Long flights at mid-latitudes in the Northern Hemisphere encounter the jet stream, an eastward airflow that can affect a plane’s speed relative to Earth’s surface. If a pilot maintains a certain speed relative to the air (the plane’s airspeed), the speed relative to the surface (the plane’s ground speed) is more when the flight is in the direction of the jet stream and less when the flight is opposite the jet stream. Suppose a round-trip flight is scheduled between two cities separated by 4000 km, with the outgoing flight in the direction of the jet stream and the return flight opposite it.The airline computer advises an airspeed of 1000 km/hr, for which the difference in flight times for the outgoing and return flights is 70.0 min.What jet-stream speed is the computer using?

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

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

Here${\overrightarrow{\mathbf{v}}}_{\mathbf{B}\mathbf{A}}$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, $t=\frac{X}{V}$ …(i)

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:

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

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:

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

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.