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An automobile, travelling at $40 \mathrm{~km} / \mathrm{h}$, can be stopped at a distance of $40 \mathrm{~m}$ by applying brakes. If the same automobile is travelling at $80 \mathrm{~km} / \mathrm{h}$, the minimum stopping distance, in metres, is (assume no skidding)
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$160 \mathrm{~m}$
$160 \mathrm{~m}$
According to question, $u_1=40 \mathrm{~km} / \mathrm{h}, v_1=0$ and $s_1=40 \mathrm{~m}$ using $v^2-u^2=2 a s ; 0^2-40^2=2 a \times 40$..(i) Again, $0^2-80^2=2$ as From eqn. (i) and (ii)
Stopping distance, $s=160 \mathrm{~m}$
Stopping distance, $s=160 \mathrm{~m}$
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