A plane flies $\mathbf{483}\mathbf{}\mathbf{km}$east from city A to city B in$\mathbf{45}\mathbf{}\mathbf{min}$and then$\mathbf{966}\mathbf{}\mathbf{Km}$south from city B to city C in$\mathbf{1}\mathbf{,}\mathbf{5}\mathbf{}\mathbf{h}$. For the total trip, what are the (a)magnitude and(b)direction of the plane’s displacement, the(c)magnitude and(d)direction of the average velocity, and(e)it’s average speed?

Distance covered from city A to B in east direction,$\mathrm{x}=483\hspace{0.33em}\mathrm{km}$

Time taken from city A to B,${\mathrm{t}}_1=45\hspace{0.33em}\min$

Distance covered from city B to C in south direction, $\mathrm{y}=966\mathrm{km}$

Time taken from city B to C, ${\mathrm{t}}_2=1.5\hspace{0.33em}\mathrm{h}$

Average velocity is the ratio of total displacement to the total time taken for the displacement whereas average speed is the ratio of total distance to the total time taken. Average velocity is a vector quantity that has magnitude and direction as well. Average speed is a scalar quantity which has magnitude only.

The expression for the average velocity is given as:

$v{⃗}_{avg}=\frac{\Delta r⃗}{\Delta t}$

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

Expression for the average speed is given as:

$\mathrm{v}=\frac{\mathrm{Total}\mathrm{distance}}{\mathrm{Total}\mathrm{time}}$ … (ii)

Assume east to be the positive x direction and north to be the positive y direction.

So, the displacement of the plane is,

$$\begin{gathered} ∆\overrightarrow{\mathrm{r}}=\mathrm{x}\hat{\mathrm{i}}-\mathrm{y}\hat{\mathrm{j}} \\ =\left(483\mathrm{km}\right)\hat{\mathrm{i}}-\left(966\mathrm{km}\right)\hat{\mathrm{j}} \end{gathered}$$

The magnitude of the displacement is calculated as:

$$\begin{gathered} \left|∆\overrightarrow{\mathrm{r}}\right|=\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2} \\ =\sqrt{{\left(483\mathrm{km}\right)}^2+{\left(-966\mathrm{km}\right)}^2} \\ =1080\mathrm{km} \end{gathered}$$

Thus, the magnitude of the displacement is $1080\mathrm{km}$.

The direction of displacement is calculated as:

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{y}{x}\right) \\ ={\tan }^{-1}\left(\frac{-966km}{483km}\right) \\ =-63.4^{\circ } \end{gathered}$$

Negative sign indicates that the angle is measured clockwise direction from the positive x-axis.

Thus, the direction of displacement is $63.4°$towards south from the east.

Total time interval for the trip is,

$$\begin{gathered} ∆\mathrm{t}={\mathrm{t}}_1+{\mathrm{t}}_2 \\ =45\min \times \left(\frac{1\mathrm{h}}{60\min }\right)+1.5\mathrm{h} \\ =0.75\mathrm{h}+1.5\mathrm{h} \\ =2.25\mathrm{h} \end{gathered}$$

Using equation (i), the average velocity is calculated as:

$$\begin{gathered} {\overrightarrow{\mathrm{V}}}_{\mathrm{avg}}=\frac{\left(483\mathrm{km}\right)\hat{\mathrm{i}}-\left(966\mathrm{km}\right)\hat{\mathrm{j}}}{2.25\mathrm{h}} \\ =\left(214.67\mathrm{km}/\mathrm{h}\right)\hat{\mathrm{i}}+\left(-429.33\mathrm{km}/\mathrm{h}\right)\hat{\mathrm{j}} \end{gathered}$$

The magnitude of the average velocity is calculated as:

Thus, the magnitude of average velocity is$480\mathrm{km}/\mathrm{h}$.

The direction of the average velocity is the same as the direction of the resultant displacement.

Thus, the direction of average velocity is$63.4°$towards south from the east.

Total distance traveled by plane is,

$$\begin{gathered} \mathrm{Total}\mathrm{distance}=483\mathrm{km}+966\mathrm{km} \\ =1449\mathrm{km} \end{gathered}$$

Using equation (ii), the average speed is calculated as:

Thus, the average speed of plane is 644 km/h.