In an orienteering class, you have the goal of moving as far (straight-line distance) from base camp as possible by making three

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In an orienteering class, you have the goal of moving as far (straight-line distance) from base camp as possible by making three

straight-line moves. You may use the following displacements in any order: (a) →a, 2.0 km due east (directly toward the east); (b) →b, 2.0 km 30° north of east (at an angle of 30° toward the north from due east); (c) →c, 1.0 km due west. Alternatively, you may substitute either −→b for →b or −→c for →c. What is the greatest distance you can be from base camp at the end of the third displacement? (We are not concerned about the direction.)

Mathematics

1 answer:

lawyer [12.5K] · Answer 1 · 2026-04-01T13:16:39+03:00

Answer:

The maximum distance from the base camp at the conclusion of the third displacement is 6.69 km

Step-by-step explanation:

Each displacement can be represented as a vector, defined by magnitude and direction.

Vectors can be expressed in terms of their x and y coordinates as shown

$\vec{t}=(x, y)$

For displacements a and c, their vector coordinates would be:

$\vec{a}=(2, 0)$

$\vec{c}=(-1, 0)$

Given that displacement b is oriented at 30° north from due east, we can determine its x and y coordinates by applying the following formulas:

$x=(magnitude)*cos(angle)$

$y=(magnitude)*sin(angle)$

Note: the angle applied in the formula is that which is formed with the east point measured counterclockwise.

Thus, the x and y coordinates for displacement b would be:

$\vec{b}=(2*cos(30), 2*sin(30))$

As vector addition is commutative, the arrangement of displacements will not influence the final position; however, any directional alteration in a displacement will affect the resultant position. To ascertain the greatest distance, we ought to compute several combinations and identify the one yielding the largest magnitude:

$\vec{R_{1}} =\vec{a}+\vec{b}+\vec{c}$

$\vec{R_{2}} =\vec{a}-\vec{b}+\vec{c}$

$\vec{R_{3}} =\vec{a}+\vec{b}-\vec{c}$

$\vec{R_{4}} =\vec{a}-\vec{b}-\vec{c}$

Each resultant vector can be determined by summing up the respective components. Next, use the following formula to find the magnitude:

$|\vec{R}|=\sqrt[ ]{(R_{x})^2 +{(R_{y})^2}}$

Now, let’s execute the calculations!

$\vec{R_{1}} =\vec{a}+\vec{b}+\vec{c}$

$R_{1_x}} =2+2*cos(30)-1=2.73$

$R_{1_y}} =0+2*sin(30)+0=1$

$\vec{R_{1}}=(2.73,1)$

$|\vec{R_{1}}|=\sqrt[ ]{(2.73)^2 +{(1)^2}}=3.86$

$\vec{R_{2}} =\vec{a}-\vec{b}+\vec{c}$

$R_{2_x}} =2-2*cos(30)-1=0.73$

$R_{2_y}} =0-2*sin(30)+0=-1$

$\vec{R_{2}}=(-0.73,-1)$ }

$|\vec{R_{2}}|=\sqrt[ ]{(-0.73)^2 +{(-1)^2}}=1.03$

$\vec{R_{3}} =\vec{a}+\vec{b}-\vec{c}$

$R_{3_x}} =2+2*cos(30)+1=4.73$

$R_{3_y}} =0+2*sin(30)-0=1$

$\vec{R_{3}}=(4.73,1)$

$|\vec{R_{3}}|=\sqrt[ ]{(4.73)^2 +{(1)^2}}=6.69$

$\vec{R_{4}} =\vec{a}-\vec{b}-\vec{c}$

$R_{4_x}} =2-2*cos(30)+1=1.26$

$R_{4_y}} =0-2*sin(30)-0=1$

$\vec{R_{4}}=(1.26,-1)$

$|\vec{R_{4}}|=\sqrt[ ]{(1.26)^2 +{(-1)^2}}=1.79$

So, after performing all calculations, we can confirm that the vector $\vec{R_{3}}$ has the maximum magnitude. Therefore, the maximum distance possible from the base camp after the third displacement is 6.69 km