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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11
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A point moves in $X-Y$ plane according to $X=4 \sin \;6t $ and $Y=4(1-\cos \;6t)$.The distance traversed by the particle in 4 seconds is (x and y are in meters)

\[(a)\;96 m \quad (b)\;48 m \quad (c)\;24 m \quad (d)\;108 m\]
Can you answer this question?
 
 

1 Answer

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$v_x =\large\frac{dx}{dy}$$=24 \cos 6t$
$v_y =\large\frac{dy}{dt}$$=24 \sin 6t$
$v=\sqrt {{v_x}^2+{v_y}^2}$
$\quad=24 m/s^2$
Since $v$- speed is constant.
distance traversed in 4 second is
$s=v \times t$
$s=24 \times 4$
$s=96 m$
Hence a is the correct answer. 

 

answered Jun 29, 2013 by meena.p
edited Jan 25, 2014 by meena.p
 

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