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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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A car is going at a speed of 6m/s when it encouters a 15 m slope of angle $30 ^{\circ}$. The frictional coefficient between road and tyre is 0.5. The driver applies the brakes. The minimum speed of car with which he can reach the bottom is $(g=10 m/s^2)$

\[(a)\;4\;m/s \quad (b)\; 3\;m/s \quad (c)\; 8.45 \;m/s \quad (d)\;7.46\;m/s \]

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1 Answer

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acceleration down the slope
$a=g \sin \theta - \mu g \cos \theta$
$a=10 \sin 30-(0.5)(10) cos 30$
$\quad=5-4.33$
$a=0.67 m/s^2$
Minimum speed while reaching the bottom
$v^2=u^2+2as$
$v^2=6^2+2(0.67)(15)$
$\quad=56.1$
$v=7.46 m/s$
Hence d is the correct answer

 

answered Jul 10, 2013 by meena.p
edited Jan 26, 2014 by meena.p
 

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