Question

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

Answer 1

Let A be the first aeroplane flied due north at a speed of 1000 km/hr and B be the second aeroplane flied due west at a speed of 1200 km/hr
Distance covered by plane A in $1\frac{1}{2}$ hours = $1000\times \frac{3}{2}=1500\mathrm{km}$
Distance covered by plane B in $1\frac{1}{2}$ hours = $1200\times \frac{3}{2}=1800\mathrm{km}$
Now, In right triangle ABC
By using Pythagoras theorem, we have
$$\begin{gathered} {\mathrm{AB}}^2={\mathrm{BC}}^2+{\mathrm{CA}}^2 \\ ={\left(1800\right)}^2+{\left(1500\right)}^2 \\ =3240000+2250000 \\ =5490000 \\ \therefore {\mathrm{AB}}^2=5490000 \\ \Rightarrow \mathrm{AB}=300\sqrt{61}\mathrm{m} \end{gathered}$$
Hence, the distance between two planes after $1\frac{1}{2}$ hours is $300\sqrt{61}$ m.