An aircraft flies at a speed of 400km/h in still air. A wind is blowing with a speed of 200√2km/h from South. The pilot wishes to travel from point A to a point B, where point B is North-East of A. Find the direction he must head if AB=1000km.
A
30∘ from North towards East
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
45∘ from East towards North
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
75∘ from North towards East
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
75∘ from East towards North
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C75∘ from North towards East
Given that speed of wind w.r.t ground vw=200√2km/h and speed of airplane w.r.t wind vaw=400km/h and
Velocity of airplane w.r.t ground →va should be along AB or in north-east direction. Thus, the direction of −−→vaw should be such that the resultant of −→vw and −−→vaw is along AB or in north-east direction.
Let −−→vaw make an angle α with AB as shown in vector triangle ABC:
Applying sine law in triangle ABC, we get: ACsin45∘=BCsinα ⇒sinα=(BCAC)sin45∘=(200√2400)1√2=12 ∴α=30∘
Therefore, the pilot should head in a direction at an angle of (45∘+α) or 75∘ from north towards east.