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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 ?

Grade:11

## 2 Answers

Shaswata Biswas
132 Points
4 years ago
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 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 $\Delta$A'BB' :
(A'B')2 = (A'B)+ (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)+ 7.29
Or, A'B' = $\dpi{120} \sqrt{(1.875t - 3.6)^{2} + 7.29}$
Now, A'B' is minimum when (1.875t - 3.6)= 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
SHAKIR
13 Points
2 years ago
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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