Question:
An automobile, travelling at \(40\ \mathrm{km/h}\), can be stopped at a distance of \(40\ \mathrm{m}\) by applying brakes. If the same automobile is travelling at \(80\ \mathrm{km/h}\), the minimum stopping distance, in metres, is (assume no skidding) :
An automobile, travelling at \(40\ \mathrm{km/h}\), can be stopped at a distance of \(40\ \mathrm{m}\) by applying brakes. If the same automobile is travelling at \(80\ \mathrm{km/h}\), the minimum stopping distance, in metres, is (assume no skidding) :
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With constant deceleration, \(v^2=2as\), so the stopping distance is proportional to the square of the initial speed.
Updated On: Nov 11, 2026
- 45 m
- 100 m
- 150 m
- 160 m
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The Correct Option is D
Solution and Explanation
For constant retardation \(a\) to rest: \(0=v^2-2as\), so \(s=\frac{v^2}{2a}\propto v^2\).
The braking force (hence \(a\)) is the same in both cases.
\(\frac{s_2}{s_1}=\left(\frac{v_2}{v_1}\right)^2=\left(\frac{80}{40}\right)^2=4\).
\(s_2=4\times40=160\ \mathrm{m}\).
Answer: (4) 160 m
The braking force (hence \(a\)) is the same in both cases.
\(\frac{s_2}{s_1}=\left(\frac{v_2}{v_1}\right)^2=\left(\frac{80}{40}\right)^2=4\).
\(s_2=4\times40=160\ \mathrm{m}\).
Answer: (4) 160 m
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