Question

A bicycle travels 6.10 km due east in 0.210 h, then 11.30 km at 15.0° east of north in 0.560 h, and finally another 6.10 km due east in 0.210 h to reach its destination. The time lost in turning is negligible. Assume that east is in the +x-direction and north is in the +y-direction. What is the direction of the average velocity for the entire trip? Enter the answer as an angle in degrees north of east

Answer 1

Step-by-step explanation:

Given

First bicycle travels 6.10 km due to east in 0.21 h

Suppose its position vector is
`r_1`

`r_1=6.10\hat{i}`

After that it travels 11.30 km at
`15^(\circ)` east of north in 0.560 h

suppose its position vector is
`r_2`

`r_(2)=11.30\left ( cos15\hat{j}+sin15\hat{i}\right )`

after that he finally travel 6.10 km due to east in 0.21 h

suppose its position vector is
`r_3`

`r_(3)=6.10\hat{i}`

so position of final position is given by

`r=r_1+r_(2)+r_(3)`

`\vec{r}=15.12\hat{i}+10.91\hat{j}`

`\vec{v_(avg)}=\frac{\vec{r}}{t}`

t=0.21+0.56+0.21=0.98 h

`\vec{v_(avg)}=15.42\hat{i}+11.13\hat{j}`

`|v_(avg)|=√(361.71)=19.01 km/hr`

For direction

`tan\theta =(11.13)/(15.42)=0.721`

`\theta =35.791^(\circ)` w.r.t to x axis