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

• A

  (1.6)d
• Correct answer
  B

  (1.4)d
• C

  d
• D

  (1.2)d

Solution

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 $\mathrm{d}^{\prime}=1.4 \mathrm{~d}$

Hence, the correct answer is option: 'B'.