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Q.

Two material particles A and B are moving in free space. How their position coordinates x, y and z vary with time t is shown in the following graphs. Determine at what instant of time the particles are closest to each other and the closest separation

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a

$t = 1 \sec$

b

$t = 0.5 \sec$

c

Closest separation = 5m

d

Closest separation = $\frac{5}{\sqrt{2}} m$

answer is B, D.

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Detailed Solution

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$l^{2} = x_{r e l}^{2} + y_{r e l}^{2} + z_{r e l}^{2}$
l increases after t = 1sec
Min separation should be between t = 0 and t = 1
$l^{2}$ is a quadratic in t
$l^{2}$ at t = 0 = $3^{2} + 4^{2} + 0^{2} = 5^{2}$
$l^{2} a t t = 1 = 0^{2} + 0^{2} + 5^{2} = 5^{2}$
$l^{2}$ is minimum at t = $0.5 \sec$
$l_{(a t t = 0.5)}^{2} = {(\frac{3}{2})}^{2} + {(\frac{4}{2})}^{2} + {(\frac{5}{2})}^{2} = \frac{5}{\sqrt{2}} m$

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