Question

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

Answer 1

Let us draw the figure first.

An aeroplane which flies due north at a speed of 1000 km/hr covers the distance AB after $1\frac{1}{2}$ hr and another aeorplane that flies due west at the speed of 1200 km/hr covers the distance BC after $1\frac{1}{2}$ hr.

We know that $Speed=\frac{Distance}{Time}$

Let us calculate AB first as shown below,

$AB=1000\times 1.5$

$\therefore AB=1500km$

Similarly we can calculate BC.

$BC=1200\times 1.5$

$\therefore BC=1800km$

Now we have find AC. To find AC we will use Pythagoras theorem,

$A{C}^2=B{C}^2+A{B}^2$

$\Rightarrow A{C}^2=(1800{)}^2+(1500{)}^2$

$\Rightarrow A{C}^2=3240000+2250000$

$\Rightarrow A{C}^2=5490000$

$\Rightarrow AC=2343.07$

Therefore, after $1\frac{1}{2}$ hr the aeroplanes will be approximately $2343km$ far apart.