Two particles instantaneously at A and B 4.5m apart are moving with uniform velocities. The former moves towards B at speed 1.5m/s and latter perpendicular to AB at 1.125m/s . The istant when they are closest is ?

Question

Shaswata Biswas · Answer

Here, velocity of A, V A = 1.5 m/s towards B. Velocity of B, V B = 1.125 m/s perpendicular to AB. Distance between them, l = 4.5 m Let they are closest after time t seconds. And A & B reaches to A ' & B' respectively. Then, distance travelled by A = 1.5×t So, A'B = (4.5 - 1.5×t) And distance travelled by B = 1.125×t So, BB' = 1.125×t At this instance, the distance between A & B is A'B' In A'BB' : (A'B') 2 = (A'B) 2 + (BB') 2 = (4.5 - 1.5t) 2 + (1.125t) 2 = 3.515625t 2 - 13.5t + 20.25 = (1.875t) 2 - 2×1.875t×3.6 + (3.6) 2 + 20.25 - (3.6) 2 = (1.875t - 3.6) 2 + 7.29 Or, A'B' = Now, A'B' is minimum when (1.875t - 3.6) 2 = 0 And, at this time, distance between them, THANKS

SHAKIR · Answer

find the minimum distance between two particles p1 and p2 moving with velocities 3 m/s & 4 m/s Initial distance between p1 and p2 at the instant as shown in the figure is d

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Two particles instantaneously at A and B 4.5m apart are moving with uniform velocities. The former moves towards B at speed 1.5m/s and latter perpendicular to AB at 1.125m/s . The istant when they are closest is ?

Two particles instantaneously at A and B 4.5m apart are moving with uniform velocities. The former moves towards B at speed 1.5m/s and latter perpendicular to AB at 1.125m/s . The istant when they are closest is ?

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