Respuesta :
Answer:
Magnitude of resultant velocity of kayaker to the nearest tenth = 10 m/s
Direction of resultant velocity of kayaker = 49.32⁰ South of west.
Explanation:
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 = [tex]\sqrt{(-4.71)^2+(-5.48)^2} = 7.23 m/s[/tex]
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.
Answer:
[tex]v = 7.27 m/s[/tex] at direction of 49.1 degree south of west
Explanation:
Initial velocity of the Kayaker paddles is given as
[tex]v_1[/tex] = 4 m/s in direction 30^0 south of west
[tex]v_1 = 4 cos30 (-\hat i) + 4 sin30 (-\hat j)[/tex]
[tex]v_1 = - 3.5\hat i - 2\hat j[/tex]
Another velocity is given as
[tex]v_2[/tex] = 3.7 m/s in direction 20 degree west of south
[tex]v_2 = 3.7 sin20 (-\hat i) + 3.7 cos20 (-\hat j)[/tex]
[tex]v_2 = -1.26 \hat i -3.5 \hat j[/tex]
now the resultant velocity is given as
[tex]v = v_1 + v_2[/tex]
[tex]v = (- 3.5 - 1.26)\hat i + (-2 - 3.5)\hat j[/tex]
[tex]v = -4.76 \hat i - 5.5 \hat j[/tex]
magnitude of the speed is
[tex]v = \sqrt{4.76^2 + 5.5^2}[/tex]
[tex]v = 7.27 m/s[/tex]
also for direction we have
[tex]\theta = tan^{-1}(\frac{5.5}{4.76})[/tex]
[tex]\theta = 49.1 degree[/tex]
