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

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Aicharya · Accepted Answer

$\mathrm{u}=15 \mathrm{~km} / \mathrm{hr}=\frac{15 \times 1000}{60 \times 60}=\frac{150}{36} \mathrm{~m} / \mathrm{s}$ $\mathrm{v}^{2}=\mathrm{u}^{2}-2 \mathrm{as}$ $\therefore \mathrm{a} \frac{u^{2}}{2 \mathrm{~s}}=\frac{150 \times 150}{36 \times 36 \times 2 \times 5} \mathrm{~m} / \mathrm{s}^{2}$ Next $\mathrm{s}=\frac{u^{2}}{2 a}=\frac{45000}{60 \times 60} \times \frac{45000}{60 \times 60} \times \frac{1}{2} \times \frac{36 \times 36}{150 \times 15}$ $=45 \mathrm{~m}$

A vehicle moving with $15 \mathrm{~km} / \mathrm{hr}$ comes to rest by covering $5 \mathrm{~m}$ distance by…

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