Chapter 11, Problem 106RE

To determine

Assume that the speed of the jet stream remains constant at the various altitudes of the plane and that the plane flies with the jet stream one way and against it the other way

Answer to Problem 106RE

  $\begin{array}{l}x=\boxed{30mph}\\ \end{array}$

Explanation of Solution

Given information:

On a flight between midway airport in Chicago and ft. Lauderdale, Florida, a Boeing $737$ jet maintains airspeed of $475$ miles per hour. If the trip from Chicago to ft Lauderdale takes $2$ hours, $30$ minutes and the return flight takes, $2$ house, $50$ minutes, what is the speed of the jet stream? (Assume that the speed of the jet stream remains constant at the various altitudes of the plane and that the plane flies with the jet stream one way and against it the other way.)

Calculation:

Let the speed of the jet-stream be $xmph$ and the distance between Chicago and Ft. Lauderdale be $ymiles$

The aircraft Boeing $737$ maintains a speed of $475moh$ . The real speed of the aircraft can be given by the expression given below. The aircraft takes lesser time when going from Chicago to Ft. Lauderdale, which men’s the speed of the aircraft is greater than the opposite direction.

Speed from Chicago to Ft. Lauderdale $=(475+x)mph$

Speed from Ft. Lauderdale to Chicago $=(475+-x)mph$

The time taken by Boeing when going from Chicago to Ft. Lauderdale is $2$ hours $30$ minutes and time taken by Boeing when going in opposite direction is $2$ hours $50$ minutes. Therefore we can obtain the following equations.

  $\begin{array}{l}Dis\tan ce=Speed\times Time\\ y=(475+x)\left(2+\frac{30}{60}\right)\mathrm{.....}\left(1\right)\\ =1187.5+2.5x\\ Dis\tan ce=Speed\times Time\end{array}$

  $\begin{array}{l}y=(475+x)\times \left(2+\frac{30}{60}\right)\mathrm{......}\left(1\right)\\ =1187.5+2.5x\\ Dis\tan ce=Speed\times Time\\ y=(475-x)\times \left(2+\frac{50}{60}\right)\mathrm{......}\left(2\right)\\ =\frac{8075}{6}-\frac{17}{6}x\end{array}$

We have obtained equation $(1)$ and $(2)$ which can be solved to obtain the speed of the jetstream. The equations can be subtracted from each other since the coefficients of $y$ in both the equations are same.

  $\begin{array}{l}1187.5+2.5x\frac{8075}{6}+\frac{17}{6}x-=0\\ 1187.5-1345.8+2.8x+2.5x\approx 0\\ 5.3x\approx 1345.8-1187.5\\ x\approx 30mph\end{array}$

Hence, therefore the speed of the jet stream is $x=\boxed{30mph}$ .