An automobile, travelling at $40 \mathrm{~km} / \mathrm{h}$, can be stopped at a distance of $40…
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)
$75 \mathrm{~m}$
$160 \mathrm{~m}$
$100 \mathrm{~m}$
$150 \mathrm{~m}$
Solution
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, $s=160 \mathrm{~m}$