Question

A kayaker needs to paddle north across a 100 -m wideharbor. The tide is going out, creating a tidal current that flowsto the east at 2.0 m/s. The Kayaker can paddle with a speed of3.0m/s.
a. In which direction shoule he paddle in order to travel straightacross the harbor ?
b. How long will it take him to cross ?

A body diagram and step by step precise solution willgive me a better understanding .. thanks for you help..

Answer 1

Answer:

a) v_Nort = 2.236 m / s

, θ = 56.3º, b)  t=  53.76 s

Explanation:

This exercise should use the addition of vectors, we have a kayak speed of v₁ = 3 m / s and an eastward speed of water v₂ = 2.0 m / s.

They ask us to cross the river that is to the north, we see that the speed of the kayak is the hypotenuse of the triangula, see attached

              v₁² = v_nort² + v₂²

              v _nort² = v₁² –v₂²

              v_nort = √ (3² - 2²)

              v_Nort = 2.236 m / s

For the angle we can use trigonometry

              tan θ = v₂ / v₁

              θ = tan⁻¹ v₂ / v₁

              θ = tan⁻¹ 2/3

              θ = 33.7º

             

This angle measured from the positive side of the x axis is

              θ = 90 - 33.7

              θ = 56.3º

b)  we look for the northward component of this speed

          sin 56.3 = $v_{y}$ / v_nort

          v_{y} = v_nort sin 56.3

          v_{y} = 2.236 sin 56.3

          v_{y} =  1.86 m/s

The time is

          v_{y} = y/t

          t = y/v_{y}

          t =100/ 1.86

          t=  53.76 s