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

Speed of north flying aeroplane$=1000km/hr$

Speed of west flying aeroplane $=1200km/hr$

Find: distance between the two planes after $1.5hrs$ i.e $BC$

$speed=\frac{distance}{time}$

$Distance=Speed\times Time$

Speed $1000km/hr$; Time $1\frac{1}{2}hrs$

Speed $1200km/hr$; Time $1\frac{1}{2}hrs$

Distance $AB=1000\times 1\frac{1}{2}=1000\times \frac{3}{2}=1500km$

Distance $AC=1200\times 1\frac{1}{2}=1200\times \frac{3}{2}=600\times 3=1800km$

Now we have to find distance $BC$

Since north and west are perpendicular

$<BAC={90}^o$

So $△ABC$ is a right angle triangle

Using pythagoras theorem

${BC}^2={AB}^2+{AC}^2$

$BC=\sqrt{{AB}^2+{AC}^2}$

$=\sqrt{{1500}^2+{1800}^2}$

$=\sqrt{2250000+3240000}$

$=\sqrt{5490000}$

$=300\sqrt{61}km$