An archiolohist begins a first trip by first walking 30km southeast from the campsite. He stops and sets up a temporal camp for the night.On the second day, he walks 60 km in direction 60degrees northeast, at which point he discovers the ancient city of gold.
(a) detaimine the components of the acheologist's desplacement for the day.
(b) detaimine the components of the acheologist desplacement R for the trip. Find the expression for R in terms of unit vectors.
a. 30km[315o,CCW] = 21.21 - 21.21i.
b. 60km[60o,CCW] = 30 + 52i.
R = 21.21-21.21i + 30+52i = 51.2 + 30.8i.
Convert to unit Vector: (51.2,30.8).
Tan A = Y/X = 30.8/51.2 = 0.60062, A = 31o.
R = = X/Cos A = 51.2/Cos31 = 59.7km[31o].
(x/59.7,y/59.7) = (51.2/59.7,30.8/59.7) = (0.86,0.52).
To determine the components of the archaeologist's displacement for the day, we need to break down the 60 km distance and the 60-degree direction into horizontal and vertical components.
(a) Components of the archaeologist's displacement for the day:
Let's consider the x-axis to represent the East-West direction, and the y-axis to represent the North-South direction.
The given direction is 60 degrees northeast, which means it forms a right-angled triangle with the horizontal and vertical axes.
The horizontal component can be determined by using the cosine function and the vertical component can be determined using the sine function.
Horizontal Component:
cos(60 degrees) = adjacent/hypotenuse
cos(60 degrees) = x/60 km (adjacent side is x)
Solving for x:
x = 60 km * cos(60 degrees)
x = 60 km * 0.5
x = 30 km
Therefore, the horizontal component of the archaeologist's displacement for the day is 30 km in the East direction.
Vertical Component:
sin(60 degrees) = opposite/hypotenuse
sin(60 degrees) = y/60 km (opposite side is y)
Solving for y:
y = 60 km * sin(60 degrees)
y = 60 km * (sqrt(3)/2)
y = 30 km * sqrt(3)
Therefore, the vertical component of the archaeologist's displacement for the day is 30 km * sqrt(3) in the North direction.
Hence, the archaeologist's displacement for the day is (30 km, 30 km * sqrt(3)).
(b) Components of the archaeologist's displacement R for the entire trip:
For this part, we need to consider the archaeologist's displacement from the campsite to the ancient city of gold. We have two displacements: 30 km southeast and (30 km, 30 km * sqrt(3)) as found in part (a).
The displacement vector R can be found by adding these two vectors.
Horizontal Component of R:
To find the horizontal component, we add the horizontal components of the two displacements.
R_x = 30 km (30 km southeast) + 30 km (East direction)
R_x = 30 km - 30 km (since east and southeast have the same direction)
R_x = 0 km
Vertical Component of R:
To find the vertical component, we add the vertical components of the two displacements.
R_y = 30 km * sqrt(3) (30 km southeast) + 30 km * sqrt(3) (North direction)
R_y = 30 km * sqrt(3) + 30 km * sqrt(3)
R_y = 60 km * sqrt(3)
Therefore, the components of the archaeologist's displacement R for the entire trip are (0 km, 60 km * sqrt(3)).
In terms of unit vectors, the expression for R would be:
R = 0 km * i + 60 km * sqrt(3) * j
R = 60 km * sqrt(3) * j, where i is the unit vector in the East-West direction and j is the unit vector in the North-South direction.