Question

A car is driven east for a distance of 48 km, then north for 29 km, and then in a direction 30° east of north for 29 km. Determine (a) the magnitude of the car's total displacement from its starting point and (b) the angle (from east) of the car's total displacement measured from its starting direction.

Answer 1

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

$T_x=48+29*sin(30)=62.5 Km\\T_y=29+29*cos(30) = 54.11 Km$

The magnitude of the vector is find using the pythagoras theorem:

$a^2=b^2+c^2$, being a, b and c the 3 sides of the triagle that forms the vector:

$T^2=T_x^2+T_y^2\\T=\sqrt{T_x^2+T_y^2}$

Replacing the values

$T=\sqrt{(62.5)^2+(54.11)^2} \\T=82.67 Km$

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$=tan^{-1}(\frac{T_y}{T_x})$

θ$=tan^{-1}(\frac{54.11}{62.5})$

θ=40.88°