Answer :

now let the aircraft be initially at a point A situated 3400m above the ground, it moves with a speed v along straight line parallel to ground and reach a point B, now let us assume a recording station is located on ground at c which is in line with midway of the journey D

Suppose the plane covered a distance S and subtended an angle of 30o on recording station as it moved from A to B


The situation has been shown in the figure



We have to find speed of the plane; the plane is moving with constant speed so we will use the relation


S = v × t or v = S/t


Where S is the distance covered by plane moving with a speed v in time t


Here we are given time t = 10.0 s


So we need to find distance covered by plane S


Now we can see distance covered = AB = AD + BD


Since D is the midpoint of AB


AD = BD or distance AB = 2 × AD


Now in Triangle ACD using trigonometry we will find AD


now


Assuming D is the midpoint of AB and C is equidistant from A and B we have


Or


AD = CD × tan15o


CD = 3400m


tan15o= 0.2679


so we get


AD = 3400m × 0.2679


= 910.86m


So total distance covered by the plane


S = AB = 2 × 910.86m = 182.17m


So the Value of S = 182.17m and t = 10 s in the equation


v = s/t ,we get


speed of the plane


so plane is flying at a speed of 18.2 m/s


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