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Per NCERT Example 2.6, the STOPPING DISTANCE of a vehicle (with initial velocity v₀ and deceleration −a) is given by which expression?
Ad_s = v₀² × a
Bd_s = 2v₀/a
Cd_s = v₀² / (2a) — proportional to the square of the initial velocity
Dd_s = v₀/a
Answer & Solution
Correct answer: C. d_s = v₀² / (2a) — proportional to the square of the initial velocity
Per NCERT Example 2.6, using v² = v₀² + 2ax with v = 0: x = v₀²/(2a) (with sign convention).
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