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Question

The displacement $$'x'$$ (in $$kg$$)  moving in one dimension under the action of a force, is related to time $$'t'$$ (in $$sec$$) by $$t=\sqrt {x}+3$$. The displacement of the particle when its velocity is zero will be :


A
2 m
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B
4 m
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C
0 m (zero)
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D
6 m
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Solution

The correct option is C $$0\ m$$ (zero)
$$ t = \sqrt{x}+3 $$
or $$ \sqrt{x} = t-3$$ ...      (i)
Squaring on both sides of equation (i),
$$ x = (t-3)^{2}$$ ...         (ii)
$$ = t^{2}+9 -6t $$          2
If v be the velocity of the particle,
Then $$ v=\dfrac{dx}{dt} = \dfrac{d}{dt}(x)$$
$$ = \dfrac{d}{dt}(t^{2}-6t+9)$$
$$ = 2t-6 $$            1
when $$ v = 0, 2t-6 = 0$$
or $$ t = 3 sec$$. ...(iii)          1
$$ \therefore $$ From equations (ii) and (iii),
$$ x = 3^{2}+9-6\times 3$$
$$ = 18-18$$
$$ x = 0$$                1

Physics

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