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

`|\vec r|=339.82\ m`

`\theta=-6.67^o`

Step-by-step 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|=√(337.52^2+(-39.48)^2)=339.82\ m`

`\boxed=339.82\ m`

`\displaystyle tan\theta=(-39.48)/(337.52)=-0.1169`

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