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Q.

A body of mass 2 kg has an initial velocity of 3 $m s^{- 1}$ along OE and it is subjected to a force of 4 newton in OF direction perpendicular to OE. The distance of the body from O after 4s will be

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a

28 m

b

48 m

c

20 m

d

12 m

answer is B.

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Detailed Solution

$a_{y} = \frac{F_{y}}{m} = \frac{4}{2} = 2$
$s_{y} = u_{y} t + \frac{1}{2} a_{y} t^{2}$
$= 0 + \frac{1}{2} (2) {(4)}^{2} = 16 m$
$s_{y} = 16$
$a_{x} = \frac{F_{x}}{m} = \frac{0}{2} = 0$
$S_{x} = 4_{x} t + \frac{1}{2} a_{x} t^{2} = (3) (4) + 0 = 12$
$S = \sqrt{16^{2} + 12^{2}} = 20 m$

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