Answer:
1. The magnitude of the vector is 26.9 m
2. The magnitude of direction of the vector is 21.8°.
Explanation:
From the question given, we obtained the following data:
Displacement along x (Bₓ) = 25 m
Displacement along y (Bᵧ) = 10 m
Magnitude of displacement (B) =?
Magnitude of direction (θ) =.?
1. Determination of the magnitude of displacement of the vector (B)
Displacement along x (Bₓ) = 25 m
Displacement along y (Bᵧ) = 10 m
Magnitude of displacement (B) =?
B² = Bₓ² + Bᵧ²
B² = 25² + 10²
B² = 625 + 100
B² = 725
Take the square root of both side
B = √725
B = 26.9 m
Therefore, the magnitude of the vector is 26.9 m
2. Determination of the magnitude of the direction of the vector.
Displacement along x (Bₓ) = 25 m
Displacement along y (Bᵧ) = 10 m
Magnitude of direction (θ) =.?
The magnitude of direction of the vector can be obtained by using Tan ratio as illustrated below:
Adjacent = Bₓ = 25 m
Opposite = Bᵧ = 10 m
Magnitude of direction (θ) =.?
Tan θ = Opposite /Adjacent
Tan θ = Bᵧ / Bₓ
Tan θ = 10 /25
Tan θ = 0.4
Take the inverse of Tan
θ = Tan¯¹ (0.4)
θ = 21.8°
Therefore, the magnitude of direction of the vector is 21.8°.
