Answer:
3.47km/h with a direction of 67.8 degrees north of west.
Explanation:
First we need to calculate the displacement on the X axis, so:
[tex]d_{x1}=50.0km/h*45min(\frac{1h}{60min})=37.5km\\d_{x2}=50.0km/h*10min(\frac{1h}{60min})*cos(45^o)=5.90km\\d_{x1}=-50.0km/h*55.0min(\frac{1h}{60min})=45.8km\\D_x=37.5+5.90-45.8=-2.4km[/tex]
then on the Y axis:
[tex]D_y=50.0km/h*10min(\frac{1h}{60min})*sin(45^o)=5.90km[/tex]
The magnitud of the displacement is given by:
[tex]D=\sqrt{D_x^2+D_y^2} \\D=6.37km[/tex]
and the angle:
[tex]\alpha =arctg(\frac{5.90}{2.41})=67.8^o[/tex]
that is 67.8 degrees north of west.
[tex]v=\frac{D}{t}\\v=\frac{6.37km}{110min*\frac{1h}{60min}}\\v=3.47km/h[/tex]
