91Ó°ÊÓ

The velocity of train,$v=60.0\mathrm{km}/\mathrm{h}$

Time of travel in east direction,$∆t_1=40.0\min$

Time of travel in 150 east of north direction,$∆t_2=20.0\min$

Time of travel in west direction,$∆t_3=50.0\min$

The average velocity may be defined as the ratio of total displacement to the total time interval for the displacement to occur.It is a vector quantity, which has magnitude as well as direction.

The expression for the average velocity is given as:

${\stackrel{→}{v}}_{avg}=\frac{∆\stackrel{→}{r}}{∆t}$… (i)

Here, $∆\stackrel{→}{r}$is the total displacement and $∆t$is the total time interval.

The displacement of train in the east direction is calculated as:

$$\begin{gathered} ∆x_1=v×∆t_1 \\ =60.0\mathrm{km}/\mathrm{h}×\left(40.0\min ×\frac{1\mathrm{h}}{60\min }\right) \\ =40.0\mathrm{km} \end{gathered}$$

The train travels for 20 min along $50°$east of north or $40°$north of east. So, the x and y component of displacement are calculated as:

$$\begin{gathered} ∆x_2=v\cos 40°×∆t_2 \\ =60.0\mathrm{km}/\mathrm{h}×\cos 40°×\left(20.0\min ×\frac{1\mathrm{h}}{60\min }\right) \\ =15.32.0\mathrm{km} \end{gathered}$$

$$\begin{gathered} ∆x_2=v\sin 40°×∆t_2 \\ =60.0\mathrm{km}/\mathrm{h}×\sin 40°×\left(20.0\min ×\frac{1\mathrm{h}}{60\min }\right) \\ =12.85\mathrm{km} \end{gathered}$$

The displacement of train in the west direction is calculated as:

$$\begin{gathered} ∆x_3=v×∆t_3 \\ =-60.0\mathrm{km}/\mathrm{h}×\left(50.0\min ×\frac{1\mathrm{h}}{60\min }\right) \\ =-50.0\mathrm{km} \end{gathered}$$

The total displacement of the train is,

$$\begin{gathered} ∆\hat{\mathrm{r}}=∆\mathrm{x}\hat{\mathrm{i}}+∆\mathrm{y}\hat{\mathrm{j}} \\ =\left(∆{\mathrm{x}}_1+∆{\mathrm{x}}_2+∆{\mathrm{x}}_3\right)\hat{\mathrm{i}}+∆{\mathrm{y}}_2\hat{\mathrm{j}} \\ =\left(40.0\mathrm{km}+15.32\mathrm{km}-50.0\mathrm{km}\right)\hat{\mathrm{i}}+\left(12.85\mathrm{km}\right)\hat{\mathrm{j}} \\ =\left(5.32\mathrm{km}\right)\hat{\mathrm{i}}+\left(12.85\mathrm{km}\right)\hat{\mathrm{j}} \end{gathered}$$

Total time interval is,

$$\begin{gathered} ∆\mathrm{t}=∆{\mathrm{t}}_1+∆{\mathrm{t}}_2+∆{\mathrm{t}}_3 \\ =40.0\min +20.0\min +50.0\min \\ =110\min ×\frac{1\mathrm{h}}{60\min } \\ =1.83\mathrm{h} \end{gathered}$$

Using equation (i), the average velocity of the train is,

$$\begin{gathered} {\stackrel{→}{v}}_{avg}=\frac{\left(5.32\mathrm{km}\right)\hat{\mathrm{i}}+\left(12.85\mathrm{km}\right)\hat{\mathrm{j}}}{1.83\mathrm{h}} \\ =\left(2.91\mathrm{km}/\mathrm{h}\right)\hat{\mathrm{i}}+\left(7.02\mathrm{km}/\mathrm{h}\right)\hat{\mathrm{j}} \end{gathered}$$

The magnitude of the average velocity is calculated as:

$$\begin{gathered} v_{avg}=\sqrt{{{\mathrm{v}}_{\mathrm{x}}}^2+{{\mathrm{v}}_{\mathrm{x}}}^2} \\ =\sqrt{{\left(2.91\mathrm{km}/\mathrm{h}\right)}^2+{\left(7.02\mathrm{km}/\mathrm{h}\right)}^2} \\ =7.60\mathrm{km}/\mathrm{h} \end{gathered}$$

Thus, the magnitude of the average velocity of train is $7.60\mathrm{km}/\mathrm{h}$.

The direction of the average velocity is calculated as:

$$\begin{gathered} \tan \mathrm{θ}=\frac{v_y}{v_x} \\ \mathrm{θ}={\tan }^{-1}\left(\frac{7.02\mathrm{km}/\mathrm{h}}{2.91\mathrm{km}/\mathrm{h}}\right) \\ =67.5° \end{gathered}$$

So, the direction of average velocity is$67.5°$north of east or$22.5°$east of due north.