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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
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A car starts with constant acceleration $ a=2 m/s^2$ at $t=0$ Two coins are released from the car at $t=3$ and $t=4$. Each coin takes 1 second to reach the ground.Then distance between two coins is (assume the coin stick to ground)

\[(a)\;9 m \quad (b)\;7 m \quad (c)\;15 m \quad (d)\;2 m\]

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Velocity of car at $t=3,v_1=6 m/s$
at $t=4,v_2=8 m/s$
Coin 1 will fall horizontally with velocity 6 m/s
Coin 2 will fall horizontally with velocity 8 m/s
Both coins will travel 6 m and 8 m horizontally before they fall from the point of release.
car moves between t=3 and t=4 $ \large\frac{6 +8}{2} $$\times 1=7 m$ (all the distance are measured from the position at time t=3)
In fourth second, position of first coin $x_1=6$ and position of 2nd coin $= 7+8=15$
Therefore distance between two coins $ =15-6=9 m$
Hence a is the correct answer.
answered Jun 27, 2013 by meena.p
edited May 24, 2014 by lmohan717

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