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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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A car of 1000 kg with velocity 18 km/hr is stopped by brake of 1000 N. distance covered by it before stopping is

\[(a)\;1\;m \quad (b)\;12.5\;m \quad (c)\;162\;m \quad (d)\;144\;m\]
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$\quad=-1 m/s^2$
$0=\bigg(\large\frac{5}{18} $$ \times 18\bigg)^2-2s$
$s=12.5\; m$
Hence b is the correct answer. 


answered Jul 15, 2013 by meena.p
edited Feb 10, 2014 by meena.p

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