An aeroplane leaves an airport and flies due north at a speed of 1,000 km/hr. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km/hr. How far apart will be the two planes after $1\frac{1}{2}$ hours?
Given:
An aeroplane leaves an airport and flies due north at a speed of 1,000 km/hr. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km/hr.
To do:
We have to find the distance between the two planes after $1\frac{1}{2}$ hours.
Solution:
Speed of the first aeroplane $=1000\ km/hr$
Time $=1\frac{1}{2}\ hr=\frac{2\times1+1}{2}\ hr=\frac{3}{2}\ hr$
Distance travelled by the first aeroplane$=Speed\times Time$
$OP=1000\times\frac{3}{2}\ km$
$=500\times3\ km$
$=1500\ km$
The final position of the first aeroplane is represented by point P in the figure.
Speed of the second aeroplane $=1200\ km/hr$
Time $=1\frac{1}{2}\ hr=\frac{2\times1+1}{2}\ hr=\frac{3}{2}\ hr$
Distance travelled by the second aeroplane$=Speed\times Time$
$OQ=1200\times\frac{3}{2}\ km$
$=600\times3\ km$
$=1800\ km$
The final position of the second aeroplane is represented by point Q in the figure.
The distance between the aeroplanes is represented by PQ in the figure.
$\triangle OPQ$ is a right-angled triangle. By Pythagoras theorem,
$PQ^2=OP^2+OQ^2$
$PQ^2=(1500)^2+(1800)^2$
$PQ^2=2250000+3240000$
$PQ^2=5490000$
$PQ=\sqrt{5490000}$
$PQ=\sqrt{90000\times61}$
$PQ=300\sqrt{61}\ km$
The two planes will be $300\sqrt{61}\ km$ apart after $1\frac{1}{2}$ hours.
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