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VITEEE 2009 Paper

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Question : 38 of 120

Marks: +1, -0

Two persons A and B are located in X-Y plane at the points (0,0) and (0,10) respectively. (The distance are measured in MKS unit). At a time t = 0, they start moving simultaneously with velocities

\vec{v}_a = \hat{2j} \mathrm{m s}^{-1}

\vec{v}_b = \hat{2i} \mathrm{m s}^{-1}

respectively. The time after which A and B are at their closest distance is

2.5 s
4 s
1 s
$\frac{10}{\sqrt{2}} \mathrm{s}$

Solution:

Let after the time (t) the position of A is

(0, v_A t)

and position of

B = (v_B t, 10)

Distance between them

y = \sqrt{(0 - v_B t)^2 + (v_A t - 10)^2}

or

y^2 = (2t)^2 + (2t - 10)^2

or

y^2 = l = 4t^2 + 4t^2 + 100 - 40t \Rightarrow l = 8t^2 + 100 - 40t

Now,

\frac{dl}{dt} = (16t - 40) = 0

or

t = \frac{40}{16} = 2.5 \mathrm{s} \because \frac{d^2 l}{dt^2} = 16 = \text{(+ve)}

So,

l

will be minimum.

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