Question

A hiker climbs a mountain. Starting at the base of the mountain, he first moved up 520m at a 32.0 degree angle. What is the final displacement of the hiker?

Answer 1

Answer:

$\displaystyle \vec{d}=<440.99\ m ,\ 275.6\ m>$

Explanation:

Displacement Vector

Suppose an object is located at a position  

$\displaystyle P_1(x_1,y_1)$

and then moves at another position at

$\displaystyle P_2(x_2,y_2)$

The displacement vector is directed from the first to the second position and can be found as

$\displaystyle \vec{d}=<x_2-x_1\ ,\ y_2-y_1>$

If the position is given as magnitude-angle data ( z , α), we can compute its rectangular components as

$\displaystyle x=z\ cos\alpha$

$\displaystyle y=z\ sin\alpha$

The question describes the situation where the initial point is the base of the mountain, where both components are zero

$\displaystyle P_1(0,0)$

The final point is given as a 520 m distance and a 32-degree angle, so  

$\displaystyle x_2=520\ cos32^o= 440.99\ m$

$\displaystyle y_2=520\ sin32^o=275.6\ m$

The displacement is

$\displaystyle \vec{d}=<440.99\ m ,\ 275.6\ m>$