A ship A is moving Westwards with a speed of 10 km/h and a ship B, 100 km South of A, is moving Northwards with a speed of 10 km/h. The time after which the distance between them becomes shortest is?
5√2h
10√2h
0 h
5 h
Solution:
→VA = -10^i →VB = 10^j →VB/A = →VB - →VA →VB/A = 10^j + 10^i = 10√2(^i + ^j/√2) The shortest distance is ac, which will be at time when B reaches point c from b, t = bc/VB/A = 100/√2 / 10√2 = 5h