Question

A ship is moving westward with a speed of 10 km per hour and the ship B 100 KM south of a is moving Northward
with a speed of 10 km per the shortest distance between them

Answer 1

Let be the time passed and x be distance covered by the two ships after which the distance between them becomes shortest.
we have,
s=√(100-x)² + x²
for s to be minimum => ds/dx=0
=>ds/dx = 1/[(100-x)+x²]- 1/2
[-2(100-x)+2x]=0
=>4x-200 = 0
=>x=50m
so after both the ships have covered 50 m distance between then becomes shortest time taken for it will t= x/10 = 50/10 = 5h