A gps tracker at their base camp shows the location of the first team as 38 km away, 23° north of west, and the second team as 34 km away, 31° 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?
using the base as the origin
1st ... 38 sin(23º) N , 38 cos(23º) W
2nd ... 34 cos(31º) N , 34 sin(31º) E
find the N/S and E/W distances , then use 1st as the origin
To find the distance and direction of the second team from the first team, we can use vector addition.
(a) Distance from the first team:
To find the distance, we need to use the concept of vector addition. We can break down the displacement of each team into its horizontal (east-west) and vertical (north-south) components.
For the first team, the displacement is given as 38 km, 23° north of west. We can represent this as a vector:
D₁ = 38 km 23° (270° - 23°) = 38 km 247°
For the second team, the displacement is given as 34 km, 31° east of north. We can represent this as a vector:
D₂ = 34 km 31°
Now, we can add these vectors to find the displacement between the two teams:
D = D₁ + D₂
To add these vectors, we first need to resolve them into their horizontal (east-west) and vertical (north-south) components.
Horizontal Component:
D₁x = 38 km * cos(247°) ≈ -30.59 km
D₂x = 34 km * sin(31°) ≈ 17.51 km
Vertical Component:
D₁y = 38 km * sin(247°) ≈ -23.50 km
D₂y = 34 km * cos(31°) ≈ 28.28 km
Now, we can add the horizontal and vertical components separately:
Dx = D₁x + D₂x ≈ -30.59 km + 17.51 km ≈ -13.08 km
Dy = D₁y + D₂y ≈ -23.50 km + 28.28 km ≈ 4.78 km
Finally, we can find the magnitude (distance) of the displacement using the Pythagorean theorem:
|D| = √(Dx² + Dy²)
|D| = √((-13.08 km)² + (4.78 km)²) ≈ 13.94 km
Therefore, the second team's distance from the first team is approximately 13.94 km.