91Ó°ÊÓ

1. Plane’s airspeed,$\left(V_{pa}\right)=1000\text{km/h}$
2. The distance covered by plane, $\left(X\right)=4000\text{km}$
3. The time difference between the flight time for outgoing and returning at 70 min or 1.165 hr.

When two frames of reference Aand Bare moving relative to each other at a constant velocity, the velocity of a particle Pas measured by an observer in frame Ausually differs from that measured in frame B. The two measured velocities are related by

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

Here${\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{B}\mathbf{A}}$is the velocity of B with respect to A.Using the relative motion concept, we can find the speed of the jet stream.

Formula:

The time taken by the jet stream, $t=\frac{X}{V}$ …(¾±)

The velocity of plane for returning jet is given as:

$V_{pa}+V_{ag}=1000+V_{ag}$

The flight time for returning is given using equation (i) as:

$\begin{array}{rcl}{t}_1& =& \frac{X}{V_{pa}+V_{ag}}\\ & =& \frac{4000  \text{km}}{\left(1000 +V_{ag}\right)  \text{km/h}}\end{array}$ …(¾±¾±)

The velocity of plane for outgoing is,

$V_{pa}-V_{ag}=1000-V_{ag}$

The flight time for outgoing is given using equation (i) as:

$\begin{array}{rcl}{t}_2& =& \frac{X}{V_{pa}-V_{ag}}\\ & =& \frac{4000  \text{km}}{\left(1000-V_{ag}\right)  \text{km/h}}\end{array}$ …(¾±¾±¾±)

So, the time difference between the flight time for outgoing and returning is given and now using equations (ii) and (iii), we get that

$\begin{array}{rcl}{t}_2-t_1& =& 1.167  \text{h}\\ \frac{4000  \text{km}}{\left(1000-V_{ag}\right)  \text{km/h}}-\frac{4000  \text{km}}{\left(1000 +V_{ag}\right)  \text{km/h}}& =& 1.167  \text{h}\\ 1.167V_{ag}^2+8000V_{ag}-1167000& =& 0\\ V_{ag}& =& 143\text{km/h}\end{array}$

Hence, the value of the speed of the jet stream is 143 km/hr.