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

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$v=at=2t$

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.

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