Answer:
a) The magnitude of the car's total displacement (T) from the starting point is T = 82.67 Km
b) The angle (θ) from east of the car's total displacement measured from the starting direction is θ = 40.88 °
Explanation:
Attached you can see a diagram of the problem.
a) Find the magnitude of the vector T that goes from point A to point D (see the diagram).
The x and y components of this vector are
[tex]T_x=48+29*sin(30)=62.5 Km\\T_y=29+29*cos(30) = 54.11 Km[/tex]
The magnitude of the vector is find using the pythagoras theorem:
[tex]a^2=b^2+c^2[/tex], being a, b and c the 3 sides of the triagle that forms the vector:
[tex]T^2=T_x^2+T_y^2\\T=\sqrt{T_x^2+T_y^2}[/tex]
Replacing the values
[tex]T=\sqrt{(62.5)^2+(54.11)^2} \\T=82.67 Km[/tex]
b) Find the angle θ that forms the vector T and the vector AB (see diagram).
To find this angle you can use the inverse tangent
θ[tex]=tan^{-1}(\frac{T_y}{T_x})[/tex]
θ[tex]=tan^{-1}(\frac{54.11}{62.5})[/tex]
θ=40.88°
