91Ó°ÊÓ

• Length of transcontinental flight, $D=4350\mathrm{km}$.
• Extra time that will be taken by the flight, $v_{flight}=966\mathrm{km}/\mathrm{h}$.
• Airspeed of the airplane, $t_1-t_2=50\min \mathrm{or}5/6\mathrm{h}$.

When there is a relative motion at a constant velocity between the two frames of reference, frameAand frame B, the velocity of a particle Pas measured by an observer in frame Ais different than the velocity of a particle P measured in frame B. The relation between two velocities is given as,

${\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{P}\mathbf{A}}\mathbf{=}{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{P}\mathbf{B}}\mathbf{+}{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{B}\mathbf{A}}$

Here${\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{BA}}$is the velocity of B with respect to A.

Use the concept of relative velocity to find the jet stream.

Formula:

The velocity of a body in motion, $v=\frac{d}{t}$ (i)

is velocity, $d$is distance, $t$and is time for the motion.

It is given that the westward plane is taking longer than that of the eastward. Hence, the time difference as given:

$t_1-t_2=\frac{5}{6}h$

Substitute the equation for time and simplify it.

$$\begin{gathered} \frac{D}{966km/h-v}-\frac{D}{966km/h+v}=\frac{5}{6} \\ \frac{4350km}{966-v}-\frac{4350km}{966+v}=\frac{5}{6} \\ 4350\left(\frac{1}{966-v}-\frac{1}{966+v}\right)=\frac{5}{6} \\ \frac{4350\left(\left(966-v\right)-\left(966-v\right)\right)}{{960}^2-v^2}=\frac{5}{6} \end{gathered}$$

Simplifying it further to calculate the value of $v$.

$$\begin{gathered} 4350×2×v=\left({966}^2-v^2\right)×\frac{5}{6} \\ {966}^2-v^2=10440v \\ {v}^2+10440v-{966}^2=0 \\ v=88.63\mathrm{km}/\mathrm{h} \end{gathered}$$

Hence, the speed of the jet stream is $88.63\mathrm{km}/\mathrm{h}$.