agnimusayoti
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- Homework Statement
- an airplane is supposed to travel from A in a direction due north to B and then return to A. The distance between and B is L. The air speed of the plane is ##v## and the wind velocity is ##v'##. Show that the time for the round trip when the wind is directed due east (or west) is
$$t_{b}=\frac{t_{a}}{\sqrt{1-\frac{v'^{2}}{v^2}}}$$
where ##t_{a}## is half of the roundtrip's time in still air.
- Relevant Equations
- Vector additon
Because of the wind, airplane was shifted to the east. Assume airplane is shifted D units long from B.
When airplane turnaround, the wind shifted airplane to the east again as far D and to the south as far as L to the A'.
Therefore,
$$2D = (v - v') t_{AA'}$$
But,
$$D = v'(t_{a}/2)$$
Thus,
$$v't_{a} = (v- v') (t_{AA'} $$.
From this relationship, I got
$$t_{AA'} = \frac{v'}{v - v'} t_{a}$$.
Time that needed for roundtrip: A to B', B' to A' and A' to A:
$$t_b = (1 + \frac{v'}{v-v'}) t_{a}$$
$$t_b = (\frac{v}{v-v'}) t_{a}$$
My answer is different from the problem at the denominator. In my answer, (v - v'), but in the problem:: ##\sqrt{v^2 - v'^2}##.
Am I right? Or I made a mistake? Thanks!
When airplane turnaround, the wind shifted airplane to the east again as far D and to the south as far as L to the A'.
Therefore,
$$2D = (v - v') t_{AA'}$$
But,
$$D = v'(t_{a}/2)$$
Thus,
$$v't_{a} = (v- v') (t_{AA'} $$.
From this relationship, I got
$$t_{AA'} = \frac{v'}{v - v'} t_{a}$$.
Time that needed for roundtrip: A to B', B' to A' and A' to A:
$$t_b = (1 + \frac{v'}{v-v'}) t_{a}$$
$$t_b = (\frac{v}{v-v'}) t_{a}$$
My answer is different from the problem at the denominator. In my answer, (v - v'), but in the problem:: ##\sqrt{v^2 - v'^2}##.
Am I right? Or I made a mistake? Thanks!