Question

An aeroplane leaves an airport and flies due north at a speed of $1,000kmperhour$. At the same time, another aeroplane leaves the same airport and flies due west at a speed of $1,200kmperhour$. How far apart will be the two planes after $1\frac{1}{2}hr$ ?

Answer 1

Step 1: Find the distance travelled by first aeroplane in the north direction

We know that,

$Dis\tan ce=speed\times time$

Now,

$OA$ is the distance travelled by aeroplane travelling towards the north

We have speed of aeroplane is $1,000kmperhour$ and time $1\frac{1}{2}hr$

Then,

$\begin{array}{rcl}OA& =& 1000\times 1\frac{1}{2}\\ & =& 1000\times \frac{3}{2}\\ & =& 500\times 3\\ & =& 1500km\end{array}$

Step 2: Find the distance travelled by the second aeroplane in the west direction.

$OB$is the distance travelled by another aeroplane travelling towards west

We have speed of aeroplane is $1,200kmperhour$ and time $1\frac{1}{2}hr$

Then,

$\begin{array}{rcl}OB& =& 1200\times 1\frac{1}{2}\\ & =& 1200\times \frac{3}{2}\\ & =& 600\times 3\\ & =& 1800km\end{array}$

Step 3: Find the distance between two aeroplanes after $1\frac{1}{2}hr$

$AB$ is the distance between two aeroplanes after $1\frac{1}{2}hr$

From the Pythagoras theorem

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Now, $In\Delta AOB,\angle AOB=90°$

$\begin{array}{rcl}A{B}^2& =& O{A}^2+O{B}^2\\ & =& {\left(1500\right)}^2+{\left(1800\right)}^2=2250000+3240000\\ & =& 5490000\end{array}$

$\begin{array}{rcl}AB& =& \sqrt{5490000}\\ & =& 300\sqrt{61}km\end{array}$

Hence, the distance between two planes after $1\frac{1}{2}hr=300\sqrt{61}km$