Answer:
Explanation:
Given
First bicycle travels 6.10 km due to east in 0.21 h
Suppose its position vector is [tex]r_1[/tex]
[tex]r_1=6.10\hat{i}[/tex]
After that it travels 11.30 km at [tex]15^{\circ}[/tex] east of north in 0.560 h
suppose its position vector is [tex]r_2 [/tex]
[tex]r_{2}=11.30\left ( cos15\hat{j}+sin15\hat{i}\right )[/tex]
after that he finally travel 6.10 km due to east in 0.21 h
suppose its position vector is [tex]r_3[/tex]
[tex]r_{3}=6.10\hat{i}[/tex]
so position of final position is given by
[tex]r=r_1+r_{2}+r_{3}[/tex]
[tex]\vec{r}=15.12\hat{i}+10.91\hat{j}[/tex]
[tex]\vec{v_{avg}}=\frac{\vec{r}}{t}[/tex]
t=0.21+0.56+0.21=0.98 h
[tex]\vec{v_{avg}}=15.42\hat{i}+11.13\hat{j}[/tex]
[tex]|v_{avg}|=\sqrt{361.71}=19.01 km/hr[/tex]
For direction
[tex]tan\theta =\frac{11.13}{15.42}=0.721[/tex]
[tex]\theta =35.791^{\circ}[/tex] w.r.t to x axis
