Question

A bird watcher meanders through the woods, walking 1.62 km due east, 0.246 km due south, and 3.08 km in a direction 50.4 ° north of west. The time required for this trip is 1.080 h. Determine the magnitudes of the bird watcher's (a) displacement and (b) average velocity.

Answer 1

Answer:

Displacement CO=2.15 Km

Average velocity = 1.99 Km/h

Explanation:

Given

Total time = 1.08 h

OA= 1.62 Km

AB=0.246 Km

BC=3.08 Km

θ=50.4°

CE=3.08 sin50.4° =2.37 Km

BE=3.08 cos50.4° =1.96 Km

CE=AB+OD

OD=CE-AB

OD=2.37-0.246

OD=2.124 Km

BE=CD+OA

CD=BE-OA

CD=1.96-1.62

CD=0.34 Km

Now CO

$CO=\sqrt{CD^2+OD^2}$

$CO=\sqrt{0.34^2+2.124^2}$

CO=2.15 Km

So the displacement CO=2.15 Km

Average velocity =Displacement/Time

Average velocity = 2.15/1.08 Km/h

Average velocity = 1.99 Km/h