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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 ?


4 years ago

Shaswata Biswas
132 Points
							Here, velocity of A, VA = 1.5 m/s towards B. Velocity of B, VB = 1.125 m/s perpendicular to AB. Distance between them, l = 4.5 mLet they are closest after time t seconds. And A & B reaches to A' & B' respectively.Then, distance travelled by A = 1.5×tSo, A'B = (4.5 - 1.5×t)And distance travelled by B = 1.125×tSo, BB' = 1.125×tAt this instance, the distance between A & B is A'B'In $\Delta$A'BB' :(A'B')2 = (A'B)2 + (BB')2              = (4.5 - 1.5t)2 + (1.125t)2            = 3.515625t2 - 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 + 7.29Or, A'B' = $\dpi{120} \sqrt{(1.875t - 3.6)^{2} + 7.29}$Now, A'B' is minimum when (1.875t - 3.6)2 = 0 $\dpi{120} \implies t = \frac {3.6}{1.875} = 1.92 s = 1\frac{23}{25}s$And, at this time, distance between them, $\dpi{120} S = \sqrt{7.29} = 2.7 m$THANKS

4 years ago
SHAKIR
13 Points
							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

one year ago
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