Given data:
* The distance traveled in east in time 0.5 h is 9.9 km
* The distance traveled in east of north in time 0.77 h is 9 km at 15 degree.
* The distance traveled in east in 0.5 h is 9.9 km.
Solution:
The diagramatic representation of the give system is
The net distance traveled in east is,
[tex]d_x=9.9+9\cos (15^{\circ})+9.9[/tex]where 9 cos( 15 degree ) terms is representing the component of 9 distance along the east direction.
Solving the value of total distance in east is,
[tex]d_x=28.49\text{ km}[/tex]The net distance traveled in the north direction is,
[tex]\begin{gathered} d_y=9.9\times\cos (90^{\circ})+9\times\sin (15^{\circ})+9.9\times\cos (90^{\circ}) \\ d_y=0+2.33+0 \\ d_y=2.33\text{ km} \end{gathered}[/tex]The total time taken to complete the trip is,
[tex]\begin{gathered} T=0.5+0.77+0.5 \\ T=1.77\text{ h} \end{gathered}[/tex]The average velocity of bicycle along the east is,
[tex]v_x=\frac{d_x}{T}[/tex]Substituting the known values,
[tex]\begin{gathered} v_x=\frac{28.49}{1.77} \\ v_x=16.1kmh^{-1} \end{gathered}[/tex]The average velocity of the bicycle along the north direction is,
[tex]\begin{gathered} v_y=\frac{2.33}{1.77} \\ v_y=1.32kmh^{-1} \end{gathered}[/tex]Thus, the magnitude of the average velocity is,
[tex]\begin{gathered} v_a=\sqrt[]{16.1^2+1.32^2} \\ v_a=16.15kmh^{-1} \end{gathered}[/tex]The direction of the average velocity is,
[tex]\begin{gathered} \tan (\theta)=\frac{v_y}{v_x} \\ \tan (\theta)=\frac{1.32}{16.1} \\ \tan (\theta)=0.082 \\ \theta=4.69^{\circ} \end{gathered}[/tex]Hence, the magnitude of the average velocity is 16.15 km per hour and direction of average velocity is at 4.69 degree east of north.
