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3-2.Motion in Plane
medium
A particle undergoes three successive displacements given by $s _1=\sqrt{2}\,m$ north-east, $s _2=2\,m$ due south and $s _3=4\,m , 30^{\circ}$ north of west, then magnitude of net displacement is
A$\sqrt{14+4 \sqrt{3}}$
B$\sqrt{14-4 \sqrt{3}}$
C$\sqrt{4}$
DNone of the above
Solution
(b)$s _1=\left(\sqrt{2} \cos 45^{\circ}\right) \hat{ i }+\left(\sqrt{2} \sin 45^{\circ}\right) \hat{ j }=\hat{ i }+\hat{ j }$
$s _2=2 \hat{ j } \text { and } s =\left(-4 \cos 30^{\circ}\right) \hat{ i }+\left(4 \sin 30^{\circ}\right) \hat{ j }$
$=-2 \sqrt{3} \hat{ i }+2 \hat{ j }$
Now $\quad s = s _1+ s _2+ s _3=(1-2 \sqrt{3}) \hat{ i }+\hat{ j }$
$=\sqrt{1-2 \sqrt{3})^2+(1)^2}$
$=\sqrt{1+12+1-4 \sqrt{3}}$
$=\sqrt{14-4 \sqrt{3}}\,m$
Standard 11
Physics
