in south direction

Write the formula to calculate $C_{\text{x}}$.

$C_{\text{x}}=C\cos \theta$ (IV)

• $C$ is the magnitude of displacement in north east direction at angle $30.0°$
• $\theta$ is the angle made by $\overrightarrow{C}$ with positive x-axis

Write the formula to calculate $C_{\text{y}}$.

$C_{\text{x}}=C\sin \theta$  (V)

Rewrite the equation (I) by substituting equation (II), equation (III), equation (IV) and equation (V).

$R=\sqrt{{\left(A_{\text{x}}+B_{\text{x}}+C\cos \theta +D_{\text{x}}\right)}^2+{\left(A_{\text{y}}+B_{\text{y}}+C\sin \theta +D_{\text{y}}\right)}^2}$

Substitute $0\text{m}$ for $A_{\text{x}}$, $75.0\text{m}$ for $A_{\text{y}}$, $2.50\times {10}^2\text{m}$ for $B_{\text{x}}$, $0\text{m}$ for $B_{\text{y}}$, $125\text{m}$ for $C$, $30.0°$ for $\theta$, $0\text{m}$ for $D_{\text{x}}$ and $-1.50\times {10}^2\text{m}$ for $D_{\text{y}}$ in the above equation to find $R$.

$\begin{array}{c}R=\sqrt{\left\{\begin{array}{l}{\left(0\text{m}+2.50\times {10}^2\text{m}+\left(125\text{m}\right)\cos 30.0°+0\text{m}\right)}^2\\ +{\left(75.0\text{m}+0\text{m}+\left(125\text{m}\right)\sin 30.0°-1.50\times {10}^2\text{m}\right)}^2\end{array}\right\}}\\ =\sqrt{{\left(250\text{m}+108.25\text{m}\right)}^2+{\left(75.0\text{m+62}\text{.5}\text{m}-150\text{m}\right)}^2}\\ =\sqrt{{\left(358.25\text{m}\right)}^{\text{2}}\text{+}{\left(-12.5\text{m}\right)}^2}\\ =358\text{m}\end{array}$

Write the formula to calculate the direction of a vector.

$\varphi ={\tan }^{-1}\left(\frac{R_{\text{y}}}{R_{\text{x}}}\right)$

• $\varphi$ is the angle made by the net displacement vector with positive x-axis measured in counter clockwise direction

Rewrite the above equation by substituting equation (II) and equation (III).

$\varphi ={\tan }^{-1}\left(\frac{\left(A_{\text{y}}+B_{\text{y}}+C\sin \theta +D_{\text{y}}\right)}{\left(A_{\text{x}}+B_{\text{x}}+C_{\text{x}}+D_{\text{x}}\right)}\right)$

Substitute $0\text{m}$ for $A_{\text{x}}$, $75.0\text{m}$ for $A_{\text{y}}$, $2.50\times {10}^2\text{m}$ for $B_{\text{x}}$, $0\text{m}$ for $B_{\text{y}}$, $125\text{m}$ for $C$, $30.0°$ for $\theta$, $0\text{m}$ for $D_{\text{x}}$ and $-1.50\times {10}^2\text{m}$ for $D_{\text{y}}$ in the above equation to find

$\begin{array}{c}\varphi ={\tan }^{-1}\left(\frac{\left(75.0\text{m}+0\text{m}+\left(125\text{m}\right)\sin 30.0°-1.50\times {10}^2{\text{m}}_{\text{y}}\right)}{\left(0\text{m}+2.50\times {10}^2\text{m}+\left(125\text{m}\right)\cos 30.0°+0\text{m}\right)}\right)\\ ={\tan }^{-1}\left(\frac{-12.5\text{m}}{358.25\text{m}}\right)\\ ={\tan }^{-1}\left(-0.0349\right)\\ \approx -2.00°\end{array}$

Conclusion:

Thus, the magnitude and direction of net displacement is $358.22\text{m}$ and $-2.00°$ respectively.

To determine

Whether the change in order of displacements of hiker can alter his final position.

Answer to Problem 14P

The final position of hiker does not change with the order of displacements.

Explanation of Solution

If the result is same even the order of elements is altered, the corresponding mathematical operation is said to be commutative. If the result is sensitive to the order of elements, mathematical operation is said to be non-commutative. Addition is a mathematical operation that satisfies law of commutation.

The law is satisfied by the addition of numbers, vectors and matrices. It is found that the magnitude and direction of resultant vector will be the same for any order of vectors. This means that the final position will be the same for any order of displacements.

Conclusion:

Thus, the final position of hiker does not change with the order of displacements