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_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
