<u>Answer:</u>
Magnitude of resultant velocity of kayaker to the nearest tenth = 10 m/s
Direction of resultant velocity of kayaker = 49.32⁰ South of west.
<u>Explanation:</u>
Let east represents positive x- axis and north represent positive y - axis. Horizontal component is i and vertical component is j.
First kayaker paddles at 4.0 m/s in a direction 30° south of west, kayaker paddles at 4.0 m/s in a direction 210° anticlockwise from positive horizontal axis.
So velocity of kayaker = 4 cos 210 i + 4 sin 210 j = -3.46 i - 2 j
He then turns and paddles at 3.7 m/s in a direction 20° west of south, kayaker paddles at 3.7 m/s in a direction 250° anticlockwise from positive horizontal axis.
So that velocity = -1.27 i - 3.48 j
So resultant velocity of kayaker = -3.46 i - 2 j +(-1.27 i - 3.48 j) = -4.71 i - 5.48 j
Magnitude of resultant velocity of kayaker =
Magnitude of resultant velocity of kayaker to the nearest tenth = 10 m/s
Direction of resultant positive horizontal axis, θ = tan⁻¹(-5.48/-4.71) = 229.32⁰ = 49.32⁰ South of west.