Question

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.)

Answer 1

Answer:

The greatest distance we can be from the base camp at the end of the third displacement is 6.69 km

Step-by-step explanation:

We can think of each displacement as a vector, with a given magnitude and direction.

A vector can be written using its x and y coordinates like this

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

So, for the displacements a and c their vector coordinates would be:

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

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

As the b displacement has an angle of 30° toward the north from due east, we can find its x and y coordinates using the following formulas:

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

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

Note: the angle in the formula is the one formed with the east measured counterclockwise.

So, the x and y coordinates for the b displacement will be:

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

As the vector additon is commutative, the order won't affect the final position. Nevertheless, any change in the direction of any displacement will change the final position. So, in order to find the combination greatest distance we should calculate the following additions and find the one with the greatest 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 found adding each component. Afterwards, the magnitude can be found using the following formula:

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

Now, let's calculate!

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

So, after all the calculation, we can know for sure that the vector $\vec{R_{3}}$ has the biggest magnitude. Then, the greatest distance we can be from the base camp at the end of the third displacement is 6.69 km