MHT CET · Physics · Motion In One Dimension
A vehicle moving with \(15 \mathrm{~km} / \mathrm{hr}\) comes to rest by covering \(5 \mathrm{~m}\) distance by applying brakes. If the same vehicle moves at \(45 \mathrm{~km} / \mathrm{hr}\), then by applying brakes, it will come to rest by covering a distance
- A \(15 \mathrm{~m}\)
- B \(45 \mathrm{~m}\)
- C \(60 \mathrm{~m}\)
- D \(30 \mathrm{~m} .\)
Answer & Solution
Correct Answer
(B) \(45 \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{u}=15 \mathrm{~km} / \mathrm{hr}=\frac{15 \times 1000}{60 \times 60}=\frac{150}{36} \mathrm{~m} / \mathrm{s}\)
\(\mathrm{v}^{2}=\mathrm{u}^{2}-2 \mathrm{as}\)
\(\therefore \mathrm{a} \frac{u^{2}}{2 \mathrm{~s}}=\frac{150 \times 150}{36 \times 36 \times 2 \times 5} \mathrm{~m} / \mathrm{s}^{2}\)
Next \(\mathrm{s}=\frac{u^{2}}{2 a}=\frac{45000}{60 \times 60} \times \frac{45000}{60 \times 60} \times \frac{1}{2} \times \frac{36 \times 36}{150 \times 15}\)
\(=45 \mathrm{~m}\)
\(\mathrm{v}^{2}=\mathrm{u}^{2}-2 \mathrm{as}\)
\(\therefore \mathrm{a} \frac{u^{2}}{2 \mathrm{~s}}=\frac{150 \times 150}{36 \times 36 \times 2 \times 5} \mathrm{~m} / \mathrm{s}^{2}\)
Next \(\mathrm{s}=\frac{u^{2}}{2 a}=\frac{45000}{60 \times 60} \times \frac{45000}{60 \times 60} \times \frac{1}{2} \times \frac{36 \times 36}{150 \times 15}\)
\(=45 \mathrm{~m}\)
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