Question

A kayaker paddles at 4.0 m/s in a direction 30° south of west. He then turns and paddles at 3.7 m/s in a direction 20° west of south. What is the magnitude of the kayaker’s resultant velocity? Round your answer to the nearest tenth. ___m/s What is the direction of the kayaker’s resultant velocity? ___ ° south of west. I need help working it out, not just the answer.

Answer 1

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 = $\sqrt{(-4.71)^2+(-5.48)^2} = 7.23 m/s$

 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 2

Answer:

$v = 7.27 m/s$ at direction of 49.1 degree south of west

Explanation:

Initial velocity of the Kayaker paddles is given as

$v_1$ = 4 m/s in direction 30^0 south of west

$v_1 = 4 cos30 (-\hat i) + 4 sin30 (-\hat j)$

$v_1 = - 3.5\hat i - 2\hat j$

Another velocity is given as

$v_2$ = 3.7 m/s in direction 20 degree west of south

$v_2 = 3.7 sin20 (-\hat i) + 3.7 cos20 (-\hat j)$

$v_2 = -1.26 \hat i -3.5 \hat j$

now the resultant velocity is given as

$v = v_1 + v_2$

$v = (- 3.5 - 1.26)\hat i + (-2 - 3.5)\hat j$

$v = -4.76 \hat i - 5.5 \hat j$

magnitude of the speed is

$v = \sqrt{4.76^2 + 5.5^2}$

$v = 7.27 m/s$

also for direction we have

$\theta = tan^{-1}(\frac{5.5}{4.76})$

$\theta = 49.1 degree$