Two cars started moving from a point, one towards north at 40 km/hr & other towards East at 30 km/hr. The shortest distance between them after 2 hours in km is ___

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Solution

In two hours first car has travelled a distance of $2 \times 40 = 80 k m$ towards north and the second car would have travelled $2 \times 30 = 60 k m$ towards east.

The shortest distance between them can be calculated using Pythagoras theorem.

$A B^{2} = O A^{2} + O B^{2}$
$A B^{2} = (80)^{2} + (60)^{2} = 6400 + 3600 = 10000$
$A B = 100 k m$ .

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Two cars started moving from a point, one towards north at 40 km/hr & other towards East at 30 km/hr. The shortest distance between them after 2 hours in km is ___

In two hours first car has travelled a distance of 2×40=80 km towards north and the second car would have travelled 2×30=60 km towards east. The shortest distance between them can be calculated using Pythagoras theorem. AB2=OA2+OB2 AB2=(80)2+(60)2=6400+3600=10000 AB=100 km.