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A car moving along a straight highway with speed $126 \mathrm{~km} /$ $h$ is brought to a stop within a distance of $200 \mathrm{~m}$. What is the retardation of the car (assumed uniform) and how long does it take for the car to stop?
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Verified Answer
Here, $u=126 \mathrm{~km} / \mathrm{h}$
$$
\begin{aligned}
&=126 \times 1000 /(60 \times 60) \mathrm{ms}^{-1}=35 \mathrm{~ms}^{-1} \\
&v=0, \mathrm{~s}=200 \mathrm{~m}
\end{aligned}
$$
We know, $v^2=u^2+2$ as
$\Rightarrow 0=(35)^2+2 a \times(200)$
or, $\quad a=-(35)^2 /(2 \times 200)$
$=-49 / 16=-3.06 \mathrm{~ms}^{-2}$
As $v=u+a t \Rightarrow 0=35+(-49 / 16) t$
$t=35 \times 16 / 49=80 / 7=11.43 \mathrm{~s}$
$$
\begin{aligned}
&=126 \times 1000 /(60 \times 60) \mathrm{ms}^{-1}=35 \mathrm{~ms}^{-1} \\
&v=0, \mathrm{~s}=200 \mathrm{~m}
\end{aligned}
$$
We know, $v^2=u^2+2$ as
$\Rightarrow 0=(35)^2+2 a \times(200)$
or, $\quad a=-(35)^2 /(2 \times 200)$
$=-49 / 16=-3.06 \mathrm{~ms}^{-2}$
As $v=u+a t \Rightarrow 0=35+(-49 / 16) t$
$t=35 \times 16 / 49=80 / 7=11.43 \mathrm{~s}$
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