question_answer

. An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after \[1\frac{1}{2}\] hours?

A)  \[300\sqrt{67}\,km\]

B)  \[400\sqrt{61}\,km\]

C) \[200\sqrt{61}\,km\]

D)  \[300\sqrt{61}\,km\]

Correct Answer: D

Solution :

Let the point A represent the airport. Let the plane-l fly towards North. \[\therefore \] Distance of the plane-l from the airport after \[1\frac{1}{2}\] hours = Speed \[\times \]Time \[=1000\times 1\frac{1}{2}\,km=1500\,km\]             Let the plane-II fly towards West. \[\therefore \] Distance of the plane-II from the airport after \[1\frac{1}{2}\] hours \[=1200\times 1\frac{1}{2}km=1800\,km\] Now, in right \[\Delta ABC,\] using Pythagoras theorem, we have \[B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}\] \[\Rightarrow \] \[B{{C}^{2}}={{(1800)}^{2}}+{{(1500)}^{2}}=5490000\] \[\Rightarrow \] \[B{{C}^{2}}=\sqrt{5490000}=300\sqrt{61}km\] Thus, after \[1\frac{1}{2}\]hours, the two planes will be \[300\sqrt{61}km,\] apart from each other.