Answer: The magnitude of the kayaker's resultant velocity = 7.23 m/s
and the direction is 49.26° south of west.
Explanation:
Initial velocity, u = 4.0 m/s 30° south of west.
Final velocity, v = 3.7 m/s 20° west of south °≈ 70° south of west.
Writing in terms of component:
u = -4.0 cos 30 i - 4.0 sin 30 j = -3.46 i -2.0 j
v = -3.7 cos 70 i - 3.7 sin 70 j = -1.26 i - 3.48 j
Resultant velocity = addition of the two velocity vectors
R = (-3.46 i -2.0 j)+(-1.26 i - 3.48 j) = -4.72 i - 5.48 j
Magnitude of the vector = √(-4.72)²+( - 5.48)² = 7.23 m/s
Direction of the resultant velocity vector, θ = tan⁻¹ (-5.48÷ (-4.72)) = 49.26° south of west.
Thus, the magnitude of the kayaker's resultant velocity = 7.23 m/s
and the direction is 49.26° south of west.
