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A vehicle without passengers is moving on a frictionless horizontal road with velocity ' $u$ ' can be stopped in a distance ' $d$ '. Now $40 \%$ of it's weight is added. If the retardation remains same the stopping distance at velocity ' $u$ ' is
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(1.4)d
Given, $\mathrm{m}_1=\mathrm{m}$ and $\mathrm{m}_2=1.4 \mathrm{~m}$
$\because \mathrm{v}^2=\mathrm{u}^2+2 \mathrm{aS}$
$\therefore 0=\mathrm{u}^2-2 \mathrm{ad}$
$\therefore \mathrm{a}=\frac{\mathrm{u}^2}{2 \mathrm{~d}}$
$\therefore$ Retarding force $\mathrm{f}=\mathrm{ma}=\frac{\mathrm{mu}^2}{2 \mathrm{~d}}$
$\therefore \frac{\mathrm{d}^{\prime}}{\mathrm{d}}=1.4$ or $\mathrm{d}^{\prime}=1.4 \mathrm{~d}$
$\because \mathrm{v}^2=\mathrm{u}^2+2 \mathrm{aS}$
$\therefore 0=\mathrm{u}^2-2 \mathrm{ad}$
$\therefore \mathrm{a}=\frac{\mathrm{u}^2}{2 \mathrm{~d}}$
$\therefore$ Retarding force $\mathrm{f}=\mathrm{ma}=\frac{\mathrm{mu}^2}{2 \mathrm{~d}}$
$\therefore \frac{\mathrm{d}^{\prime}}{\mathrm{d}}=1.4$ or $\mathrm{d}^{\prime}=1.4 \mathrm{~d}$
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