Answer:
v = 31 m, / s; θ = 1 north- east
Step-by-step explanation
One of the easiest ways to solve these problems is to work with the components of vectors.
Let's set a reference system where the x axis coincides with the West - East direction
Let's use trigonometry
cos 45 = v₁ₓ / v₁
sin 45 = v_{1y} / v₁
v₁ₓ = v1 cos 45
v_{1y} = v₁₁ sin 5
v₁ₓ = 15 cos 45 = 10.6 m / s
v _{1y} = 15 sin 45 = 10.6 m / s
sin 5 = v₂ₓ / v₂
cos 5 = v_{2y} / v₂
v₂ₓ = v2 sin 5
v_{2y}= v2 sin 5
v₂ₓ = 18 sin 5 = 1.569 m / s
v_{2yₓ} = 18 cos 5 = 17.93 m / s
Let's look for the speed in each axis
vₓ = v₁ₓ + v₂ₓ
vx = 10.6 + 1.569
vx = 12.17 m / s
v_y = v_{1y} + v_{2y}
v_y = 10.6 + 17.93
v_y = 28.53 m / s
to find the velocity let's use the Pythagorean theorem
v =
v = √ (12.17² + 28.53²)
v = 31.02 m / s
approaching the nearest ether
v = 31 m / s
b) we use trigonometry
tan θ = vy / vx
θ = tan -1
θ = tan vy / vx
θ = tan-1 17.93 / 12.17
θ= 1.47
the closeΕst integer is
θ = 1 north- east