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klio [65]
3 years ago
6

Two geological field teams are working in a remote area. A global positioning system (GPS) tracker at their base camp shows the

location of the first team as 37 km away, 21° north of west, and the second team as 32 km away, 38° east of north. When the first team uses its GPS to check the position of the second team, what does it give for the second team's (a) distance from them and (b) direction, measured from due east?
Physics
1 answer:
Masja [62]3 years ago
6 0

Answer:

distance of 2nd team from 1st team will be:  58.2

Direction of 2nd team from 1st team will be:  14.90 deg North of east

Explanation:

ASSUME Vector is R and  makes angle A with +x-axis,

therefore component of vector R is

R_x = Rcos A

R_y = Rsin A

From above relation

Assuming base camp as the origin, location of 1st team is

R_1 = 37 km away at 21 deg North of west (North of west is in 2nd quadrant, So x is -ve and y is positive)

R_{1x} = -R_1*cos A_1 = -37*cos 21 deg = -34.54 km

R_{1y} = R_1*sin A_1 = 37*sin 21 deg = 13.25 km

location of 2nd team is at

R_2 = 32 km, at 38 deg East of North = 32 km, at 58 deg North of east (North of east is in 1st quadrant, So x and y both are +ve)

R_{2x} = R_2*cos A_2 = 32*cos 58 deg = 16.95 km

R_{2y} = R_2*sin A_2 = 32*sin 58 deg = 27.13 km

Now position of 2nd team with respect to 1st team will be given by:

R_3 = R_2 - R_1

R_3 = (R_{2x} - R_{1x}) i + (R_{2y} - R_{1y}) j

Using above values:

R_3 = (16.95 - (-34.54)) i + (27.13 - 13.42) j

R_3 = 51.49 i + 13.71 j

distance of 2nd team from 1st team will be:

\left | R_3 \right | = \sqrt (51.49^2 +13.71^2)

\left | R_3 \right | = 53.28 km = 58.2 km

Direction of 2nd team from 1st team will be:

Direction = tan^{-1} \frac{R_{3y}}{R_{3x}} = tan^{-1}[ \frac{13.71}{51.49}]

Direction = 14.90 deg North of east

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A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional d
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Answer:

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  Then 3.50 km southeast, means 3.5 km 315⁰ to + ve X axis

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  Let third displacement be x i + y j

  We have final displacement = 5.80 km east = 5.80 i

  From summation we have total displacement = 2 i + 2.47 i - 2.47 j + x i + y j

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  Comparing both , we have 4.47+x = 5.80

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  So third displacement = 1.33 i + 2.47 j

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  θ = tan⁻¹(2.47/1.33) = 61.70°

  So third displacement = 2.81 m which is 61.70° north of east.


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