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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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A Car is able to stop with a uniform retardation of $8 ms^{-2} $ and the driver of the car can react to an emergency in 0.6 s. Calculate overall stopping distance of the car if the speed of car is $72 \;km\;h^{-1}$

$\begin{array}{1 1}(A)\;37\;m\\(B)\;56\;m \\(C)\;78\;m\\(D)\;28\;m\end{array} $

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Here $u= 72 km\;h^{-1}=20\;ms^{-1}$.
For the first $0.6s$, the car travels with constant velocity of $20\;ms^{-1}$ and covers a distance.
$d_1=ut=(20)(0.6) =12m$
The distance $d_2$ travelled after applying brakes is given by ,
$0^2=20^2-2(8)d_2=>d_2=25 m$
Total stopping distance $d= d_1+d_2 =12+25 =37 m$
Hence A is the correct answer.
answered Jul 18, 2014 by meena.p

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