Question

A monkey running through the jungle goes 0.198 km straight to the East, then turns 15.8° deflection from straight East toward the South, and then goes 145 m in a straight line in the diagonal direction he is now facing. What is the displacement of the monkey? (put your answer in standard units)

Answer 1

Answer:

$|\vec r|=339.82\ m$

$\theta=-6.67^o$

Explanation:

Displacement

It's a vector magnitude that measures the space traveled by a particle between an initial and a final position. The total displacement can be obtained by adding the vectors of each individual displacement. In the case of two displacements:

$\vec r=\vec r_1+\vec r_2$

Given a vector as its polar coordinates (r,\theta), the corresponding rectangular coordinates are computed with

$x=rcos\theta$

$y=rsin\theta$

And the vector is expressed as

$\vec z=<x,y>=<rcos\theta,rsin\theta>$

The monkey first makes a displacement given by (0.198 km,0°). The angle is 0 because it goes to the East, the zero-reference for angles. Thus the first displacement is

$\vec r_1=<0.198cos0^o,0.198sin0^o>=<0.198,0>\ km=<198,0>\ m$

The second move is (145 m , -15.8°). The angle is negative because it points South of East. The second displacement is

$\vec r_2=<145cos(-15.8^o),145sin(-15.8^o)>=<139.52,-39.48>\ m$

The total displacement is

$\vec r=<198,0>\ m+<139.52,-39.48>\ m$

$\vec r=<337.52,-39.48>\ m$

In (magnitude,angle) form:

$|\vec r|=\sqrt{337.52^2+(-39.48)^2}=339.82\ m$

$\boxed{|\vec r|=339.82\ m}$

$\displaystyle tan\theta=\frac{-39.48}{337.52}=-0.1169$

$\boxed{\theta=-6.67^o}$