Question

reviews the approach taken in problems such as this one. A bird watcher meanders through the woods, walking 0.916 km due east, 0.918 km due south, and 3.52 km in a direction 49.7 ° north of west. The time required for this trip is 1.750 h. Determine the magnitudes of the bird watcher's (a) displacement and (b) average velocity.

Answer 1

Answer:

Displacement: 2.230 km    Average velocity: 1.274$\frac{km}{h}$

Explanation:

Let's represent displacement by the letter S and the displacement in direction 49.7° as A. Displaement is a vector, so we need to decompose all the bird's displacement into their X-Y compoments. Let's go one by one:

• 0.916 km due east is an horizontal direction and cane be seen as  direction towards the negative side of X-axis.

• 0.928 km due south is a vertical direction and can be seen as a direction towards the negative side of Y-axis.
• 3.52 km in a direction of 49.7° has components on X and Y  axes. It is necessary to break it down using trigonometry,

First of all. We need to sum all the X components and all the Y componets.

∑$Sx = Ax -0.916$ ⇒  ∑$Sx = [tex]3.52cos(49.7) - 0.916$

∑$Sx = 1.361 km$

∑$Sy = Ay - 0.918$ ⇒ ∑$Sy = 3.52sin(49.7) - 0.918$

∑$Sy = 1.767$

The total displacement is calculated using Pythagoeran therorem:

$S_{total} =\sqrt{Sx^{2}+ Sy^{2} }$ ⇒

$S_{total} = 2.230 km$

With displacement calculated, we can find the average speed as follows:

$V = S/t$  ⇒  $V = \frac{2.230}{1.750}$

$V = 1.274\frac{km}{h}$