Question

A hiker leaves her camp and walks 3.5 km in a direction of 55° south of west to the lake. After a short rest at the lake, she hikes 2.7 km in a direction of 16° east of south to the scenic overlook.
What is the magnitude of the hiker’s resultant displacement? Round your answer to the nearest tenth.
What is the direction of the hiker’s resultant displacement? Round your answer to nearest whole degree.

Answer 1

Answer:

The magnitude of the hiker’s resultant displacement is 5.6 km

The direction of the hiker’s resultant displacement is 77 degrees

Explanation:

1) Magnitude

Let's take south as positive y-direction and east as positive x-direction. Then we have to resolve both displacements into their respective components:

d_{1x} = -(3.5 km) cos 55^{\circ}=-2.0 km

d_{1y} = (3.5 km) sin 55^{\circ}=2.87 km

d_{2x} = (2.7 km) sin 16^{\circ}=0.74 km

d_{2y} = (2.7 km) cos 16^{\circ}=2.60 km

So, the components of the total displacement are

d_x = d_{1x}+d_{2x}=-2.0 km +0.74 km=-1.26 km east (so, 1.26 km west)

d_y=d_{1y}+d_{2y}=2.87 km + 2.60 km=5.47 km south

So, the magnitude of the resultant displacement is

d=\sqrt{d_x^2+d_y^2}=\sqrt{(1.26)^2+(5.47)^2}=5.61 km

2) Direction

the direction of the hiker's displacement is

\theta= arctan(\frac{d_y}{d_x})=arctan(\frac{5.47}{1.26})=arctan(4.34)=77.0^{\circ} south of west.

Answer 2

77 degrees

Given:

A hiker leaves her camp and walks 3.5 km in a direction of 55° south of west to the lake

After a short rest at the lake, she hikes 2.7 km in a direction of 16° east of south to the scenic overlook

To find:

What is the magnitude of the hiker’s resultant displacement?

What is the direction of the hiker’s resultant displacement?

Solution:

1) The magnitude

Consider south to be a positive y-direction and east to be a positive x-direction. Then we must decompose both displacements into their constituents:

As a result, the components of total displacement are east (so, 1.26 km west)

As a result, the magnitude of the resulting displacement is to the south.

2) Direction

The hiker's displacement is to the south of west.

$d_{1x} =-(3.5km) cos55 = -2.0km\\d_{1y} = (3.5km)sin55 = 2.87km\\d_{2x} = (2.7km) sin16 = 0.74kmd_{2y} = (2.7km)cos16 = 2.60km \\d_{x} = d_{1x} + d_{2x} = -2.0km +0.74km = -1.26km d_{y} =d_{1y} + d_{2y} = 2.87km +2.60km = 5.47km \\d= \sqrt{d_{x} ^{2}+d_{y} ^{2} } \\=\sqrt{(1.26)^2 + (5.47)^2} = 5.61km$

⇒ Theta = arctan($\frac{d_{y} }{d_{x} }$)

Theta = 77° .

HENCE PROVED.

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