An airplane travels 2100 km at a speed of 720 km/h and then encounters a tailwind that boosts its speed to 990 km/h for the next 2800 km. What was the total time for the trip? What was the average speed of the plane for this trip?

[Hint: Does Eq. 2–11d apply?]

It gives the ratio of the total distance moved by airplane to the total time taken by the airplane to complete the trip.

It can be obtained by first estimating the total distance of the full trip. Then determine the total time covered in the overall trip.

The average speed expression can be given as,

$S_{avg}=\frac{\Delta x}{\Delta t}$

Here, $\Delta x$is the total distance, $\Delta t$is the total time taken.

The distance traveled by plane in the first trip is, $x_1=2100\mathrm{km}$.

The speed of the airplane on the first trip is, $S_1=720\mathrm{km}/\mathrm{h}$.

The time taken by the airplane to cover its first trip can be expressed as,

$\begin{array}{c}\mathrm{time}=\frac{\mathrm{distance}}{\mathrm{speed}}\\ t_1=\frac{x_1}{S_1}\end{array}$

Substitute the values as 2100 km for $x_1$and 720 km/h for $S_1$.

$\begin{array}{c}{t}_1=\frac{2100\mathrm{km}}{720\mathrm{km}/\mathrm{h}}\\ =2.917\mathrm{h}\end{array}$

The distance covered by plane in the second trip is, $x_2=2800\mathrm{km}$.

The airplane’s speed on the second trip is, $S_2=990\mathrm{km}/\mathrm{h}$.

The time covered by the airplane in the second trip can be given as,

$t_2=\frac{x_2}{S_2}$

Substitute the values as 2800 km for $x_2$and 990 km/h for $S_2$.

$\begin{array}{c}{t}_2=\frac{2800\mathrm{km}}{990\mathrm{km}/\mathrm{h}}\\ =2.828\mathrm{h}\end{array}$

The total time taken by the airplane to cover the whole trip can be calculated as,

$\Delta t=t_1+t_2$

Substitute the values as 2.917 h for $t_1$and 2.828 h for $t_2$.

$\begin{array}{c}\Delta t=2.917\mathrm{h}+2.828\mathrm{h}\\ =5.745\mathrm{h}\end{array}$

Thus, the total time for the trip is $5.745\mathrm{h}$.

The total distance covered by the airplane in the whole trip can be estimated as,

$\Delta x=x_1+x_2$

Substitute the values as 2100 km for $x_1$and 2800 km for $x_2$.

$\begin{array}{c}\Delta x=2100\mathrm{km}+2800\mathrm{km}\\ =4900\mathrm{km}\end{array}$

The average speed of the airplane can be expressed as,

$S_{avg}=\frac{\Delta x}{\Delta t}$

Substitute the values as 4900 km for $\Delta x$and 5.745 h for $\Delta t$.

$\begin{array}{c}{S}_{avg}=\frac{4900\mathrm{km}}{5.745\mathrm{h}}\\ =852.915\mathrm{km}/\mathrm{h}\end{array}$

Hence, the airplane’s average speed for this trip is $852.915\mathrm{km}/\mathrm{h}$.