You bicycle 3.2 km east in 0.1 h, then 3.2 km at 15.0 degrees east of north in 0.21 h, and finally another 3.2km due east in 0.1 h to reach its destination. What is the average "velocity" for the entire trip?

Dec 31, 2016

Let the direction of displacement towards EAST be along positive direction of x-axis and that of NORTH be along positive direction of y-axis. So the unit vector along these directions be $\hat{i} \mathmr{and} \hat{j}$ respectively.

First displacement ( vec(d_1) is 3.2 km east in 0.1 h

So

$\vec{{d}_{1}} = 3.2 \hat{i} k m$

Second displacement ( vec(d_2) 3.2 km at 15.0 degrees east of north in 0.21 h, This means the direction of displacement makes ${75}^{\circ}$ with $\hat{i}$

So

$\vec{{d}_{2}} = \left(3.2 \cos {75}^{\circ} \hat{i} + 3.2 \sin {75}^{\circ} \hat{j}\right) k m$

Third displacement ( vec(d_3) is 3.2 km east in 0.1 h

So

$\vec{{d}_{3}} = 3.2 \hat{i} k m$

Hence net displacement $\vec{d}$

$\vec{d} = \vec{{d}_{1}} + \vec{{d}_{2}} + \vec{{d}_{3}}$

$= \left[3.2 \hat{i} + \left(3.2 \cos {75}^{\circ} \hat{i} + 3.2 \sin {75}^{\circ} \hat{j}\right) + 3.2 \hat{i}\right] k m$

$= \left[6.4 \hat{i} + \left(0.83 \hat{i} + 3.09 j\right)\right] k m$

$= \left[\left(7.23 \hat{i} + 3.09 j\right)\right] k m$

The average velocity

${v}_{\text{average"=abs(vec(d))/"Total time}} = \frac{\sqrt{{7.23}^{2} + {3.09}^{2}}}{0.1 + 0.21 + 0.1} \frac{k m}{h r}$

$\approx 150.78 \text{km/hr}$

And the direction of average velocity along North of East is

$\theta = {\tan}^{-} 1 \left(\frac{3.08}{7.23}\right) = {23.14}^{\circ}$