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emilyyy · Oct 5, 2008, 05:43 PM

A blimp has an air speed of 10.00m/s. The pilot takes the blimp on a trip 5000m[S] directly into the wind and then 5000m[N] back with the wind. The total time as 20.0 minutes(assume that the turn-around time is negligable). Determine the velocity of the wind relative to the ground.
Any help would be great I have no idea how to do this thanks :)

Unknown008 · Oct 5, 2008, 08:52 PM

It's just a suggestion!!

Let velocity wind = $V_V$

Then, resultant velocity of the blimp is 10- $V_V$ for the south direction trip. For the north one, it'll be 10+ $V_V$ .

Then, $\frac{5000}{10-V_V}= t_1$

$\frac{5000}{10+V_V}= t_2$

$t_1 + t_2 = 20$

Then solve!

FlyYakker · Oct 6, 2008, 04:38 PM

The point being that, assuming the wind is along a North South axis you have to add or subtract the wind's ground speed from the maximum air speed of the blimp. In aircraft, the maximum rated speed (and all speed values the pilots uses for reference to control the aircraft) is relative to the surrounding air. If an airliner, say, has a maximum airspeed of 500 MPH but is flying against a wind moving at 100 MPH in the opposite direction, the air liner's speed over the ground is only 400 MPH. If the airliner turns around to go in the opposite direction, its ground speed becomes 600 MPH.