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2.Motion in Straight Line
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An automobile, travelling at $40\, km/h$, can be stopped at a distance of $40\, m$ by applying brakes. If the same automobile is travelling at $80\, km/h$, the minimum stopping distance, in metres, is (assume no skidding)..........$m$
A
$75$
B
$160$
C
$100$
D
$150$
(JEE MAIN-2018)
Solution
$\begin{array}{l}
According\,to\,question,\,{u_1} = 40\,km/h,\,{v_1} = 0\\
and\,{s_1} = 40\,m\\
{\rm{using}}\,{{\rm{v}}^2} – {u^2} = 2as;\,{0^2} – {40^2} = 2a \times 40\,\,\,\,\,\,…\left( i \right)\\
Again,\,{0^2} – {80^2} = 2as\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…\left( {ii} \right)\\
From\,eqn.\left( i \right)\,and\,\left( {ii} \right)\,\\
Stopping\,{\rm{ distance}},\,s = 160\,m\,\,
\end{array}$
Standard 11
Physics
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