(Original post by g.k.galloway)
another physics question! woo!
A microlight is a small aircraft powered by a petrol engine. The diagram represents the flight path, AB of a microlight on a short horizontal training flight.
a) on its outward journey the wind velocity is 7.5ms^-1 due north and the resultant velocity of the microlight is 20ms^-1 in the direction 68 east of north, so that it travels along AB
i) show that for the aircraft to travel along AB at 20ms^-1 it should be pointed due east.
b) after flying 10kn in the aircraft turns round and returns along the same flight path at a resultant velocity of 14ms^-1. Assuming that the turn-round time is negligible, calculate the average speed for the complete journey
for a) you can see on the diagram I used trig to find the two lengths of the triangle made under the line AB...whether you need to do this or not i have noooo idea, i really dont get itttt
I've no clue if anyone has responded to this, as I can't be bothered browsing through. Just saw your plead for help. You should quote me in future, as that way I'll notice it sooner.
Component velocity North:
20cos(68) = 7.5 ms-1
That velocity is supplied by the wind acting on the microlight, and I'm assuming some statement about the components showing that it must be pointing East will be satisfactory.
Return time = distance / speed
Distance = 10 km = 10000 m
Speed = 14 ms-1
∴ Return time = 10000 / 14 = 714 s
However, we're asked for the average speed for the complete journey. Refer to part a) and we see that the outward journey was at 20 ms-1
∴ Outward time = 10000 / 20 = 500 s
∴ Total time for complete journey = 714 + 500 = 1214 s
Average speed = total distance / total time taken
∴ Average speed = 20000 / 1214 = 16.5 ms-1
= 16 ms-1