Answer:
a) speed of the child relative to the surface of the water = 10.20 mph
b) θ = 0.197 rad
Explanation:
The speed of the child due east = 2 miles per hour
The speed of the ship due north = 10 miles per hour
a)To get the relative speed of the child to the surface of the water from the diagram drawn(attached)
[tex]v^2 = 10^2 + 2^2\\v^2 = 100 + 4\\v^2 = 104\\v = \sqrt{104} \\v = 10.20 mph[/tex]
b) Angle of the direction from the north
From the ΔABC drawn,
tan θ = opposite/Adjacent
tan θ = AB/AC
tan θ = 2/10
tan θ = 0.2
θ = tan⁻¹0.2
θ = 0.197 rad
