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

Answer:

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

$\vec{v_{avg}}=15.42\hat{i}+11.13\hat{j}$

$|v_{avg}|=\sqrt{361.71}=19.01 km/hr$

For direction

$tan\theta =\frac{11.13}{15.42}=0.721$

$\theta =35.791^{\circ}$ w.r.t to x axis