Sample 1Medium
A car moving at 20 m s⁻¹ is brought to rest by a constant retardation in a distance of 40 m. With the same retardation, the distance in which it stops when moving at 10 m s⁻¹ is
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The answer is B.
From v² = u² − 2as with v = 0, s = u²/2a, so s ∝ u². Halving u gives s = 40/4 = 10 m.
- A.
20 m assumes the stopping distance is proportional to speed, but it is proportional to the square of speed.
- C.
5 m assumes the stopping distance varies as the cube of speed.
- D.
40 m ignores that the stopping distance depends on the initial speed.
NCERT: Class 11 Physics, Chapter 2, Kinematic equations for uniformly accelerated motion