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

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

Step-by-step 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 = √(4.76^2 + 5.5^2)`

`v = 7.27 m/s`

also for direction we have

`\theta = tan^(-1)((5.5)/(4.76))`

`\theta = 49.1 degree`