MHT CET · Physics · Motion In One Dimension
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
- B (1.4)d
- C d
- D (1.2)d
Answer & Solution
Correct Answer
(B) (1.4)d
Step-by-step Solution
Detailed explanation
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 \(\mathrm{f}=\mathrm{ma}=\frac{\mathrm{mu}^2}{2 \mathrm{~d}}\)
\(\therefore d =\frac{ u ^2}{2 F}\quad\ldots(1)\)
And in the second case, \(\text m_2=1.4\text{ m}\)
And \(\text d ^{\prime}=\frac{1.4\text{mu} ^2}{2\text F}\quad\ldots(2)\)
\(\therefore \frac{\mathrm{d}^{\prime}}{\mathrm{d}}=1.4\) 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 \(\mathrm{f}=\mathrm{ma}=\frac{\mathrm{mu}^2}{2 \mathrm{~d}}\)
\(\therefore d =\frac{ u ^2}{2 F}\quad\ldots(1)\)
And in the second case, \(\text m_2=1.4\text{ m}\)
And \(\text d ^{\prime}=\frac{1.4\text{mu} ^2}{2\text F}\quad\ldots(2)\)
\(\therefore \frac{\mathrm{d}^{\prime}}{\mathrm{d}}=1.4\) or \(\mathrm{d}^{\prime}=1.4 \mathrm{~d}\)
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